Job Priority Survey USA Today reported on an Adecco Sta³ng survey of 1000 randomly selected adults. Among those respondents, 20% chose health benefits as being most important to their job.

a. What is the number of respondents who chose health benefits as being most important to their job?

b. Construct a 95% interval estimate of the proportion of all adults who choose health benefits as being most important to their job.

c. Based on the result from part (b), can we safely conclude that the true proportion is different from 1/4? Why?

The number of sampled adults is 1000(n).

The proportion of adults who chose health benefits is 0.20(p).

a.

Suppose xout of 1000 adults in the sample chose health benefits to be the most important.

The proportion of adults is expressed below:

\(\begin{array}{c}p = \frac{x}{n}\\0.2 = \frac{x}{{1000}}\\x = 200\end{array}\)

Thus, 200 out of 1000 adult respondents chose health benefits to be the most important.

b.

The standard error for sample proportion distribution is calculated below:

\(\begin{array}{c}{\sigma _p} = \sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \\ = \sqrt {\frac{{0.2\left( {1 - 0.2} \right)}}{{1000}}} \\ = 0.0126\end{array}\)

The critical z-value at 0.05 significance level obtained from the standard normal tableis \({z^*} = 1.96\).

The upper limit of the confidence interval is calculated below:

\(\begin{array}{c}U.L = \hat p + {z^*} \times {\sigma _{\hat p}}\\ = 0.2 + 1.96 \times 0.0126\\ = 0.225\end{array}\)

The lower limit of the confidence interval iscalculated below:

\(\begin{array}{c}L.L = \hat p - {z^*} \times {\sigma _{\hat p}}\\ = 0.2 - 1.96 \times 0.0126\\ = 0.175\end{array}\)

Thus, the 95% confidence interval for the population proportion of all adults who choose health benefits as the most important to their job is \(\left( {0.175,0.225} \right)\).

c.

The 95% confidence interval for all adults who choose health benefits as the most important to their job is\(\left( {0.175,0.225} \right)\).

It can be concluded with 95% confidence that the actual proportion of adults who choose health benefits as most important for their job lies between 0.175 and 0.225.

To verify if the true proportion is different from\(\frac{1}{4}\left( {0.25} \right)\), formulate the hypothesis as shown below:

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\pi = 0.25\\{{\rm{{\rm H}}}_{\rm{a}}}:\pi \ne 0.25\end{array}\)

Here, assume that the true proportion is\(\pi \).

Since 0.25 does not lie between the 95% confidence interval range, the null hypothesis is rejected at the 0.05 significance level.

Thus, there is enough evidence to prove that the true proportion is different from 0.25 at the 0.05 level of significance.