91影视

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

The sample proportion is

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

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

$z_{伪/2}=1.96$

The margin of error is

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

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$0.5943and0.7357$.

Result from exercise pat (a):

$0.5943<p<0.7257$

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$73\%$.