# Introduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 8.1 Homework Answers

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1 Introduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 8.1 Homework Answers 8.1 In each of the following circumstances state whether you would use the large sample confidence interval, the plus four method, or neither for a 95% confidence interval. What this question is asking is do we meet the criteria set forth on page 537 need at least 15 successes and 15 failures for Large Sample Confidence Interval, or on page 539, Plus four method, which requires a sample size of 10. (a) n = 20, X = 15. We could certainly use the plus-four method since n > 10. However, we could not use the large-sample confidence interval since we have 15 successes but only 5 failures (20 15). (b) n = 100, X = 15 We could certainly use the plus-four method since n > 10. We could also use the large sample confidence interval since we have 15 successes and 85 failures. (c) n = 10, X = 2 We could use the plus-four method since n = 10, but not the large sample confidence interval since we only have 2 successes and 8 failures. (d) n = 5, X = 2. We could not use the plus four method since our sample size is only 5. And of course we can not use the large sample confidence interval method since we do not have the required number of successes and failures. (e) n = 50, X = 20. We could use the plus four method since we have at least a sample size of 10, and we could also use the large-sample confidence interval method since we have 20 successes and 30 failures Explain what is wrong with each of the following: (a) An approximate 95% confidence interval for an unknown proportion p is p plus or minus its standard error. a. The error here is that it is ˆp plus or minus the margin of error, not plus or minus the standard error. (b) You can use a significance test to evaluate the hypothesis H o: ˆp = 0.3 versus the two-sided alternative. b. The mistake is the statement H o : ˆp = 0.3, it should be H o : p = 0.3. We can not make the assumption to be the sample proportion equals. This is a ridiculous statement since the sample proportion is continually changing. The population proportion p is what can assume to be a fixed value. (c) The large-sample significance test for a population proportion is based on a t statistic. c. False it is based on the z-statistic.

2 8.5 Gambling is an issue of great concern to those involved in intercollegiate athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed student-athletes concerning their gambling-related behaviors. 8 There were 5594 Division I male athletes in the survey. Of these, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. (a) Find the sample proportion and the large-sample margin of error for 95% confidence. Explain in simple terms the meaning of the 95% pˆ = This is the sample proportion of male student athletes participated in some gambling 5594 behavior. Margin of error = Z * This is referred to some times as the large sample blah, because if your pˆ(1- pˆ) n sample size is big enough it should handle the criteria mentioned in the text, at least15 successes np 15, and at least 15 failures n(1 p) ( ) Margin of error = = (b) Because of the way that the study was designed to protect the anonymity of the student-athletes who responded, it was not possible to calculate the number of students who were asked to respond but did not. Does this fact affect the way that you interpret the results? Write a short paragraph explaining your answer. 8.5 b. What this question is alluding to is if the sample is indeed representative. The problem is that we don t know how many students were polled. Suppose that 10,000 students were asked to respond. The question says that 5594 responded, a 44% non- response rate; a rather large non-response. We could then ask who are these people that did not respond. Is this set of students special in some way? We assume that the first set of 10,000 students is selected randomly, so there is great chance it is representative. But if 4406 opt out, we wonder if the remaining group of students that participate is representative. 8.7 The National Survey of Student Engagement found that 87% of students report that their peers at least "sometimes" copy information from the Internet in their reports without reporting the source. 9 Assume that the sample size is 430,000. (a) Find the margin of error for 99% confidence. 0.87(1-0.87) Margin of error = =

3 (b) Here are some items from the report that summarizes the survey. More than 430,000 students from 730 four-year colleges and universities participated. The average response rate was 43% and ranged from 15% to 89%. Institutions pay a participation fee of between \$3000 and \$7500 based on the size of their undergraduate enrollment. Discuss these as sources of error in this study. How do you think these errors would compare with the error that you calculated in part (a)? 8.7b. The error in part a only takes into account natural variability to due sampling, nothing else. The average response rate is 43%, which means people are self selecting to participate. Who are these people who want to respond and what is their motivation? We know the range is as low as 15% (highly suspect result would not be normally be admissible) to 89% (a more convincing result). The final result would be more believable if we omitted all schools that had a non-response rate lower than 70%. Or make a comparison of schools with high response to those with low response to see if there is a significant difference. Another issue is the participation fee. The problem here is that the result claims that 87% of students implying this sample is representative of all schools. But this may not be the case. Is the participation fee causing exclusion of enough schools to make the sample not to be representative Refer to Exercise Would a 99% confidence interval be wider or narrower than the one that you found in that exercise? Verify your results by computing the interval. It will be wider since we are trading certainty for accuracy. How can we go from 95% certainty to 99% certainty with exactly the same data? Make the interval wider; this gives a greater chance that p is found in the interval Yesterday, your top salesperson called on 10 customers and obtained orders for your new product from all 10. Suppose that it is reasonable to view these 10 customers as a random sample of all of her customers (a) Give the plus four estimate of the proportion of her customers who would buy the new product. Notice that we don't estimate that all customers will p% = b. (b) Give the margin of error for 95% confidence. (You may see that the upper endpoint of the confidence interval is greater than 1. In that case, take the upper endpoint to be 1.) ( ) SE p % = = ± 1.96 ( ) (0.6738, 1)

4 8.19c (c) Do the results apply to all of your sales force? Explain why or why not. The above results apply to your top salesperson only. Unless you believe that the results are typical of your entire sales force, which in that case the person would not be your top sales person The English mathematician John Kerrich tossed a coin 10,000 times and obtained 5067 heads. 8.23a. (a) Is this significant evidence at the 5% level that the probability that Kerrich's coin comes up heads is not 0.5? Use a sketch of the standard normal distribution to illustrate the P-value. The question asks if the true long run proportion that a coin will lands head is indeed 50%, which everyone agrees it should be. The number of tosses is 10000, n. And the number of successes is Let us perform a significance test ˆp = = ˆp Ho: p = 0.5 and Ha: p 0.5. How do I know that this is a two-sided test? Because, of the original question posed. that Kerrich s coin comes up heads is not 0.5? No direction indicated. Our sample proportion is ˆp = , so know we want to calculate P( ˆp >0.5067). P( ˆp >0.5067) = PZ > (1.05) = P(Z > 1.34) = Now, the p-value is 2(0.0901) = Thus, at the 5% significance level, we have found no evidence that the true proportion is some other number other than 0.5. In other words seeing a proportion of a unit difference from 0.5 is not at all uncommon for 10,000 tosses, when the population proportion is 0.5.

5 8.23b. Use a 95% confidence interval to find the range of probabilities of heads that would not be rejected at the 5% level ± 1.96(0.005) (.4969, ) Any value within the interval above would not be rejected at the 5% significance level, when performing a two-sided significance test Refer to the previous exercise. Suppose that after reviewing the results of the previous survey, you proceeded with preliminary development of the product. Now you are at the stage where you need to decide whether or not to make a major investment to produce and market it. You will use another random sample of your customers but now you want the margin of error to be smaller. What sample size would you use if you wanted the 95% margin of error to be or less? Without knowing what p might possibly be, it would be safer to assume that p = 0.5 and thus calculate for the worst case scenario (the value of p that gives the largest required sample size) n = (0.5)(0.5) = 171

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