question 11 0.1 pts use the given degree of confidence and sample data to construct a confidence interval…

Explanation:

Step1: Calculate sample proportion

Let $n = 346$ (sample size) and $x=12$ (number of defectives). The sample proportion $\hat{p}=\frac{x}{n}=\frac{12}{346}\approx0.0347$.

Step2: Determine z - value

For a 98% confidence interval, the significance level $\alpha = 1 - 0.98=0.02$. So $\alpha/2=0.01$. The $z$-value $z_{\alpha/2}=z_{0.01}\approx 2.33$.

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.0347$, $n = 346$ and $z_{\alpha/2}=2.33$ into the formula. First, $1-\hat{p}=1 - 0.0347 = 0.9653$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.0347\times0.9653}{346}\approx\frac{0.0335}{346}\approx0.00009682$. And $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx\sqrt{0.00009682}\approx0.00984$. So $E=2.33\times0.00984\approx0.0229$.

Step4: Construct confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E<p<\hat{p} + E$. Substitute $\hat{p}=0.0347$ and $E = 0.0229$ into the formula. We get $0.0347- 0.0229<p<0.0347 + 0.0229$, which simplifies to $0.0118 < p<0.0576$.

Answer:

$0.0118 < p<0.0576$