Answer:
a) The 95% confidence interval for the mean is (132.9, 140.9).
b) The 99.5% confidence interval for the mean is (131.1, 142.7).
c) Confidence level = 85.6%
Step-by-step explanation:
a) We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=136.9.
The sample size is N=123.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{22.6}{\sqrt{123}}=\dfrac{22.6}{11.091}=2.038[/tex]
The t-value for a 95% confidence interval and 122 degrees of freedom is t=1.98.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.98 \cdot 2.038=4.034[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 136.9-4.034=132.9\\\\UL=M+t \cdot s_M = 136.9+4.034=140.9[/tex]
The 95% confidence interval for the mean is (132.9, 140.9).
b) We have to calculate a 99.5% confidence interval for the mean.
The t-value for a 100% confidence interval is t=2.859.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=2.859 \cdot 2.038=5.826[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 136.9-5.826=131.1\\\\UL=M+t \cdot s_M = 136.9+5.826=142.7[/tex]
The 99.5% confidence interval for the mean is (131.1, 142.7).
c) We have a confidence interval that is (133.9, 139.9).
This corresponds to a margin of error of:
[tex]MOE=\dfrac{UL-LL}{2}=\dfrac{139.9-133.9}{2}=\dfrac{6}{2}=3[/tex]
The critical value for this margin of error is:
[tex]t=MOE/s_M=3/2.038=1.472[/tex]
This critical value of t=1.472 for 122 degrees of freedom corresponds to a confidence level of 85.6%. See picture attached.
