TutWeek11_F23

EC285 A,B Tutorial Questions Week 11 Confidence Intervals for Proportions Edda Claus EC285 A,B - Fall 2023 1 December 2023 1. Health Canada monitors the safety of smoke detectors in the marketplace and estimates that 90% of all Canadian homes have at least one smoke detector. Calgary’s fire department has been running a public safety campaign about smoke detectors consisting of posters, billboards, and ads on radio and TV and in the newspaper. The city wonders if this concerted effort has raised the local level above the 90% national rate. Building inspectors visit 400 randomly selected homes and find that 376 have smoke detectors. Is this strong evidence that the local rate is higher than the national rate? Answer Hypotheses : The null hypothesis is that Calgary is no different than anywhere else. The alternative is that the public safety campaign has had a positive effect, so the rate of compliance is higher. The hypotheses deal with the actual percentage of all homes that have smoke detectors in that city. Model : We have a simple random sample of 400 homes, and assume that’s less than 10% of all homes in this city, making them essentially independent. We expect np 0 = 400(0 . 9) = 360 successes and nq 0 = 40 failures, both at least 10, so the sample is large enough to use a Normal model. We will do a one-proportion z -test. Mechanics : n = 400 , x = 376 , ˆ p = 0 . 94 z = 0 . 94 − 0 . 90 q (0 . 90)(0 . 10) 400 = 2 . 67 P − value = P ( z > 2 . 67) = 0 . 004 1

Conclusion : The P-value is very low, making it unlikely that the observed results can be explained by sampling error. We reject the null hypothesis. There is strong evidence that, in this city, more than 90% of homes have smoke detectors. Further Comments : Interpret the P-value carefully. If in fact the rate in this city is 90% – the same as anywhere else – then only 0.4% of samples (about 1 in every 250) would find (as ours did) a proportion of 94% or more. While that could be what happened here, it’s not the most logical explanation. In fact, we are 95% confident that between 91.1% and 96.9% of the city’s homes have smoke detectors. 2. There are supposed to be 20% orange M&M’s. Suppose a bag of 122 has only 21 orange ones. Does this contradict the company’s 20% claim? Note that this is a two-tailed test. Even though the observed result looks as though there were not enough orange ones, before we had the sample we’d have been equally interested to discover that there were too many. Now the P-value is = 0 . 44 (see solution below.) That means that even if the company is correct about the 20% orange claim, 44% of all bags this size will differ from perfection by at least as much as this one did. It does not seem that this result is unusual, so we fail to reject the null hypothesis. We don’t have convincing evidence that the 20% figure is wrong. (On the other hand, we certainly did not prove it was right, so we cannot accept the null hypothesis.) Answer Hypotheses : The null hypothesis is that the proportion of M&M’s that are orange is 20%. The alternative hypothesis is that the proportion of M&M’s that are orange is something other than 20%. In symbols: H 0 : p = 0 . 20, H A := p ̸ = 0 . 20 Model : Independence Assumption: It is reasonable to think that M&M colors are mutually independent. Randomization Condition: The 122 M&M’s we have can be considered representative of all M&M’s. 10% Condition: 122 M&M’s is certainly less than 10% of all M&M’s. Success/Failure Condition: np 0 = 122 · 0 . 20 = 24 . 4 and nq 0 = 122 · 0 . 80 = 97 . 6, which are both at least 10, so the sample is large enough. 2

a. What is the critical value if H A : p ̸ = 0 . 25 and α = 0 . 05? b. What is the critical value if H A : p > 0 . 25 and α = 0 . 05? c. What is the critical value if H A : p < 0 . 25 and α = 0 . 05? d. What is the critical value if H A : p ̸ = 0 . 25 and α = 0 . 01? Answer a. ± 1 . 96; b. 1 . 645; c. − 1 . 645; d. ± 2 . 575. 5. Toronto, Montr´ eal and Vancouver accounted for 30% of Canadian retail sales in a recent year. You interview a random sample of Canadian residents this year to see whether that proportion has changed. Suppose that all assumptions are met and you test the hypotheses H 0 : p = 0 . 30 versus H A . A sample of size 500 results in a sample proportion of 0.25. a. is the standard deviation of the sample proportion? b. Compute the value of the z -statistic. c. the P-value associated with this z -statistic. d. State your conclusion at α = 0 . 05. Answer a. 0 . 0205; b. − 2 . 44; c. 0 . 0147; d. Because P − value = 0 . 0147 < 0 . 05, we reject the null hypothesis H 0 . There is strong evidence that the proportion has changed. 6. In a recent year, Canadian households in the territories and provinces other than Qu´ ebec accounted for 75% of national households. You interview a random sample of households this year to see whether the proportion of non-Qu´ ebec based households has increased. Suppose that all assumptions are met and you test the hypotheses H 0 : p = 0 . 75 vs . H A : p > 0 . 75. A sample of size 100 results in a sample proportion of 0 . 80. 4

a. What is the standard deviation of the sample proportion? b. Compute the value of the z -statistic. c. Find the P-value associated with this z statistic. d. State your conclusion at α = 0 . 05. Answer a. 0 . 0433; b. 1 . 15; c. 0 . 1251; d. The high P − value = 0 . 1251 > 0 . 05 indicates that these results could be reasonably explained by sampling error, so we fail to reject the hull hypotheses H 0 . There is no evidence to suggest that the proportion of non-Qu´ ebec based households has increased. 7. In a recent year 55% of TV viewing time in Canada was associated with foreign pro- gramming. The government has recently introduced additional grants to subsidize the production of Canadian programming. You interview a random sample of Canadi- ans this year to see whether the proportion of Canadians viewing foreign program- ming has dropped. Suppose that all assumptions are met and you test the hypotheses H 0 : p = 0 . 55 vs. H A : p < 0 . 55. A sample of size 1000 results in a sample proportion of 0 . 51. a. What is the standard deviation of the sample proportion? b. Compute the value of the z -statistic. c. Find the P -value associated with this z -statistic. d. State your conclusion at α = 0 . 05. Answer a. 0 . 0157; b. − 2 . 55; c. 0 . 0054; d. Because P − value = . 0054 < 0 . 05, we reject the null hypothesis H 0 . There is strong evidence that the proportion of Canadians viewing foreign programming has dropped. 5