STAT 145 (Notes) Al Nosedal Department of Mathematics and Statistics University of New Mexico. Fall 2013

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "STAT 145 (Notes) Al Nosedal anosedal@unm.edu Department of Mathematics and Statistics University of New Mexico. Fall 2013"

Transcription

1 STAT 145 (Notes) Al Nosedal Department of Mathematics and Statistics University of New Mexico Fall 2013

2 CHAPTER 18 INFERENCE ABOUT A POPULATION MEAN.

3 Conditions for Inference about mean We can regard our data as a simple random sample (SRS) from the population. This condition is very important. Observations from the population have a Normal distribution with mean µ and standard deviation σ. In practice, it is enough that the distribution be symmetric and single-peaked unless the sample is very small. Both µ and σ are unknown parameters.

4 Standard Error When the standard deviation of a statistic is estimated from data, the result is called the standard error of the statistic. The standard s error of the sample mean x is n.

5 Travel time to work A study of commuting times reports the travel times to work of a random sample of 1000 employed adults. The mean is x = 49.2 minutes and the standard deviation is s = 63.9 minutes. What is the standard error of the mean?

6 Solution The standard error of the mean is s n = = minutes

7 The one-sample t statistic and the t distributions Draw an SRS of size n from a large population that has the Normal distribution with mean µ and standard deviation σ. The one-sample t statistic t = x µ s/ n has the t distribution with n 1 degrees of freedom.

8 The t distributions The density curves of the t distributions are similar in shape to the Standard Normal curve. They are symmetric about 0, single-peaked, and bell-shaped. The spread of the t distributions is a bit greater than of the Standard Normal distribution. The t distributions have more probability in the tails and less in the center than does the Standard Normal. This is true because substituting the estimate s for the fixed parameter σ introduces more variation into the statistic. As the degrees of freedom increase, the t density curve approaches the N(0, 1) curve ever more closely. This happens because s estimates σ more accurately as the sample size increases. So using s in place of σ causes little extra variation when the sample is large.

9 Density curves Std. Normal t(2) t(9)

10 Density curves Std. Normal t(29)

11 Density curves Std. Normal t(49)

12 Critical values Use Table C or software to find a) the critical value for a one-sided test with level α = 0.05 based on the t(4) distribution. b) the critical value for 98% confidence interval based on the t(26) distribution.

13 Solution a) t = b) t = 2.479

14 More critical values You have an SRS of size 30 and calculate the one-sample t-statistic. What is the critical value t such that a) t has probability to the right of t? b) t has probability 0.75 to the left of of t?

15 Solution Here, df = 30 1 = 29. a) t = b) t = 0.683

16 The one-sample t confidence interval Draw an SRS of size n from a large population having unknown mean µ. A level C confidence interval for µ is x ± t s n where t is the critical value for the t(n 1) density curve with area C between t and t. This interval is exact when the population distribution is Normal and is approximately correct for large n in other cases.

17 Critical values What critical value t from Table C would you use for a confidence interval for the mean of the population in each of the following situations? a) A 95% confidence interval based on n = 12 observations. b) A 99% confidence interval from an SRS of 18 observations. c) A 90% confidence interval from a sample of size 6.

18 Solution a) df = 12 1 = 11, so t = b) df = 18 1 = 17, so t = c) df = 6 1 = 5, so t =

19 Ancient air The composition of the earth s atmosphere may have changed over time. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen: Assume (this is not yet agreed on by experts) that these observations are an SRS from the late Cretaceous atmosphere. Use a 90% confidence interval to estimate the mean percent of nitrogen in ancient air (Our present-day atmosphere is about 78.1% nitrogen).

20 Solution µ = mean percent of nitrogen in ancient air. We will estimate µ with a 90% confidence interval. With x = , s = , and t = (df = 9 1 = 8), the 90% confidence interval for µ is ( ) ± ± to

21 The one-sample t test Draw an SRS of size n from a large population having unknown mean µ. To test the hypothesis H 0 : µ = µ 0, compute the one-sample t statistic t = x µ 0 s/ n In terms of a variable T having the t(n 1) distribution, the P-value for a test of H 0 against H a : µ > µ 0 is P(T t ). H a : µ < µ 0 is P(T t ). H a : µ µ 0 is 2P(T t ). These P-values are exact if the population distribution is Normal and are approximately correct for large n in other cases.

22 Is it significant? The one-sample t statistic for testing H 0 : µ = 0 H a : µ > 0 from a sample of n = 20 observations has the value t = a) What are the degrees of freedom for this statistic? b) Give the two critical values t from Table C that bracket t. What are the one-sided P-values for these two entries? c) Is the value t = 1.84 significant at the 5% level? Is it significant at the 1% level? d) (Optional) If you have access to suitable technology, give the exact one-sided P-value for t = 1.84?

23 Solution a) df = 20-1 = 19. b) t = 1.84 is bracketed by t = (with right-tail probability 0.05) and t = (with right-tail probability 0.025). Hence, because this is a one-sided significance test, < P-value < c) This test is significant at the 5% level because the P-value < It is not significant at the 1% level because the P-value > d) From TI-84, tcdf(1.84,1000,19), P-value =

24 Is it significant? The one-sample t statistic from a sample of n = 15 observations for the two-sided test of H 0 : µ = 64 H a : µ 64 has the value t = a) What are the degrees of freedom for t? b) Locate the two-critical values t from Table C that bracket t. What are the two-sided P-values for these two entries? c) is the value t = 2.12 statistically significant at the 10% level? At the 5% level? d) (Optional) If you have access to suitable technology, give the exact two-sided P-value for t = 2.12.

25 Solution a) df = 15-1 = 14. b) t = 2.12 is bracketed by t = (with two-tail probability 0.10) and t = (with two-tail probability 0.05). Hence, because this is a two-sided significance test, 0.05 < P-value < c) This test is significant at the 10% level because the P-value < It is not significant at the 5% level because the P-value > d) From TI-84, tcdf(2.12,1000,19) = , P-value = 2(0.0261) =

26 Ancient air, continued Do the data of Exercise 18.7 (see above) give good reason to think that the percent of nitrogen in the air during the Cretaceous era was different from the present 78.1%? Carry out a test of significance at the 5% significance level.

27 Solution Is there evidence that the percent of nitrogen in ancient air was different from the present 78.1%? 1. State hypotheses. H 0 : µ = 78.1% vs H a : µ 78.1%. 2. Test statistic. t = x µ 0 σ/ n = / 9 = P-value. For df = 8, this is beyond anything shown in Table C, so P-value < (TI -84 gives ). 4. Conclusion. We have very strong evidence (P-value < 0.001) that Cretaceous air is different from nitrogen than modern air.

28 Matched pairs t procedures To compare the responses to the two treatments in a matched pairs design, find the difference between the responses within each pair. Then apply the one-sample t procedures to these differences.

29 Genetic engineering for cancer treatment Here s a new idea for treating advanced melanoma, the most serious kind of skin cancer. Genetically engineer white blood cells to better recognize and destroy cancer cells, then infuse these cells into patients. The subjects in a small initial study were 11 patients whose melanoma had not responded to existing treatments. One question was how rapidly the new cells would multiply after infusion, as measured by the doubling time in days. Here are the doubling times: a) Examine the data. Is it reasonable to use the t procedures? b) Give a 90% confidence interval for the mean doubling time. Are you willing to use this interval to make an inference about the mean doubling time in a population of similar patients?

30 Solution a) The stemplot does not show any severe evidence of non-normality, so t procedures should be safe

31 Solution b) With x = days, s = days, and t = (df = 10), the 90% confidence interval is ( ) ± ± to days

32 Solution There is no indication that the sample represents an SRS of all patients whose melanoma has not responded to existing treatments, so inferring to the population of similar patients may not be reasonable.

33 Genetic engineering for cancer treatment, continued Another outcome in the cancer experiment described above is measured by a test for the presence of cells that trigger an immune response in the body and so may help fight cancer. Here are data for the 11 subjects: counts of active cells per 100,000 cells before and after infusion of the modified cells. The difference (after minus before) is the response variable.

34 Data Before After Difference Difference/

35 Genetic engineering for cancer treatment, continued a) Examine the data. Is it reasonable to use the t procedures? b) If your conclusion in part a) is Yes, do the data give convincing evidence that the count of active cells is higher after treatment?

36 Solution a) The stem plot of differences shows an extreme right skew, and one or two high outliers. The t procedures should not be used

37 Solution b) 1. State hypotheses. H 0 : µ = 0 vs H a : µ > Test statistic. t = x µ 0 σ/ n = / = P-value. For df = 10, using Table C we have that: < t < Which implies that: < P-value < TI -84 gives Conclusion. We would have strong evidence against H 0 (P-value < 0.05). We would conclude that the count of active cells is higher after treatment.

38 Robust Procedures A confidence interval or significance test is called robust if the confidence level or P-value does not change very much when the conditions for use of the procedure are violated.

39 Using t procedures Except in the case of small samples, the condition that the data are an SRS from the population of interest is more important than the condition that the population distribution is Normal. Sample size less than 15: Use t procedures if the data appear close to Normal (roughly symmetric single peak, no outliers). If the data are clearly skewed or if outliers are present, do not use t. Sample size at least 15: The t procedures can be used except in the presence of outliers or strong skewness. Large samples: The t procedures can be used even for clearly skewed distributions when the sample is large, roughly n 40.

40 CHAPTER 19 TWO-SAMPLE PROBLEMS

41 Two-sample problems The goal of inference is to compare the responses to two treatments or to compare the characteristics of two populations. We have a separate sample from each treatment or each population.

42 Conditions for inference comparing two means We have two SRSs, from two distinct populations. The samples are independent. That is, one sample has no influence on the other. Matching violates independence, for example. We measure the same response variable for both samples. Both populations are Normally distributed. The means and standard deviations of the populations are unknown. In practice, it is enough that the distributions have similar shapes and that the data have no strong outliers.

43 The Two-Sample Procedures Draw an SRS of size n 1 from a Normal population with unknown mean µ 1, and draw and independent SRS of size n 2 from another Normal population with unknown mean µ 2. A confidence interval for µ 1 µ 2 is given by ( x 1 x 2 ) ± t s 2 1 n 1 + s2 2 n 2 Here t is the critical value for the t(k) density curve with area C between t and t. The degrees of freedom k are equal to the smaller of n 1 1 and n 2 1.

44 The Two-Sample Procedures To test the hypothesis H 0 : µ 1 = µ 2, calculate the two-sample t statistic t = x 1 x 2 s 2 1 n 1 + s2 2 n 2 and use P-values or critical values for the t(k) distribution.

45 Degrees of freedom (Option 1) Option 1. With software, use the statistic t with accurate critical values from the approximating t distribution. The distribution of the two-sample t statistic is very close to the t distribution with degrees of freedom df given by df = ( ) s n 1 + s2 2 n 2 ( ) ( ) 1 s 2 2 ( ) ( ) 1 n 1 1 n s n 2 1 n 2 This approximation is accurate when both sample sizes n 1 and n 2 are 5 or larger.

46 Degrees of freedom (Option2) Option 2. Without software, use the statistic t with critical values from the t distribution with degrees of freedom equal to the smaller of n 1 1 and n 2 1. These procedures are always conservative for any two Normal populations.

47 Logging in the rain forest Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning. These words begin a report on a statistical study of the effects of logging in Borneo. Here are data on the number of tree species in 12 unlogged forest plots and 9 similar plots logged 8 years earlier: Unlogged: Logged : Does logging significantly reduce the mean number of species in a plot after 8 years? State the hypotheses and do a t test. Is the result significant at the 5% level?

48 Solution Does logging significantly reduce the mean number of species in a plot after 8 years? 1. State hypotheses. H 0 : µ 1 = µ 2 vs H a : µ 1 > µ 2, where µ 1 is the mean number of species in unlogged plots and µ 2 is the mean number of species in plots logged 8 years earlier. 2. Test statistic.t = x 1 x 2 = ( x 1 = 17.5, x 2 = , s s2 2 n 1 n2 s 1 = , s 2 = 4.5, n 1 = 12 and n 2 = 9) 3. P-value. Using Table C, we have df = 8, and < P-value < Using TI-84, we have df = , and P-value = Conclusion. Since P-value < 0.05, we reject H 0. There is strong evidence that the mean number of species in unlogged plots is greater than that for logged plots 8 year after logging.

49 Logging in the rainforest, continued Use the data from the previous exercise to give a 99% confidence interval for the difference in mean number of species between unlogged and logged plots.

50 Solution 1. Find x 1 x 2. From what we did earlier: x 1 x 2 = = Find SE = Standard Error. We already know that: s 1 = , s 2 = 4.5, n 1 = 12 and n 2 = 9 s1 2 SE = + s = n 1 n = Find m = t SE. From Table C, we have df = 8 and 99% confidence level, then t = Hence, m = (3.355)(1.8132) = Find Confidence Interval. x 1 x 2 ± t SE = ± = to TI 84 gives a 99% confidence interval of 1.52 to species, using df =

51 Alcohol and zoning out Healthy men aged 21 to 35 were randomly assigned to one of two groups: half received 0.82 grams of alcohol per kilogram of body weight; half received a placebo. Participants were then given 30 minutes to read up to 34 pages of Tolstoy s War and Peace (beginning at chapter 1, with each page containing approximately 22 lines of text). Every two to four minutes participants were prompted to indicate whether they were zoning out. The proportion of time participants indicated they were zoning out was recorded for each subject. The table below summarizes data on the proportion of episodes of zoning out. (The study report gave the standard error of the mean s/ n, abbreviated as SEM, rather than the standard deviation s.)

52 Alcohol and zoning out Group n x SEM Alcohol Placebo a) What are the two sample standard deviations? b) What degrees of freedom does the conservative Option 2 use for two-sample t procedures for these samples? c) Using Option 2, give a 90% confidence interval for the mean difference between the two groups.

53 Solution a) Call group 1 the Alcohol group,and group 2 the Placebo group. Then, because SEM = s/ n, we have s = SEM n. Hence, s 1 = = 0.5(5) = 0.25 and s 2 = = 0.03(5) = b) Using the conservative option 2, df = min{25 1, 25 1} = min{24, 24} = 24. c) Here, with n 1 = n 2 = 25, SE = s 2 1 n 1 + s2 2 n 2 = = With df = 24, we have t = 1.711, and a 90% confidence interval for the mean difference in proportions is given by ± 1.711(0.0583) = 0.13 ± , that is, from to

54 Stress and weight in rats In a study of the effects of stress on behavior in rats, 71 rats were randomly assigned to either a stressful environment or a control (non stressful) environment. After 21 days, the change in weight (in grams) was determined for each rat. The table below summarizes data on weight gain. (The study report gave the standard error of the mean s/ n, abbreviated as SEM, rather than the standard deviation s).

55 Stress and weight in rats Group n x SEM Stress No stress a) What are the standard deviations for the two groups? b) What degrees of freedom does the conservative Option 2 use for the two-sample procedures for these data? c) Test the null hypothesis of no difference between the two group means against the two-sided alternative at the 5% significance level. Use the degrees of freedom from part b).

56 Solution a) Call group 1 the Stress group,and group 2 the No stress group. Then, because SEM = s/ n, we have s = SEM n. Hence, s 1 = 3 20 = and s 2 = 2 51 = b) Using the conservative option 2, df = min{20 1, 51 1} = min{19, 50} = 19. c) 1. Set hypotheses. H 0 : µ 1 = µ 2 vs H a : µ 1 µ 2 2. Calculate test statistic. With n 1 = 20 and n 2 = 51, s 2 SE = 1 n 1 + s2 2 n 2 = = , t = x 1 x 2 SE = P-value. With df = 19, using Table C, 0.10 < P-value < Using TI-84 (2-SampTTest), P-value = Conclusion. Since P-value > 0.05 we CAN T reject H 0. There is little evidence in support of a conclusion that mean weights of rats in stressful environments differ from those of rats without stress.

57 CHAPTER 20 INFERENCE ABOUT A POPULATION PROPORTION

58 Problem A market research consultant hired by the Pepsi-Cola Co. is interested in determining the proportion of UNM students who favor Pepsi-Cola over Coke Classic. A random sample of 100 students shows that 40 students favor Pepsi over Coke. Use this information to construct a 95% confidence interval for the proportion of all students in this market who prefer Pepsi.

59 Bernoulli Distribution x i = { 1 i-th person prefers Pepsi 0 i-th person prefers Coke µ = E(x i ) = p σ 2 = V (x i ) = p(1 p), i.e. σ = p(1 p) Let ˆp be our estimate of p. Note that ˆp = n i=1 x i = x. If n is big, by the Central Limit Theorem, we know that: x is roughly ) ( ) σ N (µ, n, that is, ˆp is roughly N p, p(1 p) n

60 Sampling distribution of a sample proportion Draw an SRS of size n from a large population that contains proportion p of successes. Let ˆp be the sample proportion of successes, Then: ˆp = number of successes in the sample n The mean of the sampling distribution of ˆp is p. The standard deviation of the sampling distribution is p(1 p) n. As the sample size increases, the sampling distribution of ˆp becomes approximately Normal. That is, for large n, ˆp has approximately the N(p, p(1 p)/n).

61 Large-sample confidence interval for a population proportion Draw an SRS of size n from a large population that contains an unknown proportion p of successes. An approximate level C confidence interval for p is ˆp ± z ˆp(1 ˆp) n where z is the critical value for the standard Normal density curve with area C between z and z. Use this interval only when the numbers of successes and failures in the sample are both 15 or greater.

62 Problem A survey of 611 office workers investigated telephone answering practices, including how often each office worker was able to answer incoming telephone calls and how often incoming telephone calls went directly to voice mail. A total of 281 office workers indicated that they never need voice mail and are able to take every telephone call. a. What is the point estimate of the proportion of the population of office workers who are able to take every telephone call? b. At 90% confidence, what is the margin of error? c. What is the 90% confidence interval for the proportion of the population of office workers who are able to take every telephone call?

63 Solution p = proportion of the population of office workers who are able to take every telephone call. a. ˆp = = 0.46 b. Margin of error = z 1.645(0.0201) = ˆp(1 ˆp) (0.46)(0.54) n = = c. ˆp ± z ˆp(1 ˆp) n 0.46 ±.0330 From to We are 90% confident that the proportion of the population of office workers who are able to take every phone call lies between 42.70% and 49.30%.

64 Computer/Internet-based crime With over 50% of adults spending more than an hour a day on the internet, the number experiencing computer-or Internet- based crime continues to rise. A survey in 2010 of a random sample of 1025 adults, aged 18 and older, reached by random digit dialing found 113 adults in the sample who said that they or a household member had been the victim of a computer or Internet crime on their home computer in the past year. Give the 99% confidence interval for the proportion p of all households that have experienced computer or Internet crime during the year before the survey was conducted.

65 Solution p = proportion of all households that have experienced computer or Internet crime during the year before the survey was conducted. Step 1. Find ˆp. ˆp = number of successes = sample size = Step 2. Find m = margin of error z ˆp(1 ˆp) ( ) n = = 2.576(0.0097) = Step 3. Find confidence interval, i.e. ˆp m and ˆp + m. ˆp m = = ˆp + m = = Based on this sample, we are 99% confident that between 8.53% and 13.51% of Internet users were victims of computer or Internet crime during the year before this survey was taken.

66 Sample Size for desired margin of error The level C confidence interval for a population proportion p will have margin of error approximately equal to a specified value m when the sample size is ( ) z 2 n = p (1 p ) m where p is a guessed value for the sample proportion. The margin of error will always be less than or equal to m if you take the guess p to be 0.5.

67 Problem The percentage of people not covered by health care insurance in 2003 was 15.6%. A congressional committee has been charged with conducting a sample survey to obtain more current information. a. What sample size would you recommend if the committee s goal is to estimate the current proportion of individuals without health care insurance with a margin of error of 0.03? Use a 95% confidence level. b. Repeat part a) assuming that you have no prior information (i.e. p = 0.5). c. Repeat part a) using a 99% confidence level. d. Repeat part c) assuming that you have no prior information (i.e. p = 0.5).

68 Solution a. n = ( ) z 2 m p (1 p ) = ( ) (0.156) ( ) = 563 b. n = ( ) z 2 m p (1 p ) = ( ) (0.5) (1 0.5) = 1068 ) 2 p (1 p ) = ( ) (0.156) ( ) = 971 c. n = ( z m c. n = ( z m 0.03 ) 2 p (1 p ) = ( ) 2 (0.5) (1 0.5) = 1844

69 Graph of p (1 p ) p*(1-p*) p*= p*

70 Canadians and doctor-assisted suicide A Gallup Poll asked a sample of Canadian adults if they thought the law should allow doctors to end the life of a patient who is in great pain and near death if the patient makes a request in writing. The poll included 270 people in Quebec, 221 of whom agreed that doctor-assisted suicide should be allowed. a) What is the margin of error of the large-sample 95% confidence interval for the proportion of all Quebec adults who would allow doctor-assisted suicide? b) How large a sample is needed to get the common ±3 percentage point margin of error? Use the previous sample as a pilot study to get p.

71 Solution a) We find that ˆp = = and SEˆp ˆp(1 ˆp) = n = Hence, the margin of error for 95% confidence is (1.96)(0.0234) = b) For a ±0.03 margin of error, we need n = ( ) (0.8185)( ) = Round this up in order to obtain a required sample size of 635.

72 Significance tests for a proportion Draw an SRS of size n from a large population that contains an unknown proportion p of successes. To test the hypothesis H 0 : p = p 0, compute the z statistic z = ˆp p 0 p 0 (1 p 0 ) n Find P-values from the standard Normal distribution. Use this test when the sample size n is so large that both np 0 and n(1 p 0 ) are 10 or more.

73 P-values To test the hypothesis H 0 : p = p 0, compute the z statistic z = ˆp p 0 p 0 (1 p 0 ) n In terms of a variable Z having the standard Normal distribution, the approximate P-value for a test of H 0 against H a : p > p 0 is P(Z > z ) H a : p < p 0 is P(Z < z ) H a : p p 0 is 2P(Z > z )

74 Problem A study found that, in 2005, 12.5% of U.S. workers belonged to unions. Suppose a sample of 400 U.S. workers is collected in 2006 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in b. If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test? c. At α = 0.05, what is your conclusion?

75 Solution p = proportion of all U.S. workers who belonged to a union in a.h 0 : p = vs H a : p > b. ˆp = = 0.13 z = ˆp p 0 p0 (1 p 0 ) n = (0.125)(0.875) 400 = 0.30 Using Table A, P-value = P(Z > z ) = P(Z > 0.30) = = c. P-value > 0.05, do not reject H 0. We cannot conclude that there was an increase in union membership.

76 Problem A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup. a. Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 64%. b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the p-value? c. At α = 0.05, what is your conclusion?

77 Solution p = proportion of all shoppers who believe supermarket brands to be as good as this particular national brand ketchup. a.h 0 : p = 0.64 vs H a : p 0.64 b. ˆp = = 0.52 z = ˆp p 0 p0 (1 p 0 ) n = (0.64)(0.36) 100 = 2.50 Using Table A, P-value = 2P(Z > z ) = 2P(Z > 2.50 ) = 2P(Z > 2.50) = 2(0.0062) = c. P-value < 0.05, reject H 0. Proportion differs from the reported 0.64.

78 Problem The National Center for Health Statistics released a report that stated 70% of adults do not exercise regularly. A researcher decided to conduct a study to see whether the claim made by the National Center for Health Statistics differed on a state-by-state basis. a. State the null and alternative hypotheses assuming the intent of the researcher is to identify states that differ from 70% reported by the National Center for Health Statistics. b. At α = 0.05, what is the research conclusion for the following states?: Wisconsin: 252 of 350 adults did not exercise regularly. California: 189 of 300 adults did not exercise regularly.

79 Solution (Wisconsin) p = proportion of all adults in the state of Wisconsin that do not exercise regularly. a.h 0 : p = 0.70 vs Ha : p 0.70 b. (Wisconsin) ˆp = = 0.72 z = ˆp p 0 p0 (1 p 0 ) n = (0.70)(0.30) 350 = 0.82 Using Table A, P-value = 2P(Z > z ) = 2P(Z > 0.82 ) = 2P(Z > 0.82) = 2(0.2061) = c. P-value > 0.05, do not reject H 0. We don t have evidence to claim that Wisconsin is one of those states where the proportion of adults that do not exercise regularly differs from 70%.

80 Solution (California) p = proportion of all adults in the state of California that do not exercise regularly. a.h 0 : p = 0.70 vs Ha : p 0.70 b. (California) ˆp = = 0.63 z = ˆp p 0 p0 (1 p 0 ) n = (0.70)(0.30) 300 = 2.65 Using Table A, P-value = 2P(Z > z ) = 2P(Z > 2.65 ) = 2P(Z > 2.65) = 2( ) = c. P-value < 0.05, we reject H 0. We have strong evidence to claim that California s proportion of adults that do not exercise regularly differs from 70%.

81 CHAPTER 23 TWO CATEGORICAL VARIABLES: THE CHI-SQUARE TEST

82 Problem A sample of 70 girls and boys was taken at random from a population of children (perhaps of a particular age of interest), and then the numbers of right-handed boys, right-handed girls, left-handed boys, and left-handed girls were recorded as shown below: Girls Boys Total Left-handed Right-handed Total Is there evidence that, in the sampled population, handedness is independent of sex? (Use α = 0.05)

83 Step 1. State Hypotheses H 0 : In the sampled population, there is NO relationship between handedness and sex. H a : In the sampled population, there is a relationship between handedness and sex.

84 Step 2. Compute test statistic. Draw an SRS from a large population and make a two-way table of the sample counts for two categorical variables. To test the null hypotheses H 0 that there is no relationship between the row and column variables in the population, calculate the chi-square statistic X 2 = ( observed count - expected count ) 2 expected count The sum is over all cells in the table. The chi-square test rejects H 0 when χ 2 is large. Sofware finds P-values from a chi-square distribution.

85 Step 2. (cont.) Expected Counts. P(left-handed) = P(right-handed) = Expected number of left-handed girls = number of girls P(left-handed) = (36)( ) = (36)(18) 70 = Expected number of right-handed girls = number of girls P(right-handed) = (36)( ) = (36)(52) 70 = Expected number of left-handed boys = number of boys P(left-handed) = (34)( ) = (34)(18) 70 = Expected number of right-handed boys = number of boys P(right-handed) = (34)( ) = (34)(52) 70 =

86 Expected Counts The expected counts in any cell of a two-way table when H 0 is true is: expected count = row total x column total table total

87 Table of Expected Counts Girls Boys Total Left-handed Right-handed Total

88 X 2 = ( ) ( ) (6 8.74) ( ) X 2 = = (Using TI-84, X 2 = ).

89 Step 3. Find P-value d.f. = (r 1)(c 1) = (2 1)(2 1) = 1 Using Table D, 2.07 < X 2 = < 2.71 which implies that 0.10 < P-value < (Using TI-84,P-value = )

90 Step 4. Conclusion Since P-value > 0.10 > 0.05 = α, we CAN T reject H 0. We conclude that there is NO relationship between handedness and sex.

91 Example Suppose the Federal Correction Agency wants to investigate the following question: Does a male released from federal prison make better adjustment to civilian life if he returns to his hometown or if he goes elsewhere to live? To put it another way, is there a relationship between adjustment to civilian life and place of residence after release from prison? The agency s psychologists interviewed 200 randomly selected former prisoners. Using a series of questions, the psychologists classified the adjustment of each individual to civilian life as being outstanding, good, fair, or unsatisfactory. The counts are given in the following contingency table. Examine the hypothesis that there is no relationship between adjustment to civilian life and place of residence after release from prison. Test at the α = 0.01 level.

92 Table of observed counts Outstanding Good Fair Unsatisfactory Total Hometown Not hometown Total

93 Step 1. State hypotheses H 0 : There is NO relationship between adjustment to civilian life and where the individual lives after being released from prison. H a : There is a relationship between adjustment to civilian life and where the individual lives after being released from prison.

94 Table of EXPECTED counts Outstanding Good Fair Unsatisfactory Total Hometown Not hometown Total

95 Step 2. Compute test statistic. X 2 = (27 24) (35 30)2 (15 20) (27 24) (25 20) (33 36) (25 30) (13 16) X 2 = X 2 = (Using TI-84, X 2 = ).

96 Step 3. Find P-value d.f. = (r 1)(c 1) = (2 1)(4 1) = 3 Using Table D, 5.32 < X 2 = < 6.25 which implies that 0.10 < P-value < (Using TI-84,P-value = )

97 Step 4. Conclusion Since P-value > 0.10 > 0.01 = α, we CAN T reject H 0. We conclude that there is NO relationship between adjustment to civilian life and where the prisoner resides after being released from prison. For the Federal Correction Agency s advisement program, adjustment to civilian life is not related to where the ex-prisoner lives after being released.

98 Cell counts required for the chi-square test You can safely use the chi-square test with P-values from the chi-square distribution when no more than 20% of the expected counts are less than 5 and all individual expected counts are 1 or greater. In particular, all four expected counts in a 2 2 table should be 5 or greater.

99 Majors for men and women in business A study of the career plans of young women and men sent questionnaires to all 722 members of the senior class in the College of Business Administration at the University of Illinois. One question asked which major within the business program the student had chosen. Here are the data from the students who responded: Female Male Total Accounting Administration Economics Finance Total This is an example of a single sample classified according to two categorical variables (gender and major).

100 Majors for men and women in business (cont.) a) Test the null hypothesis that there is no relationship between the gender of students and their choice of major. Give a P-value and state your conclusion(use α = 0.05.) b) Verify that the expected cell counts satisfy the requirement for use of chi-square. c) Which two cells have the largest terms in the chi-square statistic? How do observed and expected counts differ in these cells?

101 Solution a) Step 1. State Hypotheses. H 0 : There is NO relationship between gender and choice of major. H a : There is a relationship between gender and choice of major.

102 Solution Expected counts. Female Male Total Accounting Administration Economics Finance Total

103 Solution Step 2. Compute test statistic. X 2 = ( ) ( ) ( )2 (5 6.41) (6 4.59) ( ) ( ) ( ) X 2 = X 2 =

104 Solution Step 3. Find P-value. d.f. = (r 1)(c 1) = (2 1)(4 1) = 3 Using Table D, 9.84 < X 2 = < which implies that 0.01 < P-value < (Using TI-84,P-value = )

105 Solution Step 4. Conclusion. Since P-value < 0.05, we reject H 0. We have fairly strong evidence that gender and choice of major are related. b) One expected count is less than 5, which is only one-eight (12.5%) of the counts in the table, so the chi-square procedure is acceptable. c) The largest chi-square components are the two from the Administration row. Many more women than we expect (91 actual, expected) chose this major, whereas only 40 men chose this (54.64 expected).

106 The Chi-square distributions The chi-square distributions are a family of distributions that take only positive values and are skewed to the right. A specific chi-square distribution is specified by giving its degrees of freedom. The chi-square test for a two-way table with r and c columns uses critical values from the chi-square distribution with (r-1)(c-1) degrees of freedom. The P-value is the area under the density curve of this chi-square distribution to the right of the value of the test statistic.

107 Density curves df=1 df=4 df=

MATH 214 (NOTES) Math 214 Al Nosedal. Department of Mathematics Indiana University of Pennsylvania. MATH 214 (NOTES) p. 1/6

MATH 214 (NOTES) Math 214 Al Nosedal. Department of Mathematics Indiana University of Pennsylvania. MATH 214 (NOTES) p. 1/6 MATH 214 (NOTES) Math 214 Al Nosedal Department of Mathematics Indiana University of Pennsylvania MATH 214 (NOTES) p. 1/6 "Pepsi" problem A market research consultant hired by the Pepsi-Cola Co. is interested

More information

Name: Date: Use the following to answer questions 3-4:

Name: Date: Use the following to answer questions 3-4: Name: Date: 1. Determine whether each of the following statements is true or false. A) The margin of error for a 95% confidence interval for the mean increases as the sample size increases. B) The margin

More information

Chapter 7 Section 7.1: Inference for the Mean of a Population

Chapter 7 Section 7.1: Inference for the Mean of a Population Chapter 7 Section 7.1: Inference for the Mean of a Population Now let s look at a similar situation Take an SRS of size n Normal Population : N(, ). Both and are unknown parameters. Unlike what we used

More information

Recall this chart that showed how most of our course would be organized:

Recall this chart that showed how most of our course would be organized: Chapter 4 One-Way ANOVA Recall this chart that showed how most of our course would be organized: Explanatory Variable(s) Response Variable Methods Categorical Categorical Contingency Tables Categorical

More information

MTH 140 Statistics Videos

MTH 140 Statistics Videos MTH 140 Statistics Videos Chapter 1 Picturing Distributions with Graphs Individuals and Variables Categorical Variables: Pie Charts and Bar Graphs Categorical Variables: Pie Charts and Bar Graphs Quantitative

More information

STATISTICS 8, FINAL EXAM. Last six digits of Student ID#: Circle your Discussion Section: 1 2 3 4

STATISTICS 8, FINAL EXAM. Last six digits of Student ID#: Circle your Discussion Section: 1 2 3 4 STATISTICS 8, FINAL EXAM NAME: KEY Seat Number: Last six digits of Student ID#: Circle your Discussion Section: 1 2 3 4 Make sure you have 8 pages. You will be provided with a table as well, as a separate

More information

Class 19: Two Way Tables, Conditional Distributions, Chi-Square (Text: Sections 2.5; 9.1)

Class 19: Two Way Tables, Conditional Distributions, Chi-Square (Text: Sections 2.5; 9.1) Spring 204 Class 9: Two Way Tables, Conditional Distributions, Chi-Square (Text: Sections 2.5; 9.) Big Picture: More than Two Samples In Chapter 7: We looked at quantitative variables and compared the

More information

Online 12 - Sections 9.1 and 9.2-Doug Ensley

Online 12 - Sections 9.1 and 9.2-Doug Ensley Student: Date: Instructor: Doug Ensley Course: MAT117 01 Applied Statistics - Ensley Assignment: Online 12 - Sections 9.1 and 9.2 1. Does a P-value of 0.001 give strong evidence or not especially strong

More information

3.4 Statistical inference for 2 populations based on two samples

3.4 Statistical inference for 2 populations based on two samples 3.4 Statistical inference for 2 populations based on two samples Tests for a difference between two population means The first sample will be denoted as X 1, X 2,..., X m. The second sample will be denoted

More information

Chapter 23 Inferences About Means

Chapter 23 Inferences About Means Chapter 23 Inferences About Means Chapter 23 - Inferences About Means 391 Chapter 23 Solutions to Class Examples 1. See Class Example 1. 2. We want to know if the mean battery lifespan exceeds the 300-minute

More information

1. What is the critical value for this 95% confidence interval? CV = z.025 = invnorm(0.025) = 1.96

1. What is the critical value for this 95% confidence interval? CV = z.025 = invnorm(0.025) = 1.96 1 Final Review 2 Review 2.1 CI 1-propZint Scenario 1 A TV manufacturer claims in its warranty brochure that in the past not more than 10 percent of its TV sets needed any repair during the first two years

More information

Is it statistically significant? The chi-square test

Is it statistically significant? The chi-square test UAS Conference Series 2013/14 Is it statistically significant? The chi-square test Dr Gosia Turner Student Data Management and Analysis 14 September 2010 Page 1 Why chi-square? Tests whether two categorical

More information

Hypothesis Testing. Bluman Chapter 8

Hypothesis Testing. Bluman Chapter 8 CHAPTER 8 Learning Objectives C H A P T E R E I G H T Hypothesis Testing 1 Outline 8-1 Steps in Traditional Method 8-2 z Test for a Mean 8-3 t Test for a Mean 8-4 z Test for a Proportion 8-5 2 Test for

More information

Chapter 23. Two Categorical Variables: The Chi-Square Test

Chapter 23. Two Categorical Variables: The Chi-Square Test Chapter 23. Two Categorical Variables: The Chi-Square Test 1 Chapter 23. Two Categorical Variables: The Chi-Square Test Two-Way Tables Note. We quickly review two-way tables with an example. Example. Exercise

More information

Unit 27: Comparing Two Means

Unit 27: Comparing Two Means Unit 27: Comparing Two Means Prerequisites Students should have experience with one-sample t-procedures before they begin this unit. That material is covered in Unit 26, Small Sample Inference for One

More information

9.1 (a) The standard deviation of the four sample differences is given as.68. The standard error is SE (ȳ1 - ȳ 2 ) = SE d - = s d n d

9.1 (a) The standard deviation of the four sample differences is given as.68. The standard error is SE (ȳ1 - ȳ 2 ) = SE d - = s d n d CHAPTER 9 Comparison of Paired Samples 9.1 (a) The standard deviation of the four sample differences is given as.68. The standard error is SE (ȳ1 - ȳ 2 ) = SE d - = s d n d =.68 4 =.34. (b) H 0 : The mean

More information

Chapter 23. Inferences for Regression

Chapter 23. Inferences for Regression Chapter 23. Inferences for Regression Topics covered in this chapter: Simple Linear Regression Simple Linear Regression Example 23.1: Crying and IQ The Problem: Infants who cry easily may be more easily

More information

General Method: Difference of Means. 3. Calculate df: either Welch-Satterthwaite formula or simpler df = min(n 1, n 2 ) 1.

General Method: Difference of Means. 3. Calculate df: either Welch-Satterthwaite formula or simpler df = min(n 1, n 2 ) 1. General Method: Difference of Means 1. Calculate x 1, x 2, SE 1, SE 2. 2. Combined SE = SE1 2 + SE2 2. ASSUMES INDEPENDENT SAMPLES. 3. Calculate df: either Welch-Satterthwaite formula or simpler df = min(n

More information

Review. March 21, 2011. 155S7.1 2_3 Estimating a Population Proportion. Chapter 7 Estimates and Sample Sizes. Test 2 (Chapters 4, 5, & 6) Results

Review. March 21, 2011. 155S7.1 2_3 Estimating a Population Proportion. Chapter 7 Estimates and Sample Sizes. Test 2 (Chapters 4, 5, & 6) Results MAT 155 Statistical Analysis Dr. Claude Moore Cape Fear Community College Chapter 7 Estimates and Sample Sizes 7 1 Review and Preview 7 2 Estimating a Population Proportion 7 3 Estimating a Population

More information

9-3.4 Likelihood ratio test. Neyman-Pearson lemma

9-3.4 Likelihood ratio test. Neyman-Pearson lemma 9-3.4 Likelihood ratio test Neyman-Pearson lemma 9-1 Hypothesis Testing 9-1.1 Statistical Hypotheses Statistical hypothesis testing and confidence interval estimation of parameters are the fundamental

More information

CHAPTER 14 NONPARAMETRIC TESTS

CHAPTER 14 NONPARAMETRIC TESTS CHAPTER 14 NONPARAMETRIC TESTS Everything that we have done up until now in statistics has relied heavily on one major fact: that our data is normally distributed. We have been able to make inferences

More information

STAT 350 Practice Final Exam Solution (Spring 2015)

STAT 350 Practice Final Exam Solution (Spring 2015) PART 1: Multiple Choice Questions: 1) A study was conducted to compare five different training programs for improving endurance. Forty subjects were randomly divided into five groups of eight subjects

More information

AP Statistics 2005 Scoring Guidelines

AP Statistics 2005 Scoring Guidelines AP Statistics 2005 Scoring Guidelines The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to college

More information

5/31/2013. Chapter 8 Hypothesis Testing. Hypothesis Testing. Hypothesis Testing. Outline. Objectives. Objectives

5/31/2013. Chapter 8 Hypothesis Testing. Hypothesis Testing. Hypothesis Testing. Outline. Objectives. Objectives C H 8A P T E R Outline 8 1 Steps in Traditional Method 8 2 z Test for a Mean 8 3 t Test for a Mean 8 4 z Test for a Proportion 8 6 Confidence Intervals and Copyright 2013 The McGraw Hill Companies, Inc.

More information

Mind on Statistics. Chapter 12

Mind on Statistics. Chapter 12 Mind on Statistics Chapter 12 Sections 12.1 Questions 1 to 6: For each statement, determine if the statement is a typical null hypothesis (H 0 ) or alternative hypothesis (H a ). 1. There is no difference

More information

Chapter 8. Hypothesis Testing

Chapter 8. Hypothesis Testing Chapter 8 Hypothesis Testing Hypothesis In statistics, a hypothesis is a claim or statement about a property of a population. A hypothesis test (or test of significance) is a standard procedure for testing

More information

Mind on Statistics. Chapter 13

Mind on Statistics. Chapter 13 Mind on Statistics Chapter 13 Sections 13.1-13.2 1. Which statement is not true about hypothesis tests? A. Hypothesis tests are only valid when the sample is representative of the population for the question

More information

Two-sample t-tests. - Independent samples - Pooled standard devation - The equal variance assumption

Two-sample t-tests. - Independent samples - Pooled standard devation - The equal variance assumption Two-sample t-tests. - Independent samples - Pooled standard devation - The equal variance assumption Last time, we used the mean of one sample to test against the hypothesis that the true mean was a particular

More information

Inferential Statistics

Inferential Statistics Inferential Statistics Sampling and the normal distribution Z-scores Confidence levels and intervals Hypothesis testing Commonly used statistical methods Inferential Statistics Descriptive statistics are

More information

Chapter 7 Section 1 Homework Set A

Chapter 7 Section 1 Homework Set A Chapter 7 Section 1 Homework Set A 7.15 Finding the critical value t *. What critical value t * from Table D (use software, go to the web and type t distribution applet) should be used to calculate the

More information

Review for Exam 2. H 0 : p 1 p 2 = 0 H A : p 1 p 2 0

Review for Exam 2. H 0 : p 1 p 2 = 0 H A : p 1 p 2 0 Review for Exam 2 1 Time in the shower The distribution of the amount of time spent in the shower (in minutes) of all Americans is right-skewed with mean of 8 minutes and a standard deviation of 10 minutes.

More information

Chapter 8 Section 1. Homework A

Chapter 8 Section 1. Homework A Chapter 8 Section 1 Homework A 8.7 Can we use the large-sample confidence interval? In each of the following circumstances state whether you would use the large-sample confidence interval. The variable

More information

Unit 26: Small Sample Inference for One Mean

Unit 26: Small Sample Inference for One Mean Unit 26: Small Sample Inference for One Mean Prerequisites Students need the background on confidence intervals and significance tests covered in Units 24 and 25. Additional Topic Coverage Additional coverage

More information

Module 5 Hypotheses Tests: Comparing Two Groups

Module 5 Hypotheses Tests: Comparing Two Groups Module 5 Hypotheses Tests: Comparing Two Groups Objective: In medical research, we often compare the outcomes between two groups of patients, namely exposed and unexposed groups. At the completion of this

More information

Statistics I for QBIC. Contents and Objectives. Chapters 1 7. Revised: August 2013

Statistics I for QBIC. Contents and Objectives. Chapters 1 7. Revised: August 2013 Statistics I for QBIC Text Book: Biostatistics, 10 th edition, by Daniel & Cross Contents and Objectives Chapters 1 7 Revised: August 2013 Chapter 1: Nature of Statistics (sections 1.1-1.6) Objectives

More information

C. The null hypothesis is not rejected when the alternative hypothesis is true. A. population parameters.

C. The null hypothesis is not rejected when the alternative hypothesis is true. A. population parameters. Sample Multiple Choice Questions for the material since Midterm 2. Sample questions from Midterms and 2 are also representative of questions that may appear on the final exam.. A randomly selected sample

More information

Data Analysis Tools. Tools for Summarizing Data

Data Analysis Tools. Tools for Summarizing Data Data Analysis Tools This section of the notes is meant to introduce you to many of the tools that are provided by Excel under the Tools/Data Analysis menu item. If your computer does not have that tool

More information

Chapter 19 The Chi-Square Test

Chapter 19 The Chi-Square Test Tutorial for the integration of the software R with introductory statistics Copyright c Grethe Hystad Chapter 19 The Chi-Square Test In this chapter, we will discuss the following topics: We will plot

More information

Hypothesis Testing COMP 245 STATISTICS. Dr N A Heard. 1 Hypothesis Testing 2 1.1 Introduction... 2 1.2 Error Rates and Power of a Test...

Hypothesis Testing COMP 245 STATISTICS. Dr N A Heard. 1 Hypothesis Testing 2 1.1 Introduction... 2 1.2 Error Rates and Power of a Test... Hypothesis Testing COMP 45 STATISTICS Dr N A Heard Contents 1 Hypothesis Testing 1.1 Introduction........................................ 1. Error Rates and Power of a Test.............................

More information

Nonparametric Two-Sample Tests. Nonparametric Tests. Sign Test

Nonparametric Two-Sample Tests. Nonparametric Tests. Sign Test Nonparametric Two-Sample Tests Sign test Mann-Whitney U-test (a.k.a. Wilcoxon two-sample test) Kolmogorov-Smirnov Test Wilcoxon Signed-Rank Test Tukey-Duckworth Test 1 Nonparametric Tests Recall, nonparametric

More information

Biostatistics: Types of Data Analysis

Biostatistics: Types of Data Analysis Biostatistics: Types of Data Analysis Theresa A Scott, MS Vanderbilt University Department of Biostatistics theresa.scott@vanderbilt.edu http://biostat.mc.vanderbilt.edu/theresascott Theresa A Scott, MS

More information

An Introduction to Statistics Course (ECOE 1302) Spring Semester 2011 Chapter 10- TWO-SAMPLE TESTS

An Introduction to Statistics Course (ECOE 1302) Spring Semester 2011 Chapter 10- TWO-SAMPLE TESTS The Islamic University of Gaza Faculty of Commerce Department of Economics and Political Sciences An Introduction to Statistics Course (ECOE 130) Spring Semester 011 Chapter 10- TWO-SAMPLE TESTS Practice

More information

How to Conduct a Hypothesis Test

How to Conduct a Hypothesis Test How to Conduct a Hypothesis Test The idea of hypothesis testing is relatively straightforward. In various studies we observe certain events. We must ask, is the event due to chance alone, or is there some

More information

LAB 4 INSTRUCTIONS CONFIDENCE INTERVALS AND HYPOTHESIS TESTING

LAB 4 INSTRUCTIONS CONFIDENCE INTERVALS AND HYPOTHESIS TESTING LAB 4 INSTRUCTIONS CONFIDENCE INTERVALS AND HYPOTHESIS TESTING In this lab you will explore the concept of a confidence interval and hypothesis testing through a simulation problem in engineering setting.

More information

Introduction. Chapter 14: Nonparametric Tests

Introduction. Chapter 14: Nonparametric Tests 2 Chapter 14: Nonparametric Tests Introduction robustness outliers transforming data other standard distributions nonparametric methods rank tests The most commonly used methods for inference about the

More information

Statistiek I. t-tests. John Nerbonne. CLCG, Rijksuniversiteit Groningen. John Nerbonne 1/35

Statistiek I. t-tests. John Nerbonne. CLCG, Rijksuniversiteit Groningen.  John Nerbonne 1/35 Statistiek I t-tests John Nerbonne CLCG, Rijksuniversiteit Groningen http://wwwletrugnl/nerbonne/teach/statistiek-i/ John Nerbonne 1/35 t-tests To test an average or pair of averages when σ is known, we

More information

Inference for two Population Means

Inference for two Population Means Inference for two Population Means Bret Hanlon and Bret Larget Department of Statistics University of Wisconsin Madison October 27 November 1, 2011 Two Population Means 1 / 65 Case Study Case Study Example

More information

Statistical Inference and t-tests

Statistical Inference and t-tests 1 Statistical Inference and t-tests Objectives Evaluate the difference between a sample mean and a target value using a one-sample t-test. Evaluate the difference between a sample mean and a target value

More information

Chi Square Tests. Chapter 10. 10.1 Introduction

Chi Square Tests. Chapter 10. 10.1 Introduction Contents 10 Chi Square Tests 703 10.1 Introduction............................ 703 10.2 The Chi Square Distribution.................. 704 10.3 Goodness of Fit Test....................... 709 10.4 Chi Square

More information

Unit 26 Estimation with Confidence Intervals

Unit 26 Estimation with Confidence Intervals Unit 26 Estimation with Confidence Intervals Objectives: To see how confidence intervals are used to estimate a population proportion, a population mean, a difference in population proportions, or a difference

More information

Statistics Review PSY379

Statistics Review PSY379 Statistics Review PSY379 Basic concepts Measurement scales Populations vs. samples Continuous vs. discrete variable Independent vs. dependent variable Descriptive vs. inferential stats Common analyses

More information

Chapter 8 Introduction to Hypothesis Testing

Chapter 8 Introduction to Hypothesis Testing Chapter 8 Student Lecture Notes 8-1 Chapter 8 Introduction to Hypothesis Testing Fall 26 Fundamentals of Business Statistics 1 Chapter Goals After completing this chapter, you should be able to: Formulate

More information

Hypothesis Testing. Steps for a hypothesis test:

Hypothesis Testing. Steps for a hypothesis test: Hypothesis Testing Steps for a hypothesis test: 1. State the claim H 0 and the alternative, H a 2. Choose a significance level or use the given one. 3. Draw the sampling distribution based on the assumption

More information

MONT 107N Understanding Randomness Solutions For Final Examination May 11, 2010

MONT 107N Understanding Randomness Solutions For Final Examination May 11, 2010 MONT 07N Understanding Randomness Solutions For Final Examination May, 00 Short Answer (a) (0) How are the EV and SE for the sum of n draws with replacement from a box computed? Solution: The EV is n times

More information

AP Statistics 2010 Scoring Guidelines

AP Statistics 2010 Scoring Guidelines AP Statistics 2010 Scoring Guidelines The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in

More information

UCLA STAT 13 Statistical Methods - Final Exam Review Solutions Chapter 7 Sampling Distributions of Estimates

UCLA STAT 13 Statistical Methods - Final Exam Review Solutions Chapter 7 Sampling Distributions of Estimates UCLA STAT 13 Statistical Methods - Final Exam Review Solutions Chapter 7 Sampling Distributions of Estimates 1. (a) (i) µ µ (ii) σ σ n is exactly Normally distributed. (c) (i) is approximately Normally

More information

HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1. used confidence intervals to answer questions such as...

HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1. used confidence intervals to answer questions such as... HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1 PREVIOUSLY used confidence intervals to answer questions such as... You know that 0.25% of women have red/green color blindness. You conduct a study of men

More information

Sample Exam #1 Elementary Statistics

Sample Exam #1 Elementary Statistics Sample Exam #1 Elementary Statistics Instructions. No books, notes, or calculators are allowed. 1. Some variables that were recorded while studying diets of sharks are given below. Which of the variables

More information

THE FIRST SET OF EXAMPLES USE SUMMARY DATA... EXAMPLE 7.2, PAGE 227 DESCRIBES A PROBLEM AND A HYPOTHESIS TEST IS PERFORMED IN EXAMPLE 7.

THE FIRST SET OF EXAMPLES USE SUMMARY DATA... EXAMPLE 7.2, PAGE 227 DESCRIBES A PROBLEM AND A HYPOTHESIS TEST IS PERFORMED IN EXAMPLE 7. THERE ARE TWO WAYS TO DO HYPOTHESIS TESTING WITH STATCRUNCH: WITH SUMMARY DATA (AS IN EXAMPLE 7.17, PAGE 236, IN ROSNER); WITH THE ORIGINAL DATA (AS IN EXAMPLE 8.5, PAGE 301 IN ROSNER THAT USES DATA FROM

More information

Two-sample hypothesis testing, II 9.07 3/16/2004

Two-sample hypothesis testing, II 9.07 3/16/2004 Two-sample hypothesis testing, II 9.07 3/16/004 Small sample tests for the difference between two independent means For two-sample tests of the difference in mean, things get a little confusing, here,

More information

When σ Is Known: Recall the Mystery Mean Activity where x bar = 240.79 and we have an SRS of size 16

When σ Is Known: Recall the Mystery Mean Activity where x bar = 240.79 and we have an SRS of size 16 8.3 ESTIMATING A POPULATION MEAN When σ Is Known: Recall the Mystery Mean Activity where x bar = 240.79 and we have an SRS of size 16 Task was to estimate the mean when we know that the situation is Normal

More information

Comparing Two Populations OPRE 6301

Comparing Two Populations OPRE 6301 Comparing Two Populations OPRE 6301 Introduction... In many applications, we are interested in hypotheses concerning differences between the means of two populations. For example, we may wish to decide

More information

Summary of Formulas and Concepts. Descriptive Statistics (Ch. 1-4)

Summary of Formulas and Concepts. Descriptive Statistics (Ch. 1-4) Summary of Formulas and Concepts Descriptive Statistics (Ch. 1-4) Definitions Population: The complete set of numerical information on a particular quantity in which an investigator is interested. We assume

More information

a) Find the five point summary for the home runs of the National League teams. b) What is the mean number of home runs by the American League teams?

a) Find the five point summary for the home runs of the National League teams. b) What is the mean number of home runs by the American League teams? 1. Phone surveys are sometimes used to rate TV shows. Such a survey records several variables listed below. Which ones of them are categorical and which are quantitative? - the number of people watching

More information

Lecture 28: Chapter 11, Section 1 Categorical & Quantitative Variable Inference in Paired Design

Lecture 28: Chapter 11, Section 1 Categorical & Quantitative Variable Inference in Paired Design Lecture 28: Chapter 11, Section 1 Categorical & Quantitative Variable Inference in Paired Design Inference for Relationships: 2 Approaches CatQuan Relationship: 3 Designs Inference for Paired Design Paired

More information

Analysis of numerical data S4

Analysis of numerical data S4 Basic medical statistics for clinical and experimental research Analysis of numerical data S4 Katarzyna Jóźwiak k.jozwiak@nki.nl 3rd November 2015 1/42 Hypothesis tests: numerical and ordinal data 1 group:

More information

3. There are three senior citizens in a room, ages 68, 70, and 72. If a seventy-year-old person enters the room, the

3. There are three senior citizens in a room, ages 68, 70, and 72. If a seventy-year-old person enters the room, the TMTA Statistics Exam 2011 1. Last month, the mean and standard deviation of the paychecks of 10 employees of a small company were $1250 and $150, respectively. This month, each one of the 10 employees

More information

11. Analysis of Case-control Studies Logistic Regression

11. Analysis of Case-control Studies Logistic Regression Research methods II 113 11. Analysis of Case-control Studies Logistic Regression This chapter builds upon and further develops the concepts and strategies described in Ch.6 of Mother and Child Health:

More information

Stats Review Chapters 9-10

Stats Review Chapters 9-10 Stats Review Chapters 9-10 Created by Teri Johnson Math Coordinator, Mary Stangler Center for Academic Success Examples are taken from Statistics 4 E by Michael Sullivan, III And the corresponding Test

More information

Sampling and Hypothesis Testing

Sampling and Hypothesis Testing Population and sample Sampling and Hypothesis Testing Allin Cottrell Population : an entire set of objects or units of observation of one sort or another. Sample : subset of a population. Parameter versus

More information

Final Exam Practice Problem Answers

Final Exam Practice Problem Answers Final Exam Practice Problem Answers The following data set consists of data gathered from 77 popular breakfast cereals. The variables in the data set are as follows: Brand: The brand name of the cereal

More information

Chapter 8 Hypothesis Tests. Chapter Table of Contents

Chapter 8 Hypothesis Tests. Chapter Table of Contents Chapter 8 Hypothesis Tests Chapter Table of Contents Introduction...157 One-Sample t-test...158 Paired t-test...164 Two-Sample Test for Proportions...169 Two-Sample Test for Variances...172 Discussion

More information

Experimental Design. Power and Sample Size Determination. Proportions. Proportions. Confidence Interval for p. The Binomial Test

Experimental Design. Power and Sample Size Determination. Proportions. Proportions. Confidence Interval for p. The Binomial Test Experimental Design Power and Sample Size Determination Bret Hanlon and Bret Larget Department of Statistics University of Wisconsin Madison November 3 8, 2011 To this point in the semester, we have largely

More information

AP Statistics 2002 Scoring Guidelines

AP Statistics 2002 Scoring Guidelines AP Statistics 2002 Scoring Guidelines The materials included in these files are intended for use by AP teachers for course and exam preparation in the classroom; permission for any other use must be sought

More information

Module 9: Nonparametric Tests. The Applied Research Center

Module 9: Nonparametric Tests. The Applied Research Center Module 9: Nonparametric Tests The Applied Research Center Module 9 Overview } Nonparametric Tests } Parametric vs. Nonparametric Tests } Restrictions of Nonparametric Tests } One-Sample Chi-Square Test

More information

The Chi-Square Test. STAT E-50 Introduction to Statistics

The Chi-Square Test. STAT E-50 Introduction to Statistics STAT -50 Introduction to Statistics The Chi-Square Test The Chi-square test is a nonparametric test that is used to compare experimental results with theoretical models. That is, we will be comparing observed

More information

BA 275 Review Problems - Week 5 (10/23/06-10/27/06) CD Lessons: 48, 49, 50, 51, 52 Textbook: pp. 380-394

BA 275 Review Problems - Week 5 (10/23/06-10/27/06) CD Lessons: 48, 49, 50, 51, 52 Textbook: pp. 380-394 BA 275 Review Problems - Week 5 (10/23/06-10/27/06) CD Lessons: 48, 49, 50, 51, 52 Textbook: pp. 380-394 1. Does vigorous exercise affect concentration? In general, the time needed for people to complete

More information

Math 58. Rumbos Fall 2008 1. Solutions to Review Problems for Exam 2

Math 58. Rumbos Fall 2008 1. Solutions to Review Problems for Exam 2 Math 58. Rumbos Fall 2008 1 Solutions to Review Problems for Exam 2 1. For each of the following scenarios, determine whether the binomial distribution is the appropriate distribution for the random variable

More information

The Goodness-of-Fit Test

The Goodness-of-Fit Test on the Lecture 49 Section 14.3 Hampden-Sydney College Tue, Apr 21, 2009 Outline 1 on the 2 3 on the 4 5 Hypotheses on the (Steps 1 and 2) (1) H 0 : H 1 : H 0 is false. (2) α = 0.05. p 1 = 0.24 p 2 = 0.20

More information

Math 108 Exam 3 Solutions Spring 00

Math 108 Exam 3 Solutions Spring 00 Math 108 Exam 3 Solutions Spring 00 1. An ecologist studying acid rain takes measurements of the ph in 12 randomly selected Adirondack lakes. The results are as follows: 3.0 6.5 5.0 4.2 5.5 4.7 3.4 6.8

More information

t Tests in Excel The Excel Statistical Master By Mark Harmon Copyright 2011 Mark Harmon

t Tests in Excel The Excel Statistical Master By Mark Harmon Copyright 2011 Mark Harmon t-tests in Excel By Mark Harmon Copyright 2011 Mark Harmon No part of this publication may be reproduced or distributed without the express permission of the author. mark@excelmasterseries.com www.excelmasterseries.com

More information

BA 275 Review Problems - Week 6 (10/30/06-11/3/06) CD Lessons: 53, 54, 55, 56 Textbook: pp. 394-398, 404-408, 410-420

BA 275 Review Problems - Week 6 (10/30/06-11/3/06) CD Lessons: 53, 54, 55, 56 Textbook: pp. 394-398, 404-408, 410-420 BA 275 Review Problems - Week 6 (10/30/06-11/3/06) CD Lessons: 53, 54, 55, 56 Textbook: pp. 394-398, 404-408, 410-420 1. Which of the following will increase the value of the power in a statistical test

More information

Introduction to Hypothesis Testing. Point estimation and confidence intervals are useful statistical inference procedures.

Introduction to Hypothesis Testing. Point estimation and confidence intervals are useful statistical inference procedures. Introduction to Hypothesis Testing Point estimation and confidence intervals are useful statistical inference procedures. Another type of inference is used frequently used concerns tests of hypotheses.

More information

Good luck! BUSINESS STATISTICS FINAL EXAM INSTRUCTIONS. Name:

Good luck! BUSINESS STATISTICS FINAL EXAM INSTRUCTIONS. Name: Glo bal Leadership M BA BUSINESS STATISTICS FINAL EXAM Name: INSTRUCTIONS 1. Do not open this exam until instructed to do so. 2. Be sure to fill in your name before starting the exam. 3. You have two hours

More information

Chapter 11-12 1 Review

Chapter 11-12 1 Review Chapter 11-12 Review Name 1. In formulating hypotheses for a statistical test of significance, the null hypothesis is often a statement of no effect or no difference. the probability of observing the data

More information

Power and Sample Size Determination

Power and Sample Size Determination Power and Sample Size Determination Bret Hanlon and Bret Larget Department of Statistics University of Wisconsin Madison November 3 8, 2011 Power 1 / 31 Experimental Design To this point in the semester,

More information

Hypothesis Testing (unknown σ)

Hypothesis Testing (unknown σ) Hypothesis Testing (unknown σ) Business Statistics Recall: Plan for Today Null and Alternative Hypotheses Types of errors: type I, type II Types of correct decisions: type A, type B Level of Significance

More information

HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1. used confidence intervals to answer questions such as...

HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1. used confidence intervals to answer questions such as... HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1 PREVIOUSLY used confidence intervals to answer questions such as... You know that 0.25% of women have red/green color blindness. You conduct a study of men

More information

Math 1070 Exam 2B 22 March, 2013

Math 1070 Exam 2B 22 March, 2013 Math 1070 Exam 2B 22 March, 2013 This exam will last 50 minutes and consists of 13 multiple choice and 6 free response problems. Write your answers in the space provided. All solutions must be sufficiently

More information

AP Statistics 2011 Scoring Guidelines

AP Statistics 2011 Scoring Guidelines AP Statistics 2011 Scoring Guidelines The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. STT315 Practice Ch 5-7 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. 1) The length of time a traffic signal stays green (nicknamed

More information

NCSS Statistical Software

NCSS Statistical Software Chapter 06 Introduction This procedure provides several reports for the comparison of two distributions, including confidence intervals for the difference in means, two-sample t-tests, the z-test, the

More information

Chapter Additional: Standard Deviation and Chi- Square

Chapter Additional: Standard Deviation and Chi- Square Chapter Additional: Standard Deviation and Chi- Square Chapter Outline: 6.4 Confidence Intervals for the Standard Deviation 7.5 Hypothesis testing for Standard Deviation Section 6.4 Objectives Interpret

More information

Hypothesis Testing for a Proportion

Hypothesis Testing for a Proportion Math 122 Intro to Stats Chapter 6 Semester II, 2015-16 Inference for Categorical Data Hypothesis Testing for a Proportion In a survey, 1864 out of 2246 randomly selected adults said texting while driving

More information

Solutions to Homework 3 Statistics 302 Professor Larget

Solutions to Homework 3 Statistics 302 Professor Larget s to Homework 3 Statistics 302 Professor Larget Textbook Exercises 3.20 Customized Home Pages A random sample of n = 1675 Internet users in the US in January 2010 found that 469 of them have customized

More information

= $96 = $24. (b) The degrees of freedom are. s n. 7.3. For the mean monthly rent, the 95% confidence interval for µ is

= $96 = $24. (b) The degrees of freedom are. s n. 7.3. For the mean monthly rent, the 95% confidence interval for µ is Chapter 7 Solutions 71 (a) The standard error of the mean is df = n 1 = 15 s n = $96 = $24 (b) The degrees of freedom are 16 72 In each case, use df = n 1; if that number is not in Table D, drop to the

More information

5.1 Identifying the Target Parameter

5.1 Identifying the Target Parameter University of California, Davis Department of Statistics Summer Session II Statistics 13 August 20, 2012 Date of latest update: August 20 Lecture 5: Estimation with Confidence intervals 5.1 Identifying

More information

Outline of Topics. Statistical Methods I. Types of Data. Descriptive Statistics

Outline of Topics. Statistical Methods I. Types of Data. Descriptive Statistics Statistical Methods I Tamekia L. Jones, Ph.D. (tjones@cog.ufl.edu) Research Assistant Professor Children s Oncology Group Statistics & Data Center Department of Biostatistics Colleges of Medicine and Public

More information

9 Testing the Difference

9 Testing the Difference blu49076_ch09.qxd 5/1/2003 8:19 AM Page 431 c h a p t e r 9 9 Testing the Difference Between Two Means, Two Variances, and Two Proportions Outline 9 1 Introduction 9 2 Testing the Difference Between Two

More information

Confidence level. Most common choices are 90%, 95%, or 99%. (α = 10%), (α = 5%), (α = 1%)

Confidence level. Most common choices are 90%, 95%, or 99%. (α = 10%), (α = 5%), (α = 1%) Confidence Interval A confidence interval (or interval estimate) is a range (or an interval) of values used to estimate the true value of a population parameter. A confidence interval is sometimes abbreviated

More information