Question

educational television in a random sample of 210 people, 143 said that they watched educational television. find the 90% confidence interval of the true proportion of people who watched educational television. round intermediate answers to at least five decimal places. round your final answers to at least three decimal places.

Explanation:

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

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 143$ (number of successes) and $n=210$ (sample size). So, $\hat{p}=\frac{143}{210}\approx0.68095$

Step2: Calculate $q = 1-\hat{p}$

$q=1 - 0.68095=0.31905$

Step3: Find $z$-value for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90 = 0.10$, and $\alpha/2=0.05$. The $z$-value $z_{\alpha/2}=z_{0.05}$. From the standard normal table, $z_{0.05} = 1.645$

Step4: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E=z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$.
Substitute the values: $\hat{p}=0.68095$, $\hat{q}=0.31905$, $n = 210$, $z_{\alpha/2}=1.645$
$E=1.645\sqrt{\frac{0.68095\times0.31905}{210}}$
First, calculate $\frac{0.68095\times0.31905}{210}=\frac{0.2172}{210}\approx0.001034$
Then, $\sqrt{0.001034}\approx0.03216$
$E=1.645\times0.03216\approx0.0529$

Step5: Calculate the confidence interval

The confidence interval for the proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.68095$ and $E = 0.0529$
Lower limit: $0.68095-0.0529=0.628$
Upper limit: $0.68095 + 0.0529=0.734$

Answer:

$0.628