Question

i need help with this questiong ASAP

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to ensure that the estimate has an error of at most 0.28 words per minute. A previous study found that the mean was 26.7 words per minute. Assuming that the variance is 7.29, what is the minimum number of third graders that must be included in a sample to construct the 90 % confidence interval? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

variance = $\sigma$² = 7.29

Population standard deviation = $\sigma$ = $\sqrt{}$ $\sigma$² = $\sqrt{}$ 7.29 = 2.7

Margin of error = E = 0.28

At 90% confidence level the z is,

$\alpha$ = 1 - 90%

$\alpha$ = 1 - 0.90 = 0.10

$\alpha$/2 = 0.05

Z_$\alpha$/2 = 1.645

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [1.645 * 2.7 / 0.28]²

n = 251.61

Sample size = n = 252