Question

1. Getting a Job. Refer to Problem 9 .
   a. Determine a sample size that will ensure a margin of error of at most0.02for a95%confidence interval without making a guess for the observed value ofp^.
   b. Find a95%confidence interval forpif, for a sample of the size determined in part (a),58.7%of those surveyed say that they expect difficulty finding a job.
   c. Determine the margin of error for the estimate in part (b), and compare it to the required margin of error specified in part (a).
   d. Repeat parts (a)-(c) if you can reasonably presume that the percentage of those surveyed who say that they expect difficulty finding a job will be at least56%.
   e. Compare the results obtained in parts (a)-(c) with those obtained in part (d).

Short answer

Part a) the required sample size is $2,401$

Part b) the $95\%$confidence interval for the proportion is$(0.5674,0.6066)$

Part c) The margin of error in part (b) is slightly less than the specified margin of error in part (a)

Part d) the required sample size is $2367$

Part e) the margin of error is $0.0198$

The margin of error is the same as in the specified margin of error in part (a)

From parts (a)-(c) and part (d), it is clear that the sample size in part (a) is reduced by 34 in part (d). Moreover, the margin of error is increased$0.0196to0.0198$

Step-by-step solution

Part a) Step 1: Explanation

Obtain the sample size when the margin of error is $0.02$and the confidence level is $95\%$

From "Table II Areas under the standard normal curve" the required value of

$z_{\frac{\alpha }{2}}$with $95\%$

confidence level is $1.96$

The sample size is

$$\begin{gathered} n=\hat{p}(1-\hat{p}){\left(\frac{z_{\frac{\alpha }{2}}}{E}\right)}^2 \\ =0.5(1-0.5){\left(\frac{1.96}{0.02}\right)}^2=(0.25)(9,604) \\ =(0.25)(9,604)=2,401 \end{gathered}$$

Therefore, the required sample size is$2401$

Part b) Step 1: Explanation

Find the $95\%$confidence interval for the proportion

when $n=2,401$and $\hat{p}=0.587$

$$\begin{gathered} \hat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.587\pm 1.96\sqrt{\frac{0.587(1-0.587)}{2,401}} \\ =0.587\pm 1.96(0.0100) \\ =0.587\pm 0.0196 \\ =0.5674,0.6066 \end{gathered}$$

Thus, the $95\%$confidence interval for the proportion is$(0.5674,0.6066)$

Part c) Step 1: Explanation

From part (b), the margin of error is $0.0196$

The margin of error in part (b) is slightly less than the specified margin of error in part (a)

Part d) Step 1: Explanation

Obtain the sample size when the margin of error is $0.02$and the confidence level is $95\%$

Educated guess for $\hat{p}$can be identified by which the value in the range is closest to $0.5$

Here, the value$0.56$in the range is closest to $0.5$
Hence, educated guess $\left({\hat{p}}_g\right)$is $0.56$

The sample size is

$$\begin{gathered} n=\hat{p}(1-\hat{p}){\left(\frac{z_{\alpha }}{E}\right)}^2 \\ n=\hat{p}(1-\hat{p}){\left(\frac{z_{\alpha }}{E}\right)}^2 \\ =(0.2464)(9,604) \\ =2,366.4 \end{gathered}$$

Therefore, the required sample size is$2367$

Part e) Step 1: Explanation

Find the $95\%$confidence interval for the proportion when $n=2367$and$\hat{p}=0.587$
$$\begin{gathered} \hat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.587\pm 1.96\sqrt{\frac{0.587(1-0.587)}{2,367}} \\ =0.587\pm 1.96(0.01=0.587\pm 0.0198 \\ =(0.5672,0.6068) \\ =0.587\pm 0.0198) \end{gathered}$$

Therefore, the $95\%$confidence interval for the proportion is$(0.5672,0.6068)$

From the above result, the margin of error is$0.0198$

Comparison:

From parts (a)-(c) and part (d), it is clear that the sample size in part (a) is reduced by$34$in part (d). Moreover, the margin of error is increased$0.0196to0.0198$