# Hypothesis Tests for 1 sample Proportions

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4 Hypothesis Tests for 1 sample Proportions Answers 1. a) H 0 : p = 0.30; H A : p < 0.30 b) H 0 : p = 0.50; H A : p 0.50 c) H 0 : p = 0.20; H A : p > a) H 0 : p = 0.40; H A : p 0.40 b) H 0 : p = 0.20; H A : p < 0.20 c) H 0 : p = 0.60; H A : p < Statement d is correct. 4. Statement d is correct. 5. No, we can say only that there is a 27% chance of seeing the observed effectiveness just from natural sampling variation. There is no evidence that the new formula is more effective, but we can't conclude that they are equally effective. 6. Yes. If there is no difference, there is only a 1.7% chance of seeing such a high sample proportion just from natural sampling variation. 7. 1) Use p in hypotheses, not ˆp. 2) The question was about failing to meet the goal, so H A should be p< ) Did not check 0.04(200) = 8 or N > 10(200) ) 188/200 = 0.94: SD( p ) = ) z is incorrect; should be z ) P = P(z < -1.43) = ) There is only weak evidence that the new system does not work. 8. 1) Use p in hypotheses, not ˆp. 2) The question asks, "not accurate," so H A should be two sided: p ) The correct conditions are SRS, (0.9)(750) > 10, and (0.10)(750) > ) p = 657/750 = 0.876; SD( p ) = ) 2 is incorrect; should be z ) P = 2P(z < -2.18) = ) There is only a 2.9% chance of observing a ˆp as far from 0.90 by sampling variation, so we believe the proportion of adults who drink milk here is different from the claimed 90%. 9. a) H 0 : p = 0.30; H A : p > 0.30 b) Possibly an SRS; we don't know if the sample is less than 10% of his customers but could be viewed as less than 10% of all possible customers; (0.3)(80) > 10 and (0.7)(80) > 10. Wells are independent only if customers don't have farms on the same underground springs. c) z = 0.73; P-value = d) If his dowsing is no different from standard methods, there is more than a 23% chance of seeing results as good as those of the dowsers, or better, by natural sampling variation. e) These data provide no evidence that the dowser's chance of finding water is any better than normal drilling. 10. a) H 0 : p = 0.05; H A : p > 0.05 b) SRS (not clear from information provided), <10% of all children, (0.05)(384) > 10, and (0.95)(384) > 10. c) z = 6.28, P = 2 X d) If the autism rate has not increased, the chance of observing at least 46 autistic children in a sample of 384 is 2 x (almost 0). e) Reject H 0. These data show the rate of autism is now more than 5%. f) We do not know that chemicals cause autism, only that the rate is higher now than in the past. 11. a) Based on these data, we are 95% confident the proportion of adults in 1995 who had never smoked cigarettes is between 48.7% and 55.3%. b) H 0 : p = 0.44; H A : p Since 44% is not in the confidence interval, we will reject Hg and conclude the proportion of adults in 1995 who had never smoked was more than in the 1960s. 12. a) Based on these data, we are 95% confident the proportion of dissatisfied customers is between 1.1% and 4.6%. b) H 0 : p = 0.05; H A : p < Since 5% is not in the 95% confidence interval, we will reject H 0 at = and conclude the proportion of dissatisfied customers is less than 5%, so the company has met its goal. 13. H 0 : p = 0.20; H A : p > SRS (not clear from information provided); 22 is more than 10% of the population of 150; (0.20)(22) < 10. Do not proceed with a test. 14. H 0 : p = 0.02; H A : p > SRS; less than 10% of all washers and dryers made by the company; (0.02)(60) < 10. Do not proceed with a test. 15. H 0 : p = 0.03; H A : p < p = One mother having twins will not impact another, so observations are independent; not an SRS; sample is less than 10% of all births. However, the mothers at this hospital may not be representative of all teenagers; (0.03)(469) = > 10; (0.97)(469) > 10. z = -1.91; P-value = With a P-value this low, reject H 0. These data show evidence that the rate of twins born to teenage girls at this hospital is less than the national rate of 3%. It is not clear whether this can be generalized to all teenagers. 16. H 0 :p = 0.50; H A : p > Results of one game should not impact another, so games are independent; data are all results for one season,

5 which should be representative of all seasons; sample is less than 10% of all games; (0.50)(240) > 10; (0.50)(240) > 10. z = 2.32; P-value = With a P-value this low, reject H 0. These data show strong evidence that the home team does have an advantage; they win more than 50% of games at home. 17. H 0 : p = 0.25; H A : p > SRS; sample is less than 10% of all potential subscribers; (0.25)(500) > 10; (0.75)(500) > 10. z = 1.24; P-value = The P- value is high, so do not reject H 0. These data do not show that more than 25% of current readers would subscribe; the company should not go ahead with the WebZine on the basis of these data. 18. H 0 : p = 0.92; H A : p < Seeds in a single packet may not be independent of each other. This is a cluster sample of all seeds in the packet. We may view this cluster as representative of all year-old seeds, in which case the sample is less than 10% of all seeds; (0.92)(200) > 10; (0.08)(200) > 10. z = ; P-value = Because the P-value is very low, we reject H 0. There is strong evidence that these seeds have lost viability; their germination rate is less than 92%. 19. H 0 : p = 0.40; H A : p < Data are for all executives in this company and may not be able to be generalized to all companies; (0.40)(43) > 10; (0.60)(43) > 10. z = -1.31; P-value = Because the P-value is high, we fail to reject H 0. These data do not show that the proportion of women executives is less than the 40% women in the company in general. 20. H 0 : p = 0.19; H A : p < ˆp = z = -1.41; P- value = Because the P-value is high, we fail to reject H 0. These data do not show convincing evidence that Hispanics are represented in the jury pool at less than their 19% proportion in the population in general, but indicates that it is below what was expected. 21. H 0 : p = 0.109; H A : p > ˆp = 0.118; z = 1.198; P-value = Because the P-value is high, we fail to reject H 0. These data do not show the dropout rate has increased from 10.9%. 22. H 0 : p = 0.15; H A : p > ˆp = 0.25; z = 2.80; P- value = The 95% confidence interval is (0.165, 0.335). We must assume the trees sampled are a SRS of the trees in the area and are less than 10% of all trees in the forest. The results are generalizable only to the Hopkins forest (or nearby if the forest is viewed as representative). Because the P-value is so low, we reject H 0. There is strong evidence that the proportion of trees damaged by acid rain in the Hopkins forest is higher than the 15% average for the Northeast. 23. H 0 : p = 0.90; H A : p < ˆp = 0.844; z = -2.05; P- value = Because the P-value is so low, we reject H 0. There is strong evidence that the actual rate at which passengers with lost luggage are reunited with it within 24 hours is less than the 90% claimed by the airline. 24. H 0 : p = 0.40; H A : p > ˆp = 0.431; z = 1.29; P- value = Because the P-value is high, we fail to reject H 0. These data do not show that at least 40% of the public recognizes the brand; I would not recommend they continue to advertise during Super Bowls on the basis of these data.

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