Answer:
(15,595, 16,805)
Step-by-step explanation:
We have to:
m = 16.2, sd = 3.75, n = 150
m is the mean, sd is the standard deviation and n is the sample size.
the degree of freedom would be:
n - 1 = 150 - 1 = 149
df = 149
at 95% confidence level the t is:
alpha = 1 - 95% = 1 - 0.95 = 0.05
alpha / 2 = 0.05 / 2 = 0.025
now well for t alpha / 2 (0.025) and df (149) = t = 1,976
the margin of error = E = t * sd / (n ^ (1/2))
replacing:
E = 1,976 * 3.75 / (150 ^ (1/2))
E = 0.605
The 95% confidence interval estimate of the popilation mean is:
m - E <u <m + E
16.2 - 0.605 <u <16.2 + 0.605
15,595 <u <16,805
(15,595, 16,805)
