yesharabaskoro
21.12.2020 •
Mathematics
A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of $60.52 and standard deviation of $23.29 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.
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Ответ:
53.15≤x≤67.89
Step-by-step explanation:
The formula for calculating confidence interval is expressed as;
CI = xbar±(z×s/√n)
xbar is the sample mean
z is the z score at 90% CI
s is the standard deviation
n is the sample size
Given
xbar = $60.52
s = $23.29
n = 27
z = 1.645
Substitute into the formula
CI = xbar±(z×s/√n)
CI = 60.52±(1.645×23.29/√27)
CI = 60.52±(1.645×23.29/5.196)
CI = 60.52±(1.645×4.482)
CI = 60.52±7.373
CI = [60.52-7.373, 60.52+7.373]
CI = [53.15, 67.89]
Hence the 90% confidence interval for the mean repair cost for the refrigerators is 53.15≤x≤67.89
Ответ: