Question

MUST USE EXCEL

A recent study asserts that, over the past five decades, the number of hours that the average college student studies each week has been steadily dropping. In 1961, students invested 24 hours per week in their academic pursuits, whereas today's students study an average of 14 hours per week (The Boston Globe, July 4, 2010). Your aunt, Susan, happens to be the dean of a large university in California. She wonders if the study trend is reflective of students at her university. She knows that you are taking Statistics this summer, so she turns to you for help. Based on a random sample of 35 students collected by Susan, she finds out that the students surveyed study an average of 16.3714 hours per week with a sample standard deviation of 7.2155 hours. Using these results to help Susan answer the following questions. Write the complete hypothesis testing process. 1. Determine if the mean study time of students at her university is below the 1961 national average of 24 hours per week. 2. Determine if the mean study time of students at her university differs from today's national average of 14 hours per week. 3. Based on questions 1 and 2, what can you conclude about students' average study time per week at Susan's university? Does it surprise you? Why or why not?

Answer 1

Here we are given that

n= 35, T = 16.3714 and s= = 7.2155

1. We need to check whether the mean study time of students is less than the national average (24 hours/week).

Null and alternate hypothesis:

$H_{0}:\mu=24$

H, :μ< 24

Level of significance: $\alpha =0.05$

Test Statistic: $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ follow a t-distribution with n-1 df

Critical region: $R=\left \{ t:t<-1.6909 \right \}$

$t=\frac{16.3714-24}{7.2155/\sqrt{35}}$

$t=\frac{-7.6286}{7.2155/5.9161}$

$t=\frac{-45.1314}{7.2155}$

$t=-6.2548$

Since the calculated value of t falls in rejection region, we reject the null hypothesis. Hence we conclude that there is enough evidence that the mean study time of students at her University is below the 1961 national average.

2. We need to check whether the mean study time of students is less than the today's national average (14 hours/week).

$H_{0}:\mu=14$

$H_{1}:\mu<14$

Level of significance: $\alpha =0.05$

Test Statistic: $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ follow a t-distribution with n-1 df

Critical region: $R=\left \{ t:t<-1.6909 \right \}$

$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$

$t=\frac{16.3714-14}{7.2155/\sqrt{35}}$

$t=\frac{2.3714}{7.2155/5.9161}$

$t=\frac{14.0294}{7.2155}=1.9443$

Since the calculated value of t doesn't fall in the rejection region, we fail to reject the null hypothesis.

Hence, we conclude that there is no evidence to claim that the mean study time of students at her University is below the today's average(14 hours)

3. Based on the above answers, we can say that the students at Susan's University study on an average more than today's average but less than the 1961 national average (24 hours)..