Question

# A researcher is interested in determining the average SAT score to be expected of a group...

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.

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