Statistics help please

Question

A certain instructor was interested in the average amount of out of class time the "A" students spent per week on the course. a survey conducted among 20 such students showed an average of 8.3 hours with a variance of 71.3.

Calculate the 95% interval for the average number of hours "A" students spend per week on the course. Interpret your confidence interval.

Answer 1

To find the 95% confidence interval, first we determine that this is a t-distribution since the population standard deviation is not known, so we will use the sample standard deviation in the formula.

x-bar +/- t*(s/square root(n))

x-bar is the sample mean of 8.3

s is the sample standard deviation, which is the square root of the variance. Square root of 71.3 is approximately 8.444

n is the sample size, which is 20.

To find the t critical value for the formula, Look up in any t distribution table under two-tailed area of .05 and 19 degrees of freedom. Remember the degrees of freedom needed is (n-1). The t critical value is 2.093.

Putting it all together we get

8.3 +/- 2.093(8.444/square root(20))

8.3 +/- 3.95

(4.35, 12.25)

We are 95% confident that the population average number of hours "A" students per week in the course is between 4.35 and 12.25.

Answer 2

95% confidence ---> 1.96 constant

stddev = sqrt(variance) = sqrt(71.3) = 8.44393273303382018....

mean +or- 1.96*stddev/sqrt( n)

8.3 +or - 1.96* 8.44393273303382018/sqrt(20)

8.3 +or- 3.7

(4.6, 13)