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Mathematical Design: HWName:_________________________________Practice: One-Proportion Z-Test – Binge Drinking ExampleDate___________________________After hearing the national result that 44% of students engage in binge drinking (5 drinks at a sitting for men, 4 for women), a professor initiated an anti-binge drinking campaign. After a month of running this campaign, he surveyed a random sample of 250 students and only 100 of those surveyed said that they were involved in binge drinking. Is this enough evidence that the proportion of binge drinkers at his university decreased from the national proportion? That is, can the professor claim that his campaign worked?State null and alternative hypothesis.Describe the sampling distribution. You must check the conditions here.How extreme is the sample outcome? Find the P-value.Explain your conclusions in nontechnical language.Mathematical Design: HWName:KEYPractice: One-Proportion Z-Test – Binge Drinking ExampleDate___________________________After hearing the national result that 44% of students engage in binge drinking (5 drinks at a sitting for men, 4 for women), a professor initiated an anti-binge drinking campaign. After a month of running this campaign, he surveyed a random sample of 250 students and only 100 of those surveyed said that they were involved in binge drinking. Is this enough evidence that the proportion of binge drinkers at his university decreased from the national proportion? That is, can the professor claim that his campaign worked?State null and alternative hypothesis.Ho: p = 0.44 (the proportion of binge drinkers at the university is equal to 44%)Ha: p < 0.44 (the proportion of binge drinkers at the university is less than 44%)Describe the sampling distribution.The following conditions were satisfied: np = 250(.44) = 110 > 10 and nq = 250(.56) = 140 >10,there are over 2500 college students at the university (sample is less than 10% of population), the sample was taken randomly, and,the drinking habits of college students are independent.Therefore, we can use the normal model for the sampling distribution of the p-hats (sample proportions). Its mean is p = 0.44 and its standard deviation is sqrt(.44*.56/250) = 0.0314. How extreme is the sample outcome? Find the P-value.Using the calculator: STAT, TESTS, 5:1-PropZTest: po: .44; x: 100; n: 250; prop < po; Calculate: z = -1.274; p = .1013P-value = P(p-hat < .4) = P(z < -1.274) = .101 = 10.1%This P-value is not so extreme (quite big) being more than the default alpha value of .05 (5%).Explain your conclusions in nontechnical language.Meaning of P-value: If the true proportion of college binge drinkers at the university is 44% (.44), then we expect to see a sample proportion of 40% or less 10.1% of the time, i.e. in 101 out of 1000 samples. This is not sufficient evidence that the true proportion of binge drinkers is less than 44%. We fail to reject the null hypothesis that the true proportion of binge drinkers at the university is equal to 44%. We do not have evidence for the alternative hypothesis that the true proportion is less than 44%, that is, we do not have evidence that the campaign has worked after one month. ................
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