Question

latoya wants to estimate ( p ), the proportion of all students at her large boarding high school that like the cafeteria’s food. she interviews an srs of 50 of the students at the school and finds that 14 like the cafeteria’s food. calculate a 99% confidence interval for the proportion of students at latoya’s school who like the cafeteria’s food. (round each value to 3 decimal places.)

Explanation:

Step1: Calculate sample proportion

$\hat{p} = \frac{14}{50} = 0.28$

Step2: Find critical z-value

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

Step3: Compute standard error

$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.28(1-0.28)}{50}} = \sqrt{\frac{0.2016}{50}} \approx 0.0634$

Step4: Calculate margin of error

$ME = z^* \times SE = 1.645 \times 0.0634 \approx 0.1043$

Step5: Find interval bounds

Lower bound: $\hat{p} - ME = 0.28 - 0.1043 = 0.176$
Upper bound: $\hat{p} + ME = 0.28 + 0.1043 = 0.384$

Answer:

0.176 to 0.384