When the 2000 General Social Survey asked subjects whether t

When the 2000 General Social Survey asked subjects whether they would be willing to accept cuts in their standard of living to protect the environment, 344 of 1170 subjects said ^\'\'yes.^\'\' a. Estimate the population proportion who would say ^\'\'yes.?^\'\' b. Conduct a significance test to determine whether a majority or minority of the population would say ^\'\'yes.?^\'\' Report and interpret the P-value. c. Construct and interpret a 99percent confidence interval for the population proportion who would say ^\'\'yes.^\'\'

Solution

(a) p=344/1170 =0.2940171

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(b) The test hypothesis
Ho: p=0.5 (i.e null hypothesis)

Ha: p not equal to 0.5 (i.e. alternative hypothesis)

The test statistic is

Z=(phat-p)/sqrt(p*(1-p)/n)

=(0.2940171-0.5)/sqrt(0.5*0.5/1170)

=-14.09

It is a two-tailed test.

So the p-value= 2*P(Z<-14.09) =0 (from standard normal table)

Assume that the signficant level a=0.01

Since the p-value is less than 0.01, we reject Ho.

So we can not conclude that a majority or minority of the population would say yes

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(c) Given a=1-0.99=0.01, Z(0.005) = 2.58 (from standard normal table)

So the lower bound is

p -Z*sqrt(p*(1-p)/n) =0.2940171 - 2.58*sqrt(0.2940171*(1-0.2940171)/1170) =0.2596526

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.2940171 + 2.58*sqrt(0.2940171*(1-0.2940171)/1170)=0.3283816

We have 99% confident that the population proportion will be within this interval.