Question

A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit survey was conducted with a random sample of 9 seniors. The sample mean was 7 novels with standard deviation 2.29 novels. Assuming that all conditions for conducting inference have been met, which of the following is a 94.645% confidence interval for the population mean number of novels read by all seniors?

A. 7 + or - 1.960 (2.29/underroot 8).
B. 7 + or - 1.960 (2.29/underroot 9).
C. 7 + or - 2.262 ((2.29/underroot 9).
D. 7 + or - 2.306 (2.29/underoot 8).
E. 7 + or - 2.306 (2.29/underoot 9).

Answer 1

E

Explanation:

on ap classroom

Answer 2

Population mean = 7 ± 2.306 ×
`(2.29)/(√(9) )`

Explanation:

Given - A university researcher wants to estimate the mean number

of novels that seniors read during their time in college. An exit

survey was conducted with a random sample of 9 seniors. The

sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

been met, which of the following is a 94.645% confidence

interval for the population mean number of novels read by

all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

= n - 1

= 9 - 1

= 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

=1 - 0.94645

=0.05355 ≈ 0.05

⇒α = 0.5

Now,

`(\alpha)/(2) = (0.05)/(2)`

= 0.025

Then,

`t_{(\alpha)/(2), df }` = 2.306

∴ we get

Population mean = x⁻ ±
`t_{(\alpha)/(2), df }` ×
`(s)/(√(n) )`

= 7 ± 2.306 ×
`(2.29)/(√(9) )`

⇒Population mean = 7 ± 2.306 ×
`(2.29)/(√(9) )`