Help please. Thanks

Question

1. A sample of size 14 is drawn from a normally distributed population. The sample mean is x = 0.33 and the sample standard deviation is s = 18. Construct an 80% confidence interval.
2. In a random sample of five people, the mean driving distance to work was 22.2 miles and the standard deviation was 5.8 miles. Assume the driving time to work is normally distributed. Construct a 95% confidence interval.

Answer 1

1. Confidence interval = sample mean +/- t-critical value * standard error

sample mean = 0.33

t critical value for 80% confidence and 14 - 1 = 13 degrees of freedom = 1.35

standard error = standard deviation/square root(14) = 18/sqrt(14) = 4.8

2.

sample mean = 22.2

t critical value for 95% confidence and 4 - 1 = 3 degrees of freedom = 2.776

standard error = standard deviation/square root(4) = 5.8/2 = 2.9