On May $23,2013$, Gallup reported that of the $1005$people surveyed, $76\%$of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at $95\%$with a $\pm 3\%$margin of error.

a. Determine the estimated proportion from the sample.

b. Determine the sample size.

c. Identify CL and .

d. Calculate the error bound based on the information provided.

e. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the values.

f. Create a confidence interval for the results of this study.

g. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience?

According to Gallup, $76$percent of US workers expect they will work into retirement age, based on a study of $1005$persons.

a) The estimated proportion of the sample is,

$$\begin{gathered} p^{\text{'}}=76\% \\ =0.76 \end{gathered}$$

b) The sample size is,

$n=1005$people

c) The level of confidence is,

$CL=95\%$

and therefore,

$$\begin{gathered} \alpha =1-0.95 \\ =0.05 \end{gathered}$$

We've come to the conclusion that,

$$\begin{gathered} \frac{\alpha }{2}=\frac{1-0.95}{2} \\ =\frac{0.05}{2} \\ =0.025 \end{gathered}$$

and

$z_{\frac{\alpha }{2}}=1.96$

d) The error bound is,

$$\begin{gathered} EBM=z_{\alpha /2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}} \\ =1.96\sqrt{\frac{0.76(1-0.76)}{1006}} \\ =0.0264. \end{gathered}$$

e) The difference between the error bound of $\pm 2.64$percent and the margin of error stated by Gallup of $\pm 3$percent is insignificant.

e) The confidence interval for the result of study is

$p^{\text{'}}-EBM\le p$$\le p^{\text{'}}+EBM$,

$0.76-0.0264\le p\le 0.76+0.0264,$

$0.7336\le p\le 0.7864$

f) With $95\%$confidence the true population proportion of U.S workers who believe that they will continue working past retirement age is between $73.36\%$and $78.64\%$