In: Statistics and Probability
1. A random sample of 40 college student students shows that the score of a College Statistics is normally distributed with its mean, 81 and standard deviation, 8.4. Find 99 % confidence interval estimate for the true mean.
Solution :
Given that,
= 81
s = 8.4
n = 40
Degrees of freedom = df = n - 1 = 40 - 1 = 39
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005.39 = 2.708
Margin of error = E = t/2,df * (s /n)
= 2.708 * (8.4 / 40)
= 3.6
The 99% confidence interval estimate of the population mean is,
- E < < + E
81 - 3.6 < < 81 + 3.6
77.4 < < 84.6
(77.4,84.6)
The 99% confidence interval 77.4 to 84.6