Question

In a survey of 2230 U.S. adults, 1272 think that air travel is much more reliable than taking cruises. Construct a 95% confidence interval for the population proportion of U.S. adults who think that air travel is much more reliable than taking cruises.

Answer 1

Answer:

(0.5499, 0.5909)

Step-by-step explanation:

Given that in a survey of  2230 U.S. adults, 1272 think that air travel is much more reliable than taking cruises.

Sample proportion p = $\frac{1272}{2230} \\=0.5704$

Standard error of proportion $=\sqrt{pq/n} \\=\sqrt{\frac{0.5704(1-0.5704)}{2230} } \\= 0.010483$

We can use z critical value as sample size is very large

For 95% z critical is 1.96

Margin of error = 1.96*(0.0104)

= 0.02055

Confidence interval = p-margin of error, p + margin of error

=$(0.5704-0.0205, 0.5704+0.0205)\\=(0.5499, 0.5909)$