TutWeek10_F23

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Jan 9, 2024

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EC285 A,B Tutorial Questions Week 10 Confidence Intervals for Proportions Edda Claus EC285 A,B - Fall 2023 24 November 2023 1. Your local newspaper polls a random sample of 330 voters, finding 144 who say they will vote “yes” on the upcoming school budget. Create a confidence interval for actual sentiment of all voters. Just use 2 SDs and concentrate on the conditions and the interpretation. Answer Plan – We want a confidence interval for the proportion of all voters who will vote “yes” on the upcoming school budget, based on a random sample of 330 voters. In our sample, 144 respondents stated that they would vote “yes”. We want to be reasonably confident in our results, so we will construct a confidence interval of ± 2 standard errors around our sample proportion, or approximately 95% confidence. Independence Assumption : It is reasonable to think that the responses were mu- tually independent, provided good surveying techniques were used. Randomization Condition : The voters were sampled randomly. 10% Condition : Provided there are more than 3300 eligible voters, 330 are less than 10% of the population of voters. Success/Failure Condition : n ˆ p = 144 and n ˆ q = 186, which are both at least 10, so the sample is large enough. The conditions are satisfied, so we can use the Normal model to find a one proportion z - interval. 1

Do – Given n = 330 and ˆ p = 144 330 ≈ 0 . 436, so SE (ˆ p ) = q ˆ p ˆ q n = q 0 . 436 · 0 . 564 330 ≈ 0 . 027 The margin of error is ME = z ⋆ SE (ˆ p ) = 2 · 0 . 027 ≈ 0 . 054 The confidence interval is 0 . 436 ± 0 . 054 or (0 . 382 , 0 . 490) Report – We are (approximately) 95% confident that between 38.3% and 49.0% of voters will vote “yes” on the upcoming budget. 2. An experiment finds that 27% of 53 subjects report improvement after using a new medicine. Create a 95% confidence interval for the actual cure rate. Use z ⋆ = 1 . 96. Why is this interval so wide? Make it narrower – 90% confidence. What are the advantages and disadvantages? What sample size would we need in a follow-up study if we want a margin of error of 5% with 98% confidence? Answer Plan – We want a confidence interval for the proportion of all people who will improve after using a new medication, based on an experiment involving 53 subjects. In our sample, 27% of the subjects improved. Independence Assumption : One patient responding to the medication shouldn’t have an influence on other patients responding to the medication. Randomization Condition : The patients were part of an experiment, which hopefully included ran- dom assignment of volunteers to treatment groups. 10% Condition : The sample is certainly less than 10% of the population. Success/Failure Condition: n ˆ p = 53 · 0 . 27 = 14 . 31 and n ˆ q = 53 · 0 . 73 = 38 . 69, which are both at least 10, so the sample is large enough. The conditions are satisfied, so we can use the Normal model to find a one proportion z interval. Do – Given n = 53 and ˆ p = 27, so SE (ˆ p ) = q ˆ p ˆ q n = q 0 . 27 · 0 . 73 53 ≈ 0 . 061 The margin of error is ME = z ⋆ SE (ˆ p ) = 1 . 96 · 0 . 061 ≈ 0 . 12 The confidence interval is 0 . 27 ± 0 . 12 or (0 . 15 , 0 . 39) Report – We are 95% confident that between 15% and 39% of people will improve after using the new medication. This interval is quite wide for a couple of reasons. The sample size of 53 people doesn’t provide a great deal of accuracy in our estimate. The standard error is still quite large. Also, the more confident we are that we have succeeded in capturing the true proportion, the less precise our interval becomes. 2

90% Confidence Interval The margin of error is ME = z ⋆ SE (ˆ p ) = 1 . 645 · 0 . 061 ≈ 0 . 10 The confidence interval is 0 . 27 ± 0 . 10 or (0 . 17 , 0 . 37) We are 90% confident that between 17% and 37% of people will improve after using the new medication. This interval has the advantage of being more precise, but we are less confident in our ability to capture the true proportion within our interval. Finding the sample size required for a margin of error of 5% with 98% confidence Consider the formula for margin of error. We believe the improvement rate to be 0 . 27 from our preliminary study. The value of z ⋆ for 98% confidence is 2 . 326. Using ME = q ˆ p ˆ q n , we get 0 . 05 = 2 . 326 0 . 27 · 0 . 73 n 0 . 05 2 = 2 . 326 2 ( 0 . 27 · 0 . 73 n ) n = 2 . 326 2 · 0 . 27 · 0 . 73 0 . 05 2 = 426 . 5462 To be safe, always round the estimate up. We need to run an experiment with at least 427 people receiving the new medication in order to have a margin of error of 5% with 98% confidence. 3. What sample size does it take to estimate the outcome of an election with a margin of error of 3%? Answer We will do what polling organizations usually do, using 95% confidence and choosing the most cautious proportion, 50%. So, ˆ p = 0 . 5, ˆ q = 0 . 5, ME = 0 . 03 and z ⋆ = 1 . 96. Using ME = z ⋆ SE (ˆ p ), we get 0 . 03 = 1 . 96 q 0 . 5 · 0 . 5 n 0 . 03 2 = 1 . 96 2 ( 0 . 5 · 0 . 5 n ) n = 1 . 96 2 · 0 . 5 · 0 . 5 0 . 03 2 = 1067 . 1 To be safe, always round the estimate up. A sample size of at least 1068 voters is required. 4. Suppose a sample is selected from a population in order to estimate the population proportion p. Based on the sample data, the proportion is calculated to be 0 . 22. a. What is the standard error if the sample size is 50? 3

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b. What is the standard error if the sample size is 100? c. What is the standard error if the sample size is 1000? d. How is the standard error affected by the sample size? Answer a. 0.0586; b. 0.0414; c. 0.0131; d. As the sample size increases, the standard error decreases. 5. Suppose that you are constructing a confidence interval for the population proportion p . Find the appropriate z ⋆ value for each of the following levels of confidence. a. 90%; b. 95%; c. 99%; d. How is the confidence interval affected by the stated level of confidence? Answer a. 1.645; b. 1.960; c. 2.576; d. As the stated level of confidence increases, the width of the interval increases. 6. Suppose that you are constructing a confidence interval for the population proportion p . You have selected a sample and based on the data calculate the proportion as 0 . 45. a. What is the margin of error if the sample size is 50 and you want to be 90% confident? b. What is the margin of error if the sample size is 1000 and you want to be 90% confident? c. What is the margin of error if the sample size is 50 and you want to be 99% confident? d. What is the margin of error if the sample size is 1000 and you want to be 99% confident? 4

Answer a. 0.1157; b. 0.0259; c. 0.1812; d. 0.0405. 7. A random sample of 800 individuals provided 544 “no” responses. a. What is the sample proportion? b. Is the success/failure condition satisfied? Explain. c. Construct a 95% confidence interval for the true proportion of “no” responses. d. nterpret this confidence interval. Answer a. 0.68; b. Yes, n ˆ p = 544 > 10 and n ˆ q = 256 > 10; c. The 95% confidence interval is 0 . 68 ± 0 . 032, i.e. , between 0 . 648 and 0 . 712; d. We are 95% confident that the true proportion of “no” responses is between 0 . 648 and 0 . 712. 8. Suppose you want to estimate the proportion of college students who own their own car. a. What sample size is needed if you wish to be 95% confident that your estimate is within 0.02 of the true proportion? b. What sample size is needed if you wish to be 99% confident that your estimate is within 0.02 of the true mean? c. What sample size is needed if you wish to be 95% confident that your estimate is within 0.05 of the true mean? d. Suppose that based on past studies, you have an idea that the proportion is 0.25. What sample size is needed if you wish to be 95% confident that your estimate is within 0.02 of the true proportion? Answer 5

a. n = 2401; b. n = 4148; c. n = 385; d. n = 1801. 9. Suppose that it was found that 75 out of 300 students enrolled in the accounting program failed at least one course, and that a similar study found that 80 out of 400 students enrolled in finance courses failed at least one course. Find a 95% confidence interval for the difference in proportions of failures between accounting and finance students. Answer The 95% confidence interval is 0 . 05 ± 0 . 062751, i.e. , between − 0 . 013 and 0 . 113. 6

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TutWeek10_F23

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