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

Answer:

Population mean = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

Step-by-step 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,

$\frac{\alpha}{2} = \frac{0.05}{2}$

  = 0.025

Then,

$t_{\frac{\alpha}{2}, df }$  = 2.306

∴ we get

Population mean = x⁻ ± $t_{\frac{\alpha}{2}, df }$ ×$\frac{s}{\sqrt{n} }$

                           = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

⇒Population mean = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

Answer 2

Answer:

E

Step-by-step explanation:

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