Chapter 8. Hypothesis Testing


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1 Chapter 8 Hypothesis Testing
2 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 a claim about a property of a population. If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probably not correct.
3 Null Hypothesis: H 0 The null hypothesis (denoted by ) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We test the null hypothesis directly. Either reject or fail to reject H. H0 0 Example H 0 : p = 0.5 H 0
4 Alternative Hypothesis: H 1 The alternative hypothesis (denoted by ) is the statement that the parameter has a value that somehow differs from the null hypothesis. H or H or H 1 a A The symbolic form of the alternative hypothesis must use one of these symbols:. Example H 0 : p = 0.5,, H 1 : p < 0.5
5 Nonstatistical Hypothesis Testing A criminal trial is an example of hypothesis testing without the statistics. In a trial a jury must decide between two hypotheses. The null hypothesis is H 0 : The defendant is innocent The alternative hypothesis or research hypothesis is H 1 : The defendant is guilty The jury does not know which hypothesis is true. They must make a decision on the basis of evidence presented. 11.5
6 Nonstatistical Hypothesis Testing In the language of statistics convicting the defendant is called rejecting the null hypothesis in favor of the alternative hypothesis. That is, the jury is saying that there is enough evidence to conclude that the defendant is guilty (i.e., there is enough evidence to support the alternative hypothesis). If the jury acquits it is stating that there is not enough evidence to support the alternative hypothesis. Notice that the jury is not saying that the defendant is innocent, only that there is not enough evidence to support the alternative hypothesis. That is why we never say that we accept the null hypothesis. 11.6
7 Nonstatistical Hypothesis Testing There are two possible errors. Type I error and Type II error. A Type I error occurs when we reject a true null hypothesis. That is, a Type I error occurs when the jury convicts an innocent person. We would want the probability of this type of error to be very small for a criminal trial where a conviction results in the death penalty. A Type II error occurs when we don t reject a false null hypothesis. That occurs when a guilty defendant is acquitted. In practice, this type of error is by far the most serious mistake we normally make. For example, if we test the hypothesis that the amount of medication in a heart pill is equal to a value which will cure your heart problem and accept the null hypothesis that the amount is ok. Later on we find out that the average amount is WAY too large and people die from too much medication. 11.7
8 Nonstatistical Hypothesis Testing The probability of a Type I error is denoted as α (Greek letter alpha). The probability of a type II error is β (Greek letter beta). The two probabilities are inversely related. Decreasing one increases the other, for a fixed sample size. In other words, you can t have and β both real small for any old sample size. You may have to take a much larger sample size, or in the court example, you need much more evidence. 11.8
9 Types of Errors A Type I error occurs when we reject a true null hypothesis (i.e. Reject H 0 when it is TRUE) H 0 T F Reject I Reject II A Type II error occurs when we don t reject a false null hypothesis (i.e. Do NOT reject H 0 when it is FALSE) 11.9
10 Nonstatistical Hypothesis Testing The critical concepts are these: 1. There are two hypotheses, the null and the alternative hypotheses. 2. The procedure begins with the assumption that the null hypothesis is true. 3. The goal is to determine whether there is enough evidence to infer that the alternative hypothesis is true, or the null is not likely to be true. There are two possible decisions: Conclude that there is enough evidence to support the alternative hypothesis. Reject the null. Conclude that there is not enough evidence to support the alternative hypothesis. Fail to reject the null
11 Example 1 Claim: The XSORT method of gender selection increases the likelihood of birthing a girl. Let p denote the proportion of girls born. The claim is equivalent to p > 0.5 The null hypothesis must say equal to : H 0 : p = 0.5 The alternative hypothesis states the difference: H 1 : p > 0.5
12 Example 2 Claim: For couples using the XSORT method, the likelihood of having a girl is 50% Again, let p denote the proportion of girls born. The claim is equivalent to p=0.5 The null hypothesis must say equal to : H 0 : p = 0.5 The alternative hypothesis states the difference: H 1 : p 0.5
13 Example 3 Claim: For couples using the XSORT method, the likelihood of having a girl is at least 50% Again, let p denote the proportion of girls born. The claim is equivalent to p 0.5 The null hypothesis must say equal to : H 0 : p = 0.5 The alternative hypothesis states the difference: H 1 : p < 0.5 we can t use or in the alternative hypothesis, so we test the negation
14 Example Consider the claim that the mean weight of airline passengers (including carryon baggage) is at most 195 lb (the current value used by the Federal Aviation Administration). Identify the null hypothesis and the alternative hypothesis. Step 1: Express the given claim in symbolic form. The claim that the mean is at most 195 lb is expressed in symbolic form as. 195lb Step 2: If is false, then must be true. 195lb 195lb 195lb 195lb Step 3: Of the two symbolic expressions and, we see that does not contain equality, so we let the alternative hypothesis be. Also, the null hypothesis must be a statement that the mean equals 195 lb, so we let H0 be 195lb. 195lb H 1 195lb Note that the original claim that the mean is at most 195 lb is neither the alternative hypothesis nor the null hypothesis. (However, we would be able to address the original claim upon completion of a hypothesis test.)
15 Test Statistic  Formulas Test statistic for proportion z ˆp p pq n Test statistic for mean z x n or t x s n Test statistic for standard deviation 2 n1 s 2 2
16 Basic Methods of Testing Claims about a Population Proportion p Notation n = number of trials pˆ x n (sample proportion) p = population proportion (used in the null hypothesis) q 1 p
17 ˆp ˆp Obtaining ˆp sometimes is given directly 10% of the observed sports cars are red is expressed as ˆ 0.10 p sometimes must be calculated 96 surveyed households have cable TV and 54 do not is calculated using x 96 pˆ 0.64 n (96 54) (determining the sample proportion of households with cable TV)
18 Requirements for Testing Claims About a Population Proportion p 1) The sample observations are a simple random sample. 2) The conditions for a binomial distribution are satisfied. np 5 3) The conditions and nq 5 are both satisfied, so the binomial distribution of sample proportions can be approximated by a normal distribution with and. Note: p is the assumed proportion not the sample proportion. npq np
19 Example: Body Weight The problem: In the 1970s, year old men in the U.S. had a mean (μ) body weight of 170 pounds. Standard deviation σ was 40 pounds. We test whether mean body weight in the population now differs. Null hypothesis H 0: μ = 170 ( no difference ) The alternative hypothesis can be either H a: μ > 170 (overweight problem)(onesided test) or H a: μ 170 (twosided test)
20 Example: z statistic For the example, μ 0 = 170 We know σ = 40 Take an Simple Random Sampling of n = 64. Therefore If we found a sample mean of 173, then z stat SE x x SE x 0 n If we found a sample mean of 185, then z stat x SE x
21 Onesided Pvalue for z stat of 0.6
22 Onesided Pvalue for z stat of 3.0
23 TwoSided PValue Onesided H a Area under curve in tail beyond z stat Twosided H a consider potential deviations in both directions double the onesided Pvalue Examples: If onesided P = , then twosided P = = If onesided P = , then twosided P = =
24 Example A department store manager determines that a new billing system will be costeffective only if the mean monthly account is more than $170. A random sample of 400 monthly accounts is drawn, for which the sample mean is $178. The accounts are approximately normally distributed with a standard deviation of $65. Can we conclude that the new system will be costeffective?
25 Solution The system will be cost effective if the mean account balance for all customers is greater than $170. We express this belief as a our research hypothesis, that is: H 1 : > 170 (this is what we want to determine) Thus, our null hypothesis becomes: H 0 : = 170
26 What we want to show: H 1 : > 170 H 0 : = 170 (we ll assume this is true) Normally we put H o first. We know: n = 400, = 178, and = 65 = 65/SQRT(400) = 3.25 = 0.05 (significance level)
27 Rejection Region The rejection region is a range of values such that if the test statistic falls into that range, we decide to reject the null hypothesis in favor of the alternative hypothesis. is the critical value of to reject H 0.
28 At a 5% significance level (i.e = 0.05), we get [all in one tail] Z = Z 0.05 = Therefore, Upper critical value = *3.25 = Since our sample mean (178) is greater than the critical value we calculated (175.35), we reject the null hypothesis in favor of H 1 OR >1.645) Reject null hypothesis OR pvalue = P( > 178) = P(Z > 2.46) = < 0.05 Reject null hypothesis.
29 H 1 : > 170 H 0 : = 170 = =178 Reject H 0 in favor of
30 Interpreting the pvalue The smaller the pvalue, the more statistical evidence exists to support the alternative hypothesis. If the pvalue is less than 1%, there is overwhelming evidence that supports the alternative hypothesis. If the pvalue is between 1% and 5%, there is a strong evidence that supports the alternative hypothesis. If the pvalue is between 5% and 10% there is a weak evidence that supports the alternative hypothesis. If the pvalue exceeds 10%, there is no evidence that supports the alternative hypothesis. We observe a pvalue of.0069, hence there is overwhelming evidence to support H 1 : > 170.
31 One tail test with rejection region on right The last example was a one tail test, because the rejection region is located in only one tail of the sampling distribution: More correctly, this was an example of a right tail test. H 1 : μ > 170 H 0 : μ < 170
32 One tail test with rejection region on left The rejection region will be in the left tail.
33 Two tail test with rejection region in both tails The rejection region is split equally between the two tails.
34 Example Let s again consider the claim that the XSORT method of gender selection increases the likelihood of having a baby girl. Preliminary results from a test of the XSORT method of gender selection involved 14 couples who gave birth to 13 girls and 1 boy. Use the given claim and the preliminary results to calculate the value of the test statistic. Use the format of the test statistic given above, so that a normal distribution is used to approximate a binomial distribution.
35 The claim that the XSORT method of gender selection increases the likelihood of having a baby girl results in the following null and alternative hypotheses and H : p We work under the assumption that the null hypothesis is true with p = 0.5. The sample proportion of 13 girls in 14 births results in. Using p = 0.5, and n = 14, we find the value of the test statistic as follows: pˆ pˆ H : p 0.5 0
36 z pˆ p pq n We know from previous chapters that a z score of 3.21 is unusual (because it is greater than 3). It appears that in addition to being greater than 0.5, the sample proportion of 13/14 or is significantly greater than 0.5. The figure on the next slide shows that the sample proportion of does fall within the range of values considered to be significant because
37 they are so far above 0.5 that they are not likely to occur by chance (assuming that the population proportion is p = 0.5). Sample proportion of: or Test Statistic z = 3.21 ˆp 0.929
38 Critical Region The critical region (or rejection region) is the set of all values of the test statistic that cause us to reject the null hypothesis. For example, see the redshaded region in the previous figure.
39 Significance Level The significance level (denoted by ) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. This is the same introduced in Section 72. Common choices for are 0.05, 0.01, and 0.10.
40 Critical Value A critical value is any value that separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to rejection of the null hypothesis. The critical values depend on the nature of the null hypothesis, the sampling distribution that applies, and the significance level. See the previous figure where the critical value of z = corresponds to a significance level of. 0.05
41 PValue The Pvalue (or pvalue or probability value) is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. Critical region in the left tail: Pvalue = area to the left of the test statistic Critical region in the right tail: Critical region in two tails: Pvalue = area to the right of the test statistic Pvalue = twice the area in the tail beyond the test statistic
42 PValue The null hypothesis is rejected if the P value is very small, such as 0.05 or less. Here is a memory tool useful for interpreting the Pvalue: If the P is low, the null must go. If the P is high, the null will fly.
43 Procedure for Finding PValues Figure 85
44 Example Consider the claim that with the XSORT method of gender selection, the likelihood of having a baby girl is different from p = 0.5, and use the test statistic z = 3.21 found from 13 girls in 14 births. First determine whether the given conditions result in a critical region in the right tail, left tail, or two tails, then use Figure 85 to find the Pvalue. Interpret the Pvalue.
45 Solution The claim that the likelihood of having a baby girl is different from p 0.5 can be expressed as p 0.5 so the critical region is in two tails. Using Figure 85 to find the Pvalue for a twotailed test, we see that the Pvalue is twice the area to the right of the test statistic z = We refer to Table A2 (or use technology) to find that the area to the right of z = 3.21 is In this case, the Pvalue is twice the area to the right of the test statistic, so we have: Pvalue = = The Pvalue is (or if greater precision is used for the calculations). The small Pvalue of shows that there is a very small chance of getting the sample results that led to a test statistic of z = This suggests that with the XSORT method of gender selection, the likelihood of having a baby girl is different from 0.5.
46 Types of Hypothesis Tests: Twotailed, Lefttailed, Righttailed The tails in a distribution are the extreme regions bounded by critical values. Determinations of Pvalues and critical values are affected by whether a critical region is in two tails, the left tail, or the right tail. It therefore becomes important to correctly characterize a hypothesis test as twotailed, lefttailed, or righttailed.
47 Twotailed Test H 0 : H 1 : is divided equally between the two tails of the critical region Means less than or greater than
48 Lefttailed Test H 0 : H 1 : the left tail Points Left
49 Righttailed Test H 0 : H 1 : Points Right
50 Conclusions in Hypothesis Testing We always test the null hypothesis. The initial conclusion will always be one of the following: 1. Reject the null hypothesis. 2. Fail to reject the null hypothesis.
51 Decision Criterion Pvalue method: Using the significance level : If Pvalue, reject. H 0 0 If Pvalue, fail to reject. H
52 Decision Criterion Traditional method: If the test statistic falls within the critical region, reject. H 0 If the test statistic does not fall within the critical region, fail to reject. H 0
53 Example In a study 57 out of 104 pregnant women correctly guessed the sex of their babies. Use these sample data to test the claim that the success rate of such guesses is no different from the 50% success rate expected with random chance guesses. Use a 0.05 significance level. Requirements are satisfied: simple random sample; fixed number of trials (104) with two categories (guess correctly or do not); and np (104)(0.5) 52 5 (104)(0.5) 52 5 Step 1: original claim is that the success rate is no different from 50%: Step 2: opposite of original claim is nq p 0.50 p 0.50 p Step 3: does not contain equality so it is. H H : p H : p null hypothesis and original claim alternative hypothesis Step 4: significance level is α = 0.05 Step 5: sample involves proportion so the relevant statistic is the sample proportion, ˆp
54 Step 6: calculate z: 57 ˆ 0.50 p p z pq n 104 twotailed test, Pvalue is twice the area to the right of test statistic Table A2: z = 0.98 has an area of to its left, so area to the right is = , doubles yields Step 7: the Pvalue of is greater than the significance level of 0.05, so fail to reject the null hypothesis. Here is the correct conclusion: There is not sufficient evidence to warrant rejection of the claim that women who guess the sex of their babies have a success rate equal to 50%.
55 Requirements for Testing Claims About a Population Mean (with Known) 1) The sample is a simple random sample. 2) The value of the population standard deviation is known. 3) Either or both of these conditions is satisfied: The population is normally distributed or n 30.
56 Example People have died in boat accidents because an obsolete estimate of the mean weight of men was used. Using the weights of the simple random sample of men from Data Set 1 in Appendix B, we obtain these sample statistics: n = 40 and = lb. Research from several other sources suggests that the population of weights of men has a standard deviation given by = 26 lb. Use these results to test the claim that men have a mean weight greater than lb, which was the weight in the National Transportation and Safety Board s recommendation M Use a 0.05 significance level, and use the Pvalue method to claim your hypothesis. x
57 Solution Requirements are satisfied: simple random sample, sample size is 40 ( n 30) Step 1: Express claim as 166.3lb is known (26 lb), Step 2: alternative to claim is 166.3lb Step 3: 166.3lb does not contain equality, it is the alternative hypothesis: H : 166.3lb 0 null hypothesis H : 166.3lb 1 Step 4: significance level is alternative hypothesis and original claim 0.05 Step 5: claim is about the population mean, so the relevant statistic is the sample mean ( lb), is known (26 lb), sample size greater than 30 Step 6: calculate z x x z righttailed test, so Pvalue is the area is to the right of z = 1.52; n 40
58 Table A2: area to the left of z = 1.52 is , so the area to the right is = The Pvalue is Step 7: The Pvalue of is greater than the significance level of, we fail to reject the null hypothesis Pvalue = = or z = 0 x or z = lb The Pvalue of tells us that if men have a mean weight given by, there is a good chance (0.0643) of getting a sample mean of lb. A sample mean such as lb could easily occur by chance. There is not sufficient evidence to support a conclusion that the population mean is greater than lb, as in the National Transportation and Safety Board s recommendation.
59 The traditional method: Use z = instead of finding the Pvalue. Since z = 1.52 does not fall in the critical region, again fail to reject the null hypothesis. Confidence Interval method: Use a onetailed test with, so construct a 90% confidence interval: The confidence interval contains lb, we cannot support a claim that is greater than Again, fail to reject the null hypothesis.
60 Requirements for Testing Claims About Mean (with a Population Not Known) 1) The sample is a simple random sample. 2) The value of the population standard deviation is not known. 3) Either or both of these conditions is satisfied: The population is normally distributed or. n 30
61 Test Statistic for Testing a Claim About a Mean (with Not Known) t x x s n Pvalues and Critical Values Found in Table A3 Degrees of freedom (df) = n 1
62 Example: People have died in boat accidents because an obsolete estimate of the mean weight of men was used. Using the weights of the simple random sample of men from Data Set 1 in Appendix B, we obtain these sample statistics: n 40 and x lb, and 26.33lb. Do not assume that the value of is known. Use these results to test the claim that men have a mean weight greater than lb, which was the weight in the National Transportation and Safety Board s recommendation M Use a 0.05 significance level, and the traditional method outlined in Figure 89.
63 Example: Requirements are satisfied: simple random sample, population standard deviation is not known, sample size is 40 Step 1: Express claim as ( n 30) 166.3lb Step 2: alternative to claim is 166.3lb 166.3lb Step 3: does not contain equality, it is the alternative hypothesis: null hypothesis H : 166.3lb 1 alternative hypothesis and original claim H : 166.3lb 0
64 Example: Step 4: significance level is Step 5: claim is about the population mean, so the relevant statistic is the sample mean, lb Step 6: calculate t 0.05 t x x s n 40 df = n 1 = 39, area of 0.05, onetail yields t = 1.685;
65 Example: Step 7: t = does not fall in the critical region bounded by t = 1.685, we fail to reject the null hypothesis. = or z = 0 x or t = 1.52 Critical value t = 1.685
66 Example: Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support a conclusion that the population mean is greater than lb, as in the National Transportation and Safety Board s recommendation.
67 Normal Distribution Versus Student t Distribution The critical value in the preceding example was t = 1.782, but if the normal distribution were being used, the critical value would have been z = The Student t critical value is larger (farther to the right), showing that with the Student t distribution, the sample evidence must be more extreme before we can consider it to be significant.
68 Question: Let X represent Weschler Adult Intelligence scores (WAIS),typically, X ~ N(100, 15) Take SRS of n = 9 from population Data {116, 128, 125, 119, 89, 99, 105, 116, 118}. Calculate: xbar = Does sample mean provide strong evidence that population mean μ > 100? A. Hypotheses: H 0 : µ = 100 versus H a : µ > 100 (onesided) H a : µ 100 (twosided) B. Test statistic: SE x n z stat x SE x
69 C. Pvalue: P = Pr(Z 2.56) = P =.0052 it is unlikely the sample came from this null distribution strong evidence against H 0
70 H a : µ 100 Considers random deviations up and down from μ 0 tails above and below ±z stat Thus, twosided P = = TwoSided Pvalue
71 Wording of Final Conclusion Figure 87
72 Accept Versus Fail to Reject Some texts use accept the null hypothesis. We are not proving the null hypothesis. Fail to reject says more correctly The available evidence is not strong enough to warrant rejection of the null hypothesis (such as not enough evidence to convict a suspect).
73 Type I Error A Type I error is the mistake of rejecting the null hypothesis when it is actually true. The symbol (alpha) is used to represent the probability of a type I error.
74 Type II Error A Type II error is the mistake of failing to reject the null hypothesis when it is actually false. The symbol (beta) is used to represent the probability of a type II error.
75 Type I and Type II Errors
76 Example Assume that we are conducting a hypothesis test of the claim that a method of gender selection increases the likelihood of a baby girl, so that the probability of a baby girls is p 0.5. Here are the null and alternative hypotheses:, and. H : p 0.5 : a) Identify a type I error. b) Identify a type II error. H p
77 Example a) A type I error is the mistake of rejecting a true null hypothesis, so this is a type I error: Conclude that there is sufficient evidence to support p 0.5, when in reality p 0.5. b) A type II error is the mistake of failing to reject the null hypothesis when it is false, so this is a type II error: Fail to reject (and therefore fail to support ) when in reality. p 0.5 p 0.5 p 0.5
78 Controlling Type I and Type II Errors For any fixed, an increase in the sample size n will cause a decrease in For any fixed sample size n, a decrease in will cause an increase in. Conversely, an increase in will cause a decrease in. To decrease both and, increase the sample size.
79 Comprehensive Hypothesis Test PValue Method
80 Comprehensive Hypothesis Test Traditional Method
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82 Basics of Hypothesis Testing. Definitions. Rare Event Rule for Inferential Statistics. Null Hypothesis
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