Can someone help with this story problem?
2 answers:
You might be interested in
Answer:
Lower confidence limit = 20.191
Step-by-step explanation:
Sample mean; x' = 22
Sample standard deviation; s = 5.5
Sample size; n = 50
Confidence interval = 99%
Since sample size > 30, we will use the formula;
CI = x' ± zs/√n
Where;
x' is sample mean
s is sample standard deviation
n is sample size
z is z-value of the confidence interval
From the table attached, z for 99% = 2.3263
Thus;
CI = 22 ± (2.3263 × 5.5)/√50
CI = 22 ± 1.809
CI = 23.809 or 20.191
Lower confidence limit = 20.191
