Question

in a recent survey of 1080 students, 975 of them would like to recommend www.helpyourmath.com to their friends.
what is the point estimate for the proportion of students who would recommend the site?

construct a 95% confidence interval to estimate the proportion of all students who would recommend www.helpyourmath.com to their friends. give your answers to three decimals:
( , )

Explanation:

Step1: Calculate point estimate

The point estimate for the population proportion is the sample proportion, calculated as the number of successes divided by the sample size.
$\hat{p} = \frac{975}{1080} \approx 0.9028$

Step2: Find critical z-value

For a 95% confidence interval, the critical standard normal z-value is $z_{\alpha/2} = 1.96$.

Step3: Compute standard error

The standard error for a proportion is calculated using the sample proportion and sample size.
$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.9028(1-0.9028)}{1080}} \approx \sqrt{\frac{0.9028 \times 0.0972}{1080}} \approx 0.0090$

Step4: Calculate margin of error

Margin of error is the product of the critical z-value and standard error.
$ME = z_{\alpha/2} \times SE = 1.96 \times 0.0090 \approx 0.0176$

Step5: Compute confidence interval

Lower bound: $\hat{p} - ME = 0.9028 - 0.0176 \approx 0.885$
Upper bound: $\hat{p} + ME = 0.9028 + 0.0176 \approx 0.920$

Answer:

Point estimate: $0.903$
95% confidence interval: $(0.885, 0.920)$