Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ.

The data below consists of the test scores of 32 students. Construct a 99% confidence interval for the population

80	74	61	93	96	70	80	64
51	98	93	87	72	77	84	96
100	67	71	79	99	85	66	70
57	75	86	92	94	61	80	99

Answer 1

The given data is inputted in EXCEL here as:

S.No.	X	X - Mean(X)	(X - Mean(X))^2
1		80	0.09375	0.008789063
2	74	-5.90625	34.88378906
3	61	-18.90625	357.4462891
4	93	13.09375	171.4462891
5	96	16.09375	259.0087891
6	70	-9.90625	98.13378906
7	80	0.09375	0.008789063
8	64	-15.90625	253.0087891
9	51	-28.90625	835.5712891
10	98	18.09375	327.3837891
11	93	13.09375	171.4462891
12	87	7.09375	50.32128906
13	72	-7.90625	62.50878906
14	77	-2.90625	8.446289063
15	84	4.09375	16.75878906
16	96	16.09375	259.0087891
17	100	20.09375	403.7587891
18	67	-12.90625	166.5712891
19	71	-8.90625	79.32128906
20	79	-0.90625	0.821289063
21	99	19.09375	364.5712891
22	85	5.09375	25.94628906
23	66	-13.90625	193.3837891
24	70	-9.90625	98.13378906
25	57	-22.90625	524.6962891
26	75	-4.90625	24.07128906
27	86	6.09375	37.13378906
28	92	12.09375	146.2587891
29	94	14.09375	198.6337891
30	61	-18.90625	357.4462891
31	80	0.09375	0.008789063
32	99	19.09375	364.5712891
	2557	0	5890.71875

For the above data, the sum of the columns is shown in the last row.  

The mean of the above data is computed here as:

Now the sample standard deviation is computed here as:

For n - 1 = 31 degrees of freedom, we get from the t distribution tables:

P( t31 < 2.744) = 0.995

Therefore, due to symmetry, we get here:
P( -2.744 < t31 < 2.744) = 0.99

Therefore the confidence interval for the mean here is obtained as:

This is the required 99% confidence interval for the population mean here.