Question

a recent poll of 738 randomly selected customers of a major u.s. cell - phone carrier found that 170 of them had walked into something or someone while talking on a cell phone.
(a) show that the conditions for calculating a confidence interval for a proportion are satisfied.
random? the cell - phone carrier randomly selected 738 customers.
large counts? (nhat{p}=170) and (n(1 - hat{p}) = 568) are both at least 10.
(b) calculate a 90% confidence interval for the proportion of all customers who have walked into something or someone while talking on a cell phone. (round each value to 3 decimal places.)

Explanation:

Step1: Calculate sample proportion

$\hat{p} = \frac{170}{738} \approx 0.230$

Step2: Find critical z-value

For 90% confidence, $z^* = 1.645$

Step3: Compute standard error

$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.230(1-0.230)}{738}} \approx \sqrt{\frac{0.1771}{738}} \approx 0.0155$

Step4: Calculate margin of error

$ME = z^* \times SE = 1.645 \times 0.0155 \approx 0.0255$

Step5: Compute confidence interval

Lower bound: $\hat{p} - ME = 0.230 - 0.0255 = 0.2045$
Upper bound: $\hat{p} + ME = 0.230 + 0.0255 = 0.2555$

Answer:

0.205 to 0.256