Answer:
a. CI=[128.79,146.41]
b. CI=[122.81,152.39]
c. As the confidence level increases, the interval becomes wider.
Explanation:
a. -Given the sample mean is 137.6 and the standard deviation is 20.60.
-The confidence intervals can be constructed using the formula;

where:
is the sample standard deviation
is the s value of the desired confidence interval
we then calculate our confidence interval as:
![\bar X\pm z(s)/(√(n))\\\\=137.60\pm z_(0.05/2)*(20.60)/(√(21))\\\\=137.60\pm1.960* (20.60)/(√(21))\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]](https://img.qammunity.org/2021/formulas/mathematics/college/vntrplw1gh9i4ya6lee4otsgb87cmrynrl.png)
Hence, the 95% confidence interval is between 128.79 and 146.41
b. -Given the sample mean is 137.6 and the standard deviation is 20.60.
-The confidence intervals can be constructed using the formula in a above;
![\bar X\pm z(s)/(√(n))\\\\=137.60\pm z_(0.01/2)*(20.60)/(√(21))\\\\=137.60\pm3.291* (20.60)/(√(21))\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]](https://img.qammunity.org/2021/formulas/mathematics/college/qydlrydv7rg7ox8dloskh66w3gghmofnsf.png)
Hence, the variable's 99% confidence interval is between 122.81 and 152.39
c. -Increasing the confidence has an increasing effect on the margin of error.
-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.
-The 99% Confidence interval is the most appropriate to use in such a case.