graduation gifts ninety - five percent of students said that cash is their top choice for a graduation gift…

Explanation:

Step1: Identify sample proportion and sample size

Let $\hat{p}$ be the sample - proportion and $n$ be the sample size. Given $\hat{p}=0.95$ and $n = 175$.

Step2: Find the z - value for 95% confidence interval

For a 95% confidence interval, the significance level $\alpha=1 - 0.95=0.05$. Then $\alpha/2=0.025$. The $z$ - value $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.95$, $n = 175$, and $z_{\alpha/2}=1.96$ into the formula: [ \begin{align*} E&=1.96\sqrt{\frac{0.95\times(1 - 0.95)}{175}}\ &=1.96\sqrt{\frac{0.95\times0.05}{175}}\ &=1.96\sqrt{\frac{0.0475}{175}}\ &=1.96\sqrt{0.000271429}\ &=1.96\times0.0165\ &=0.03234 \end{align*} ]

Step4: Calculate the confidence interval

The confidence interval for a proportion is given by $\hat{p}-E<p<\hat{p} + E$. Substitute $\hat{p}=0.95$ and $E = 0.03234$: $0.95-0.03234<p<0.95 + 0.03234$ $0.918<p<0.982$

Answer:

$0.918<p<0.982$