hw4soln - PADP 8120 Fertig Homework 4 Solutions Spring 2011...

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Unformatted text preview: PADP 8120 Fertig Homework 4 Solutions Spring 2011 UGA 1. According to data on older adults from the United States, the distribution of Y=total number of children has a mean of 2.9 and a standard deviation of 2.0. a. Is Y normally distributed? Explain. No. Since people cannot have a negative number of children, this distribution is truncated at 0. If it were normally distributed, 95% of the sample would be between 2 standard deviations of the mean. But 2 standard deviations below the mean is 2.9-4=-1.1, which is not possible with this variable. So, the distribution must be skewed and not normal. b. Assuming that this mean and standard deviation were for a random sample of 100 people, calculate the standard error of the sampling distribution of Ybar. SE(Ybar)=s/sqrt(n)=2.0/sqrt(100) = 0.2 c. Based on the information in part b, report the 95% confidence interval for the mean. CI = estimate z*SE = 2.9 1.96*0.2 = 2.9 0.39. So, the 95% CI is from 2.51 to 3.29. 2. A city council votes to appropriate funds for a new civic auditorium. The mayor of the city threatens to veto this decision unless it can be shown that a majority of citizens would use it at least twice a year. The council commissions a poll of city residents. For a random sample of 400 residents, 230 say they would use the facility at least twice a year. a. Find a 95% confidence interval for the proportion of all residents of the town who would say they would use the proposed auditorium at least twice a year. Interpret the interval. P=230/400=0.575 P(1- P) 0.575(1- 0.575) SE = = = 0.0247 n 400 CI = P 1.96*SE = 0.575 1.96*0.0247 = 0.575 0.048. So, the 95% CI is from 0.527 to 0.623. Thus, we are 95% confident that between 52.7% and 62.3% of the residents will use the auditorium at least twice a year. 1 We want to know whether more than half of the residents would use the auditorium at least twice a year. So, the null hypothesis is that P=0.5 and the alternative hypothesis is the P0.5. y - 0.575 - 0.5 t= = = 3.03 SE 0.0247 Using an online calculator, I find that the corresponding p-value for a two-tailed test is 0.0026. Thus, I can say with 99.74% confidence that more than half of the residents will use the auditorium at least twice a year. c. Compare and contrast the information you obtained in parts a. and b. The results from the CI and the t-test are consistent. They are basically two ways of saying the same thing. d. What would you advise the mayor to do? I would advice the mayor to not veto the auditorium since it has met her criteria. b. Perform a significance test that would help the mayor make her decision and interpret your z- and p-values. 2 ...
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This note was uploaded on 01/18/2012 for the course PADP 8120 taught by Professor Fertig during the Summer '11 term at UGA.

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