Question

cor 8.5/6, 1.11 of 6 pt Question Help The random sample shown below was selected from a normal distribution 8, 4, 4, 6, 5, 90 Complete parts a and b. a. Construct a 99% confidence interval for the population mean 2.55.9.45) (Round to two decimal places as needed.) b. Assume that sample mean x and sample standard deviation s remain exactly the same as those you just caloulated but that are based on a sample ofn25 observations. Repeat part a What is the effect of increasing the sample size on the width of the confidence intervals? D. (Round to two decimal places as needed.) The confidence interval is

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	8	4
	4	4
	4	4
	6	0
	5	1
	9	9
Total	36	22

Mean X̅ = Σ Xi / n
X̅ = 36 / 6 = 6
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 22 / 6 -1 ) = 2.0976

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
Critical value t(α/2, n-1) = t(0.01 /2, 6- 1 ) = 4.032 ( From t table )
6 ± t(0.01/2, 6 -1) * 2.0976/√(6)
Lower Limit = 6 - t(0.01/2, 6 -1) 2.0976/√(6)
Lower Limit = 2.55
Upper Limit = 6 + t(0.01/2, 6 -1) 2.0976/√(6)
Upper Limit = 9.45
99% Confidence interval is ( 2.55 , 9.45 )
Width = 9.45 - 2.55 = 6.91

Part b)
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
Critical value t(α/2, n-1) = t(0.01 /2, 25- 1 ) = 2.797 ( From t table )
6 ± t(0.01/2, 25 -1) * 2.0976/√(25)
Lower Limit = 6 - t(0.01/2, 25 -1) 2.0976/√(25)
Lower Limit = 4.83
Upper Limit = 6 + t(0.01/2, 25 -1) 2.0976/√(25)
Upper Limit = 7.17
99% Confidence interval is ( 4.83 , 7.17 )
Width = 7.17 - 4.83 = 2.35

As sample size increases, width of confidence interval decreases.