Question

Refer to Exercise $52$.

a. Construct and interpret a $95\%$confidence interval for the true proportion p of all first year students at the university who would identify being very well-off as an important personal goal. Assume that the conditions for inference are met.

b. Explain why the interval in part (a) provides more information than the test in Exercise 52.

Short answer

Part a. $0.5943<p<0.7257$

Part b. The confidence interval is not having $73\%$(or $0.73$) which is the national value. Therefore there is sufficient proof to help the claim that the proportion is different (less) than the national value of$73\%$.

Step-by-step solution

Part a. Step 1. Given information

$$\begin{gathered} n=200 \\ p=73\%=0.73 \\ x=132 \end{gathered}$$

Part a. Step 2. Explanation

The sample proportion is

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{132}{200} \\ =0.66 \end{gathered}$$

For confidence level $11-\alpha =0.95$, determine $z_{\alpha /2}=z_{0.025}$using table II (look up$0.025$in the table, the z-score is then they found z-score with opposite sign):

$z_{\alpha /2}=1.96$

The margin of error is

$$\begin{gathered} E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =1.96\times \sqrt{\frac{0.66(1-0.66)}{200}} \\ =0.0657 \end{gathered}$$

The confidence interval then becomes:

$$\begin{gathered} \hat{p}-E<p<\hat{p}+E=0.66-0.0657<p<0.66+0.657 \\ =0.5943<p<0.7257 \end{gathered}$$

There is $95\%$confident that the true proportion of all first- year students at the university who would identify being very well-off as an important personal goal is between$0.5943and0.7357$.

Part b. Step 1. Given information

Result from exercise pat (a):

$0.5943<p<0.7257$

Part b. Step 1. Explanation

The confidence interval is not having $73\%$(or $0.73$) which is the national value. Therefore there is sufficient proof to help the claim that the proportion is different (less) than the national value of$73\%$.