Question

cyber monday shopping a survey of 1050 u.s. adults found that 34% of people said that they would get no work done on cyber monday since they would spend all day shopping online. find the 90% confidence interval of the true proportion. round intermediate answers to at least five decimal places. round your final answers to at least three decimal places.

Explanation:

Step1: Identify given values

$n=1050$, $\hat{p}=0.34$, confidence level=90%

Step2: Find z-score for 90% confidence

$z_{\alpha/2}=1.645$ (from standard normal table)

Step3: Calculate standard error

$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.34×0.66}{1050}}≈\sqrt{0.000212571}≈0.01458$

Step4: Compute margin of error

$ME=z_{\alpha/2}×SE=1.645×0.01458≈0.02398$

Step5: Find confidence interval

Lower bound: $0.34 - 0.02398≈0.316$; Upper bound: $0.34 + 0.02398≈0.364$

Answer:

0.316 < p < 0.364