Question

A sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2. Assuming that the population has a normal distribution, use this sample data to construct a 95% confidence interval for the population mean μ of all test scores.

A. (77.29, 85.71)

B. (56.12, 78.34)

C. (87.12, 98.32)

D. (66.35, 69.89)

Answer 1

Answer:

Given,

n = 25

Mean = 81.5

Standard deviation = 10.2

To give the 95% confidence interval

Degree of freedom = n - 1

= 25 - 1

= 24

Alpha = 0.05

alpha/2 = 0.025

Here for 95% confidence interval , corresponding alpha 0.025 & degree of freedom 24 the critical value t 0.025,24 is 2.064

So consider,

Interval = Xbar +/- t*s/sqrt(n)

substitute the values

= 81.5 +/- 2.064*10.2/sqrt(25)

= 81.5 +/- 4.21

= (81.5 - 4.21 , 81.5 + 4.21)

= (77.29 , 85.71)

Interval = (77.29 , 85.71)

So Option A is right answer.