Question

A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is:________

a. 20.0 to 30.0
b. 20.5 to 26.5
c. 23.0 to 27.0
d. 24.4 to 25.6

Answer 1

The correct option is;

d. 24.2 to 25.6

Explanation:

Here we have a sample with unknown population standard deviation, we therefore apply the student t distribution at 64 - 1 degrees of freedom

Therefore, we have

`CI=\bar{x}\pm t(s)/(√(n))`

Where:

`\bar x` = Mean = 25

σ = Standard deviation = 2

n = Sample size = 64

t = T value at 98% =
`\pm 2.387`

Which gives

`CI=25\pm t_(63)(2)/(√(64))`

That is the value is from

24.40325 to 25.59675 which gives,by rounding to one decimal place, is

24.4 to 25.6.