a random sample of adults was surveyed about whether they shop online. of the 90 adults surveyed, 72 stated…

Explanation:

Step1: Calculate sample proportion $\hat{p}$

$\hat{p}=\frac{72}{90}=0.8$

Step2: Determine $z$-score for 99% confidence interval

For 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$, and $\alpha/2=0.005$. Looking up in the z - table, $z_{\alpha/2}=2.576$.

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Here, $n = 90$, $\hat{p}=0.8$, and $z_{\alpha/2}=2.576$. So $E=2.576\sqrt{\frac{0.8\times(1 - 0.8)}{90}}=2.576\sqrt{\frac{0.8\times0.2}{90}}=2.576\sqrt{\frac{0.16}{90}}\approx2.576\times0.042=0.108$.

Step4: Calculate the confidence interval

The confidence interval is $\hat{p}-E<p<\hat{p} + E$. Substituting the values, we get $0.8 - 0.108<p<0.8+0.108$, which simplifies to $0.692 < p<0.908\approx(0.69,0.91)$.

Answer:

C. $(0.69, 0.91)$