In: Statistics and Probability
Solution :
Given that,
= 7.987
s = 2.459
n = 10
Degrees of freedom = df = n - 1 = 10 - 1 = 9
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,9 = 2.262
Margin of error = E = t/2,df * (s /n)
= 2.262 * (2.459 / 10)
= 1.759
The 95% confidence interval estimate of the population mean is,
- E < < + E
7.987 - 1.759 < < 7.987 + 1.759
6.228 < < 9.746
(6.228,9.746)
The 95% confidence interval is 6.228 to 9.746
Given that,
= 6.930
s = 3.748
n = 10
Degrees of freedom = df = n - 1 = 10 - 1 = 9
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,9 = 2.262
Margin of error = E = t/2,df * (s /n)
= 2.262 * (3.748 / 10)
= 2.681
The 95% confidence interval estimate of the population mean is,
- E < < + E
6.930 - 2.681 < < 6.930 + 2.681
4.249 < < 9.611
(4.249,9.611)
The 95% confidence interval is 4.249 to 9.611