Question

Below is the data for the exam scores of a sample of 9 students. The population is normally distributed.

85, 70, 66, 95, 77, 94, 65, 79, 84

a)Find the margin of error for a 95% confidence interval for the mean. Round to 3 decimal places.  

b)Find the confidence interval for the mean. Round to 3 decimal places.  

Answer 1

For given sample, mean is $\overline{x}=\frac{\sum x}{n}=79.4444$

Create the following table.

data	data-mean	(data - mean)2
85	5.5556	30.86469136
70	-9.4444	89.19669136
66	-13.4444	180.75189136
95	15.5556	241.97669136
77	-2.4444	5.97509136
94	14.5556	211.86549136
65	-14.4444	208.64069136
79	-0.4444	0.19749136
84	4.5556	20.75349136

Find the sum of numbers in the last column to get. $\sum (xi-\overline{x})^2=990.2222$

So standard deviation is $s=\sqrt{\frac{\sum (xi-\overline{x})^2}{n-1}}=11.1255$

a. t table value for 8df is TINV(0.05,8)=2.306

So Margin of Error is $E=t*\frac{s}{\sqrt{n}}=2.306*\frac{11.1255}{3}=8.552$

b. Hence CI is $CI=\overline{x} \pm E=79.444\pm 8.552=(70.892, 87.996)$