{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw4soln

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online