Find upper bound and lower bound? (statistics)

Question

A simple random sample of size n is drawn. The sample​ mean, x​, is found to be 19.2​, and the sample standard deviation, s, is found to be 4.8.

Construct a 95​% confidence interval about μ if the sample​ size, n, is 35.

Lower​ bound: ?

Upper​ bound: ?

​(Use ascending order. Round to two decimal places as​ needed.)

Answer 1

The sampling distribution for the standard score in this case is student's t-distribution with n-1 degrees of freedom.

We need to cover the middle 0.95 probability, so the cdf must have F(-t^*)=0.025 and F(t^*) = 0.975, which we can look up in the t-table with df=34 or use software.

I used R-Studio by calling the routine qt(0.975,df=34).

t^* = 2.032245

Now we also need the standard error:

SE = s / √n = 4.8 / √35 = 0.8113481

So the confidence bounds are x̄ ± t^*SE.

x̄ + t^*SE = 19.2 + 2.032245 * 0.8113481 = 20.84886

x̄ - t^*SE = 19.2 - 2.032245 * 0.8113481 = 17.55114

The 95% confidence interval is approximately (17.55, 20.85)