Question

Dean Halverson recently read that full-time college students study 20 hours each week. She decides to do a study at her university to see if there is evidence that students study an average of less than 20 hours each week. A random sample of 35 students were asked to keep a diary of their activities over a period of several weeks. It was found that he average number of hours that the 35 students studied each week was 18.5 hours. The sample standard deviation of 4.3 hours.
Find the p-value.
PS: If you know how to do this in excel, please explain how to do so but if you don't I would still greatly appreciate the answer you provide.

Answer 1

Answer: The p-value is 0.154.

Step-by-step explanation:

Since we have given that

We claim that

Null hypothesis :

$H_0:\mu=20$

Alternate hypothesis :

$H_1:\mu<20$

Population mean = 20 hours

Sample mean = 18.5 hours

Sample standard deviation = 4.3 hours

Sample size n = 35

So, test statistic would be

$z=\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{18.5-20}{\dfrac{4.3}{\sqrt{35}}}\\\\z=\dfrac{-1.5}{0.726}\\\\z=-2.066$

So, the  p value would be 0.154.

Hence, the p-value is 0.154.

Answer 2

Based on the sample size, the sample mean, and the sample standard deviation, the p-value would be 0.0467.

What is the p-value?

First, find the test statistic:

= (Sample mean hours - Null hypothesis) / (Standard deviation / √Sample)

= (18.5 - 20) / (4.3 / √35)

= -2.0637

Using a df = n - 1

= 35 - 1  

= 34

And a 2 tail test, we find:

P (t = - 2.0637) x 2

= 2 x 0.02335

= 0.0467

Find out more on the p-value at https://brainly.com/question/4621112.