# Math 251, Review Questions for Test 3 Rough Answers

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1 Math 251, Review Questions for Test 3 Rough Answers 1. (Review of some terminology from Section 7.1) In a state with 459,341 voters, a poll of 2300 voters finds that 45 percent support the Republican candidate, where in reality, unknown to the pollster, 42 percent support the Republican candidate. (a) What is the value of the statistic of interest? Answer. 45% (a statistic is a numerical property of the sample) (b) What is the value of the parameter of interest? Answer. 42% (a parameter is a numerical property of the population) (c) Describe the population of interest. Answer. The population is all voters in the state. (d) In general, is it true that given a certain population, the parameter of interest will not change under repeated sampling? xplain. Answer. True, the parameter does not depend on a specific sample, so it doesn t change when the sample changes. 2. (a) A produce company claims that the mean weight of peaches in a large shipment is 6.0 oz with a standard deviation of 1.0 oz. Assuming this claim is true, what is the probability that a random sample of 1000 of these peaches would have a mean weight of 5.9 oz or less? Answer: This is a central limit theorem type problem on sampling distributions (see Section 7.2 for more) z = P (z < 3.16) = =.0008 Thus, there is approximately a.0008 probability of obtaining such a sample assuming that the mean is 6.0 oz. (b) If the store manager randomly selected 1000 peaches and found that the mean weight of those 1000 peaches was 5.9 oz, should she be suspicious of the produce company s claim that the mean weight of peaches in the shipment is 6.0 oz? xplain. Answer. Yes, she is about 99.9% sure that the true mean weight is less than 6.0 oz. 3. (From p. 349 #6). The heights of 18 year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches. 1

2 (a) What is the probability that an 18 year-old man selected at random is between 67 and 69 inches tall. ( Answer. P (67 < x < 69) = P 1 3 < z < 1 ) = = (b) If a random sample of nine 18 year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? Answer. P (67 < x < 69) = P ( < z < (c) Why was the probability in (b) higher than that in (a)? ) = P ( 1 < z < 1) = Answer. The sampling distribution standard deviation is smaller than the standard deviation for the original distribution, so (b) should be higher. In general, one expects that a randomly chosen group will have an average much closer to the population mean than a randomly selected individual, the larger the group, the closer one would expect its sample mean to be to the population mean. (d) Would you expect the probability that an 18 year-old man selected at random is more than 74 inches tall to be lower, the same as, or higher than the probability of selecting a random sample of nine eighteen year-old men with a mean height of more than 74 inches? Answer. Because the sampling distribution has a smaller standard deviation than the original population distribution, there is a much higher probability of finding individuals far away from the mean, than random samples whose sample mean is far away from the population mean. Thus, we expect that it is much more likely to randomly an individual more than 74 inches tall, than a group of 9 whose mean height is more than 74 inches. 4. (From p. 349 # 18). The taxi and takeoff time for commercial jets is a random variable x with mean 8.5 minutes and standard deviation of 2.5 minutes. You may assume the jets are lined up on the runway so that one taxis and takes off immediately after the other, and they take off one at time on a given runway. What is the probability that for 36 jets on a given runway total taxi and take off time will be (a) less than 320 minutes? Answer. P ( x < 320 ) ( = P z < 36 (b) more than 275 minutes? Answer. P ( x > 275 ) ( = P z > 36 ) = P (z <.93) = ) = P (z > 2.07) = =

6 9. (From p. 385 #6) The Roman Arches is an Italian restaurant. The manager wants to estimate the average amount a customer sends on lunch Monday through Friday. A random sample of 115 customers lunch tabs gave a mean of x = \$9.74 with a standard deviation s = \$2.93. (a) Find a 95% confidence interval for the average amount spent on lunch by all customers. Answer. The sample size is n = , and so we use the formula for large sample means, i.e. x ± z c σ x. In this case, σ x = =.2732 and z.95 = 1.96 and so the 95% confidence interval is (\$9.20, \$10.28). (b) For a day when the Roman Arches has 115 lunch customers, use part (a) to estimate the range of dollar values for the total lunch income that day. Answer. We are 95% confident that total would be in the range (115 \$9.20, 115 \$10.28), that is (\$1058, \$1182). 10. (From p. 398 #12) The number of calories for 3 ounces of french fries at eight popular fast food chains are as follows Use these data to find a 99% confidence interval for the mean calorie count in 3 ounces of french fries obtained from fast-food restaurants. Answer. First, one needs to compute the sample mean and sample standard deviation for the 8 numbers above. For this, one finds x = 1956 x 2 = SS x = = Thus x = = and s = = This is a small sample (n < 30) with unknown standard deviation, and we ll assume the population is normal, or nearly so. We use the t-distribution with 7 degrees of freedom, thus s t.99 = 3.499, and the endpoints of the confidence interval are given by x ±t c. Thus the 99% n confidence interval has endpoints ± which yields the interval (217.6, 271.4). 11. (a) Under are the conditions necessary for finding a confidence interval for the mean from a small sample? Answer. We should use the t-distribution with n 1 degrees of freedom when the sample size, n, is less than 30 provided the population is normal (or nearly normal) and the population standard deviation is not known. 6

9 Answer. We would reject H 0 for levels of significance α =.03 and α =.05 because the Pvalue is less than these α s, we would not reject H 0 for α = (a) If a population has a standard deviation of 80, what sample size would be necessary in order for a 95% confidence interval to estimate the population mean within 10? ( zc σ ) ( ) Answer. We use the formula n = and we find that n = = The 10 sample size must be the next larger whole number which is n = 246. (b) What size of sample is needed by the Gallup organization to estimate a population proportion within ±.02 with 95% confidence. In your calculation assume that there is no preliminary estimate for p. Answer. We use the formula n = 1 4 ( zc ) 2 1 = 4 ( ) which gives us a sample size of (c) Suppose you are to construct a 99% confidence interval for the mean using a sample of size n = 12 from a normal population with unknown standard deviation. What value of t c would s you use in the formula x ± t c? n Answer. We use a t-distribution with n 1 = 11 degrees of freedom, and look on the table in the back cover to find t.99 = (d) Find the critical region if a two-tailed test on a mean is conducted using a large sample at a level of significance of.01. Answer. With the help of the normal table in the front cover, we find that the critical region is z 2.58 or z (e) Find the critical region for a left-tailed test on a proportion at a level of significance α =.05? Answer. With the help of the normal table in the front cover, we find that the critical region is z (f) xplain what type I and type II errors are in hypothesis tests. Answer. See text Section 9.1. (g) What is the probability with which we are willing to risk a type I error called? Answer. The level of significance, which is denoted by α. 16. (a) A developer wishes to test whether the mean depth of water below the surface in a large development tract was less than 500 feet. The sample data was as follows: n = 32 test 9

10 holes, the sample mean was 486 feet, and the standard deviation was s = 53 feet. Complete the test by computing the Pvalue, and report the conclusion for a 1% level of significance. Answer: Null Hypothesis: µ = 500 Alternative Hypothesis: µ < 500 Using the sample data, we compute z = = 1.49 Thus the Pvalue is P (z < 1.49) =.0681 We would not reject the null hypothesis at a level of significance of.01, because the P-value is larger than (b) What would the conclusion of the test be for a level of significance of α =.05. Answer. Do not reject H 0 since the Pvalue is bigger than.05. (c) What type of error was possibly made in (b)? Answer. A type II error. 17. A vendor was concerned that a soft drink machine was not dispensing 6 ounces per cup, on average. A sample size of 40 gave a mean amount per cup of 5.95 ounces and a standard deviation of.15 ounce. (a) Find the Pvalue for an hypothesis test to determine if the mean is different from 6 ounces. Answer. This is a two-tailed test with Null Hypothesis: µ = 6 Alternative Hypothesis: µ 6. Using the data, we compute z = = The Pvalue is: P (z < 2.11) + P (z > 2.11) = 2 P (z < 2.11) = 2(.0174) =.0348 (b) For which of the following levels of significance would the null hypothesis be rejected? (i) α =.10 (ii) α =.05 (iii ) α =.04 (iv) α =.01 Answer. Reject the null hypothesis in (i), (ii) and (iii) since the Pvalue is smaller than α; do not reject the null hypothesis in (iv). (c) For each case in part (b), what type of error has possibly been committed? 10

11 Answer. Possible Type I error may occur in (i), (ii) and (iii) while a Type II error may occur in (iv). (d) Find a 96% confidence interval for the mean amount of soda dispensed per cup. Answer. For c =.96, z c = 2.05 (approximately), look at z value corresponding to an area of.98 on table. Thus the confidence interval, using the large sample method (n is at least 30) yields endpoints: ± and, so the confidence interval is: (5.901, 5.999). (e) Is your interval in (d) consistent with the test conclusion in (b)(iii)? xplain. Answer. Yes. In (b)(iii) we have a confidence level of (at least) c = 1 α =.96 that the mean is different from 6, while in (d) we had a 96% confidence interval that did not contain 6, and so we were (at least) 96% certain that the mean is different from 6. Notice also the confidence interval comes very close to containing 6, this is reflected in the Pvalue being very close to 0.04 in the hypothesis test. (f) Based on your answer to (b)(iv) would you expect a 99% confidence interval to contain 6? Answer. Yes, because we were not 99% confident that the mean was different from 6, we would expect the corresponding confidence interval to contain 6. (g) Suppose that the population standard deviation is σ =.15, what sample size would be needed so that the maximum error in a 96% confidence interval is =.01? ( zc σ ) 2. Answer. The formula to use is: n = So we compute ( ) n = = ,.01 thus we should use a sample size of n = (a) Suppose that a February Gallup poll of 1200 randomly selected voters found that 53 percent support George W. Bush s energy policy. Conduct an hypothesis test at a level of significance of α =.01 to test whether the true voter population support for George W. Bush s energy policy in February was less than 56 percent. Answer. The null hypothesis is H 0 : p =.56, and the alternative hypothesis is H 1 : p <.56. The critical region is z Since n = 1200 and p =.56 and q =.44, we clearly have np > 5 and nq > 5. Thus we compute z = (.56)(.44)

12 Because 2.09 does not fall in the critical region, we conclude there is not sufficient evidence to reject the null hypothesis at a level of significance of 1%. (b) Report the Pvalue of the test in (a) and give a practical interpretation of it. Answer. The Pvalue is P (z < 2.09) = Thus we are roughly 98% certain that less than 56% of all voters at the time of the poll supported President G.W. Bush s energy policy. 19. (From p. 536 #8) A reading test is given to both a control group and an experimental group (which received special tutoring). The average score for the 30 subjects in the control group was with a standard deviation of The average score for the 30 subjects in the experimental group was with a standard deviation of Use a 4% level of significance to test the claim that the experimental group performed better than the control group. Answer. We wish to conduct the test H 0 : µ 1 = µ 2 versus H 1 : µ 1 > µ 2 where µ 1 is the population mean test score for all people who received the special tutoring. We use the formula z = ( x 1 x 2 ) (µ 1 µ 2 ) σ1 2 where σ x x2 = + σ2 2 Thus σ x1 x 2 n 1 n σ x x2 = = ( ) 0 Therefore, z = = The Pvalue is P (z > 1.52) = = Because the Pvalue is larger than α =.04, we do not reject the null hypothesis at a 4% level of significance. There is not sufficient evidence to show that the tutoring increases the mean score. 20. (From p. 505 #14) Nationally about 28% of the population believes that NAFTA benefits America. A random sample of 48 interstate truck drivers showed that 19 believe NAFTA benefits America. Conduct an hypothesis test to determine whether the population proportion of interstate truckers who believe NAFTA benefits America is higher than 28%. Test at a level of significance of α =.05. (a) State the null and alternative hypotheses. Is this a right-tailed, left-tailed or two-tailed test? Answer. The null hypothesis is H 0 : p =.28, the alternative hypothesis is H 1 : p >.28. This is a right-tailed test. (b) Find the Pvalue for the test. Answer. For this data, we have n = 48, p =.28 and q =.72. Clearly np > 5 and nq > 5 and so we compute ˆp = 19 =.39583; then z = 1.79 (0.28)(0.72) 48 12

13 Therefore, the Pvalue is P (z > 1.79) = = Because the Pvalue is less than α =.05, we reject H 0. The results are statistically significant; they indicate that the proportion of interstate truck drivers that believe NAFTA benefits America is higher than (c) Would you reject the null hypothesis at a level of significance of α =.04? Answer. Yes, because the Pvalue is less than (d) Would you reject the null hypothesis at a level of significance of α =.01? Answer. No, because the Pvalue is more than (e) Do you think the proportion of interstate truckers who believe NAFTA benefits America is higher than 28%? xplain your answer. Answer. Yes, and I m quite sure of this, in fact, I m roughly 96% sure of it. 21. (From p. 539 #18) A random sample of n 1 = 288 voters registered in the state of California showed that 141 voted in the last general election. A random sample of n 2 = 216 registered voters in Colorado showed that 125 voted in the last general election. Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a 5% level of significance. Answer. Let p 1 and p 2 represent the population proportions of voter turnout in California and Colorado respectively. We will test H 0 : p 1 = p 2 versus H 1 : p 1 < p 2. Since we are assuming p 1 = p 2, we use the pooled estimate for proportions (see p. 530), that is ˆp = = Then we use this for an estimate of p 1 and p 2 in the formula for σˆp1 ˆp 2, so we obtain σˆp1 ˆp 2 = From this, we compute (0.5278)(0.4722) 288 z = (ˆp 1 ˆp 2 ) (p 1 p 2 ) σˆp1 ˆp 2 = + (0.5278)(0.4722) 216 ( ) = = The Pvalue is P (z < 1.98) = Because the Pvalue is less than.05, we reject H 0 at the 5% level of significance. 13

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