Question

question 9 (1 point)
in a random sample of 145 people, 112 said that they watched educational tv. find the 93% confidence interval of the true proportion of people who watched educational tv.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 112$ (number of successes) and $n=145$ (sample size). So, $\hat{p}=\frac{112}{145}\approx0.7724$.

Step2: Find $z -$ value

For a 93% confidence interval, the significance level $\alpha=1 - 0.93 = 0.07$. Then $\alpha/2=0.035$. Looking up in the standard - normal table, $z_{\alpha/2}=z_{0.035}\approx1.81$.

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.7724$, $n = 145$, and $z_{\alpha/2}=1.81$ into the formula.
First, $1-\hat{p}=1 - 0.7724 = 0.2276$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.7724\times0.2276}{145}\approx\frac{0.1757}{145}\approx0.001212$. And $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx\sqrt{0.001212}\approx0.0348$. So, $E = 1.81\times0.0348\approx0.063$.

Step4: Calculate confidence interval

The confidence interval is $\hat{p}-E

Answer:

$(0.7094,0.8354)$