Question

Assuming that the population is normally​ distributed, construct a 99​% confidence interval for the population mean for each of the samples below. Explain why these two samples produce different confidence intervals even though they have the same mean and range.

Sample A: 1,3,3,3,6,6,6,8

Sample B: 1,2,3,4,5,6,7,8

Q1. Construct a 99​% confidence interval for the population mean for sample A.

____ <_ u <_ _____

Q2.

Construct a 99​% confidence interval for the population mean for sample B.

____ <_ u <_ _____

Q3. Explain why these two samples produce different confidence intervals even though they have the same mean and range.

A.The samples produce different confidence intervals because their standard deviations are different.

B.The samples produce different confidence intervals because their critical values are different.

C.The samples produce different confidence intervals because their sample sizes are different.

D.The samples produce different confidence intervals because their medians are different.

Answer 1

Let us calculate the confidence interval for sample A

Number	Sample A
1	1	12.25 (1 - 4.5)2
2	3	2.25  (3 - 4.5)2
3	3	2.25  (3 - 4.5)2
4	3	2.25  (3 - 4.5)2
5	6	2.25  (6 - 4.5)2
6	6	2.25  (6 - 4.5)2
7	6	2.25  (6 - 4.5)2
8	8	12.25  (8 - 4.5)2
Total	36	38

Mean = 36 / 8 = 4.5

Standard Deviation S =   = 38 / 7 =  5.43 = 2.33

Confidence Interval

= 3.499

4.5 3.499 * 2.33 /

Lower Limit = 1.618

Upper Limit = 7.382

( 1.618 < < 7.382 )

Question 2

For Sample B

Number	Sample A
1	1	12.25 ( 1 - 4.5)2
2	2	6.25 ( 2 - 4.5)2
3	3	2.25 ( 3 - 4.5)2
4	4	0.25 ( 4 - 4.5)2
5	5	0.25  ( 5 - 4.5)2
6	6	2.25  ( 6 - 4.5)2
7	7	6.25  ( 7 - 4.5)2
8	8	12.25  ( 8 - 4.5)2
Total	36	42

Mean = 36 / 8 = 4.5

Standard Deviation S =   = 42/ 7 =  6= 2.45

Confidence Interval

= 3.499

4.5 3.499 * 2.45/

Lower Limit = 1.469

Upper Limit = 7.531

( 1.469 < < 7.531 )

Question 3

	Standard Deviation	Lower Limit	Upper Limit
Sample A	2.33	1.618	7.382
Sample B	2.45	1.469	7.531

We can observe that lower limit and upper limit are not sample for sample A and B even though their means are equal.

But due to sample standard deviation are different so their confidence interval differ, hence option A is correct.