Question

A researcher is interested in determining the average SAT score to be expected of a group of test takers. They collect data from 12 randomly selected high school seniors and construct a 98% confidence interval. Assume the SAT scores are normally distributed. [If you do not have a calculator: Px = 21170, Px2 = 38047900] 1700 1940 1510 2000 1430 1870 1990 1650 1820 1670 2210 1380
(a) To create the desired confidence interval, should we be using a T or Z distribution? Why? (b) Find the sample mean, sample standard deviation and the degrees of freedom if necessary. (c) Give the margin of error and construct the confidence interval.

Answer 1

a. As sample size is n<30 so we will use t distribution to find CI

b.

Create the following table.

data	data-mean	(data - mean)2
1700	-64.1667	4117.36538889
1940	175.8333	30917.34938889
1510	-254.1667	64600.71138889
2000	235.8333	55617.34538889
1430	-334.1667	111667.38338889
1870	105.8333	11200.68738889
1990	225.8333	51000.67938889
1650	-114.1667	13034.03538889
1820	55.8333	3117.35738889
1670	-94.1667	8867.36738889
2210	445.8333	198767.33138889
1380	-384.1667	147584.05338889

Find the sum of numbers in the last column to get.

So

df is 12-1=11

c. t value for 11 df for 98% CI is TINV(0.02,11)=2.718

So Margin of Error is

Hence CI is