When the 2000 General Social Survey asked subjects whether they would be willing

Question

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.^''

Explanation / Answer

(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.

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