Math 251, Review Questions for Test 3 Rough Answers

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). 5.9 6.0 z = 1.0 3.16 1000 P (z < 3.16) =.5.4992 =.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

(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 ) = 0.6293 0.3707 = 0.2586. 3 (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 ( 1 3 9 < z < 1 3 9 (c) Why was the probability in (b) higher than that in (a)? ) = P ( 1 < z < 1) = 0.6826 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 ) 8.39 8.5 2.5 = P (z <.93) = 0.8238. 36 ) 7.639 8.5 2.5 = P (z > 2.07) = 1 0.0192 = 0.9808. 36 2

(c) between 275 and 320 minutes? Answer. From solutions (a) and (b), the probability is.8238.0192 =.8046. 5. In a Gallup poll it was reported that 47% of adult Americans surveyed approve of the way the United States has handled the war in Iraq. Moreover, Gallup reported their methods were as follows: These results are based on telephone interviews with a randomly selected national sample of 1,006 adults, aged 18 and older, conducted Oct. 24-26, 2003. For results based on this sample, one can say with 95% confidence that the maximum error attributable to sampling and other random effects is 3 percentage points. In addition to sampling error, question wording and practical difficulties in conducting surveys can introduce error or bias into the findings of public opinion polls. (a) What is the confidence interval that Gallup is suggesting for the proportion of adult Americans favoring the way the U.S. has handled war in Iraq? Answer. Gallup is claiming that the 95% confidence interval for the proportion is (.44,.50). (b) Find a 99% confidence interval for the proportion of adult Americans that approve of the way the United States has handled the war in Iraq. Answer. We are given ˆp =.47, n = 1006. Now, ˆq =.53 and clearly nˆp > 5 and nˆq > 5 so we pq can use the formulas ˆp ± z c σˆp where σˆp = n. Using the table we find z.99 = 2.58. Therefore, (.47)(.53) the 99% confidence interval is.47 ± 2.58 which yields the interval (0.4294, 0.5106). 1006 (c) What sample size would be needed to estimate the proportion of adult Americans that approve of the way the United States has handled the war in Iraq to within ±.01 with 95% confidence? ( zc ) 2 Answer. If we use p =.47 as an estimate for p in the formula n = pq we get n = (.47)(.53) ( ) 1.96 2.01 = 9569.43 and so we would use n = 9570. If we don t assume an initial estimate for p, we get n = 1 ( zc ) 2 = 9605. 4 6. (a) What sample size would be needed to estimate the mean from a population with standard deviation of 100 with a maximum error of 5 with 95% confidence? Answer. We use the formula n = use n = 1537. ( zc σ ) ( ) 2, 2 1.96 100 and so n = = 1536.64 and so we 5 (b) What sample size from the same population would be needed to estimate the mean with a 3

maximum error of 10 with 95% confidence? ( zc σ ) ( ) 2, 2 1.96 100 Answer. We use the formula n = and so n = = 384.16 (which is 10 one-fourth of the answer in (a)). Going to up to the next whole number, we get n = 385. (c) In general, what effect would increasing a sample size by a factor of 16 have on the maximum error? Answer. Multiplying the sample size by 16 reduces the error to one-fourth of what it was σ with the original sample size. This is verified with the following calculation. If = z c n the error with the original sample size, then the error with a sample 16 times the size of the original is σ z c = 1 16n 4 z σ c = 1 n 4. (d) In general, how much must a sample size be increased to cut the maximum error by one-half? Answer. ( The sample size must be quadrupled. This is verified with the following calculation. zc σ ) 2 If n = is the original sample size, then sample size needed to halve is ( ) 2 z c σ ( zc σ ) 2 = 4 = 4n. 2 7. In a 1999 survey of 80 Computer Science graduates and 110 lectrical ngineering graduates, it was found that the Computer Science graduates had a mean starting salary of $48, 100 with a standard deviation of $7, 200, while the lectrical ngineering graduates had a mean starting salary of $52, 900 with a standard deviation of $5, 300. (a) Find a point estimate for the difference in average starting salaries for Computer Science and lectrical ngineering graduates. Answer. The reasonable point estimate is x 1 x 2 = 48, 100 52, 900 = 4800. (b) Let µ 1 be the population mean starting salary for the Computer Science graduates and let µ 2 be the population mean salary for the lectrical ngineering graduates. Find a 96% confidence interval for µ 1 µ 2. Answer: Since each sample size is at least 30, we use the formula: σ1 2 x 1 x 2 ± z c σ x1 x 2 where σ x1 x 2 = + σ2 2 n 1 n 2 4

For this, c =.96 and so z.96 = 2.05 (which is found from the normal table as the z-value for which P (z < z c ) =.98 since.98 =.5 +.96/2). Now compute 7, 200 2 5, 3002 σ x1 x 2 = + = 950.45 80 110 and so we compute 4800 ± 2.05 950.45 which yields the interval ( $6, 748.43, $2, 851.57). This means that we are 96% confident that the true difference in salaries is between $6, 748.43 and $2, 851.57, i.e. we are very confident that the Computer Science graduates starting mean salary is from $2,851.57 to $6,748.43 less than lectrical ngineering graduates mean starting salaries. (c) Based on the interval in (b), would you be comfortable saying that the mean starting salary for Computer Science graduates is less than that for lectrical ngineering graduates? xplain. Answer: Yes, we are very confident that it is at least $2,851.57 less than the mean for lectrical ngineering graduates. 8. (a) A Gallup News Release (see http://www.gallup.com/poll/releases/pr031030.asp) reported that in July 1996, 69% of those surveyed supported assisted suicide for terminally ill, while in May 2003, 72% of those surveyed supported assisted suicide for the terminally ill. Assume the 1996 poll surveyed 1103 adult Americans, while the May 2003 poll surveyed 1009 adult Americans. Find a 99% confidence interval for p 1 p 2 where p 1 is the proportion of adult Americans supporting assisted suicide in July 1996, and p 2 is the proportion of adult Americans supporting assisted suicide in May 2003. Answer: First, because n 1ˆp 1 = (1103)(.69) > 5, n 1ˆq 1 = (1103)(.31) > 5, n 2ˆp 2 = (1009)(.72) > 5 and n 2ˆq 2 = (1009)(.28) > 5, we can use the formula p1 q 1 ˆp 2 ˆp 2 ± z c σˆp1 ˆp 2 where σˆp1 ˆp 2 = + p 2q 2. n 1 n 2 With c =.99 we find z.99 = 2.58 from the table, and we compute ˆp1ˆq 1 σˆp1 ˆp 2 = + ˆp 2ˆq 2 (.69)(.31) = + (.72)(.28) n 1 n 2 1103 1009 =.0198426 The 99% confidence interval has endpoints (.69.72) ± 2.58.0198436, thus the interval is (.0812,.0212). We are 99% confident in July 1996 the percentage of adult Americans that supported assisted suicide was 8.12% less to 2.12% more than the percentage in May 2003. (b) Based on your interval in (a) would you be comfortable in saying that the proportion of adult Americans supporting assisted suicide was higher in May 2003 than in July 1996? xplain. Answer: No, I could not say that with 99% confidence because the interval found in (a) included the possibility that the support was up to 2.12% higher in July 1996 than in May 2003. 5

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 = 115 30, and so we use the formula for large sample means, i.e. x ± z c σ x. In this case, σ x = 2.93 115 =.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. 222 255 254 230 249 222 237 287 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 = 481548 SS x = 481548 19562 8 = 3306. Thus x = 1956 8 = 244.5 and s = 3306 7 = 21.73. 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 244.5 ± 3.499 21.73 8 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

(b) Under what conditions can you find a confidence interval for the mean using a large sample? Answer. If the sample size is 30 or larger. In some asymmetric distributions a larger sample size may be necessary. However, for many practical problems, n 30 is a good rule of thumb. (c) What are the conditions necessary for finding a confidence interval for a population proportion? Answer. We need np > 5 and nq > 5, for this we check that nˆp > 5 and nˆq > 5, i.e. there are more than 5 successes and more than 5 failures in the sample. (Notice if p is very close to 0, or very close to 1 the binomial distribution is highly nonsymmetric about its mean, and so values of n much larger than 30 will be needed to guarantee np > 5 and nq > 5, so do not use the rule of thumb n 30 when dealing with binomial distributions.) 12. A recent Gallup poll of 1006 adult Americans found that 37% of those asked oppose cloning of human organs. (a) Find a 98% confidence interval for the proportion of adult Americans that oppose cloning of human organs. Answer. First z.98 = 2.33 (use table), ˆp =.37 and n = 1006. Now nˆp > 5 and nˆq > 5 and so we can use the confidence interval formula for proportions since the sampling distribution will be approximately normal. Therefore, the 98% confidence interval has endpoints (.37)(.63).37 ± 2.33 1006 That gives a confidence interval of (.3345,.4055). (b) Based on the answer to (a), would it be appropriate for the Gallup organization to state that less than 38% of adult Americans oppose cloning of human organs? xplain. Answer. Because the upper end of the confidence interval is 40.55% which is well above 38%, it would not be appropriate to state that less that 38% of all adult Americans oppose cloning of human organs using this confidence interval. 13. On June 7, 1999 a poll on the USA Today website showed that out of 2000 respondents, 71% felt that Andre Agassi deserved to be ranked among the greatest tennis players ever. (a) Assuming that the 2000 respondents form a random sample of the population of tennis fans, construct a 95% confidence interval for the proportion of all tennis fans who feel that Andre Agassi should be ranked among the greatest tennis players ever. Answer. The confidence interval is (.6901,.7299). To find this confidence interval we found z.95 = 1.96, ˆp =.71, ˆq =.29 and n = 2000. Clearly nˆp > 5 and nˆq > 5 and so we computed: (0.71)(0.29).71 ± 1.96 =.71 ±.0199 to find the endpoints of the interval. 2000 7

(b) Based on (a), would you be comfortable in saying that the poll is accurate to within plus or minus 2 percent 19 times out of 20? xplain. Answer. Yes, the 95% confidence interval is.71 ±.0199, hence intervals based on this size of random sample with the given proportion should have an accuracy of ±2% on average 19 times out of 20. (c) In actuality, the survey was based on voluntary responses from readers of the USA Today sports website. Do you think the 2000 respondents actually formed random sample? xplain. Answer. No the readership of the website is limited to those who have access to the site and choose to visit it; moreover, the survey was not based on a random selection of even those users of the website, but on those who chose to respond to the poll. 14. A recent survey of 104 randomly selected gas stations in California found that the mean price for unleaded gasoline in the sample was $1.52 per gallon with a sample standard deviation of $.10. (a) Conduct an hypothesis test at a 1% level of significance to determine whether the mean price for unleaded gasoline in California is more than $1.50 per gallon. (i) State the null and alternative hypotheses. Answer. The null hypothesis is H 0 : µ = $1.50, and the alternative hypothesis is H 1 : µ > $1.50. (ii) State or draw the critical region, report the conclusion of your test. Answer. The critical region is z 2.33. Now we compute z = 1.52 1.50 0.1 = 2.039 104 Because z does not fall in the critical region, we conclude that there is not sufficient evidence to reject the null hypothesis at a level of significance of.01. (b) Report the Pvalue of the test in (a). Answer. P (z > 2.04) =.5.4793 =.0207 (c) xplain what the Pvalue in (b) means. Answer. There is about a 98% probability that the true mean price for unleaded gasoline in California is more that $1.50 per gallon. (d) If you were to repeat the test in (a) for levels of significance α =.02, α =.03 and α =.05, what would the conclusion of the test be? 8

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 =.02. 15. (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 ) ( ) 2 2 1.96 80 Answer. We use the formula n = and we find that n = = 245.86. 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 ( ) 2 1.96 which gives us a sample size of 2401..02 (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 = 3.106 (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 2.58. (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 1.645. (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

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 = 486 500 53 32 = 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 0.01. (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 = 5.95 6.15 40 = 2.11. 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

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:.15 5.95 ± 2.05 40 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 ( ) 2 2.05.15 n = = 945.5625,.01 thus we should use a sample size of n = 946. 18. (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 2.33. Since n = 1200 and p =.56 and q =.44, we clearly have np > 5 and nq > 5. Thus we compute z =.53.56 (.56)(.44) 1200 11 2.09

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) =.0183. 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 349.2 with a standard deviation of 56.6. The average score for the 30 subjects in the experimental group was 368.4 with a standard deviation of 39.5. 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 2 39.5 2 σ x x2 = 30 + 56.62 30 = 12.6013 (368.4 349.2) 0 Therefore, z = = 1.52. The Pvalue is P (z > 1.52) = 1.9357 = 0.0643 12.6013 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 48.39583.28 z = 1.79 (0.28)(0.72) 48 12

Therefore, the Pvalue is P (z > 1.79) = 1.9633 =.0367. 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 0.28. (c) Would you reject the null hypothesis at a level of significance of α =.04? Answer. Yes, because the Pvalue is less than 0.04. (d) Would you reject the null hypothesis at a level of significance of α =.01? Answer. No, because the Pvalue is more than 0.01. (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 = 141 + 125 288 + 216 =.5278. 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 (.4896.5787) 0 0.04494 = 0.04494. = 1.98. The Pvalue is P (z < 1.98) = 0.0239. Because the Pvalue is less than.05, we reject H 0 at the 5% level of significance. 13