Question

0 The state test scores for 12 randomly selected high school seniors are shown on the right Complete parts (a) through (c) below. Assume the population is normally distributed 1420 697 721 622 1225 722 745 1445 984 834 542 943 (a) Find the sample mean. (Round to one decimal place as needed.) (b) Find the sample standard deviation. (Round to one decimal place as needed.) (c) Construct a 99% confidence interval for the population mean . A 99% confidence interval for the population mean is (CD. (Round to one decimal place as needed.) S AM

Answer 1

a. $\overline{x}=\frac{\sum x}{n}=908.3$

b. Create the following table.

data	data-mean	(data - mean)2
1420	511.6667	261802.81188889
697	-211.3333	44661.76368889
721	-187.3333	35093.76528889
622	-286.3333	81986.75868889
1225	316.6667	100277.79888889
722	-186.3333	34720.09868889
745	-163.3333	26677.76688889
1445	536.6667	288011.14688889
984	75.6667	5725.44948889
834	-74.3333	5525.43948889
542	-366.3333	134200.08668889
943	34.6667	1201.78008889

Find the sum of numbers in the last column to get. $\sum (xi-\overline{x})^2=1019884.7$

So standard deviation is $s=\sqrt{\frac{\sum (xi-\overline{x})^2}{n-1}}=304.5$

c. t value for 99% CI is TINV(0.01,11)=3.1

So Margin of Error is $E=t*\frac{s}{\sqrt{n}}=3.1*\frac{304.5}{\sqrt{12}}=272.5$

So CI is

CI = TIE = 908.3 + 272.5 = (635.8, 1180.8)