Question

The dean at a large university in California wonders if students study less than the 1961 national average of 24 hours per week. She randomly selects 25 students and asks their average study time per week (in hours). From their responses, she calculates a sample mean of 19 hours and a sample standard deviation of 10 hours. Assume the study times are normally distributed.
1. Specify the null and alternative hypotheses.
Select one:
a. H(0): μ≤24 Versus H(a): μ>24
b. H(0): μ≥24 Versus H(a): μ<24
2. The standard error (SE) of X¯X¯ is
Select one:
a. 2
b. 0.4
c. 10
d. 3.8
3. The test statistics value is
Select one:
a. -1.25
b. -1.67
c. -2.5
d. -2
4. The p-value is
Select one:
a. 0.11
b. 0.05
c. 0.01
d. 0.03
5. At α=0.05 and based on p-value
Select one:
a. We reject H(0) in favor of H(a)
b. We do not reject H(0)

Answer 1

Answer:

1. H(0) : μ≥24 Versus H(a) : μ<24

2. The standard error (SE) of $\bar X$ is 2.

3. The test statistics value is -2.5

4. The p-value is 0.01.

5. At α=0.05 and based on p-value ; We reject H(0) in favor of H(a).

Step-by-step explanation:

We are given that the dean at a large university in California wonders if students study less than the 1961 national average of 24 hours per week.

She randomly selects 25 students and asks their average study time per week (in hours). From their responses, she calculates a sample mean of 19 hours and a sample standard deviation of 10 hours.

Let $\mu$ = true mean hours students study at a large university in California.

1. Null Hypothesis, $H_0$ : $\mu \geq$ 24 hours   {means that the students study more than or equal to the 1961 national average of 24 hours per week}

Alternate Hypothesis, $H_A$ : $\mu$ < 24 hours   {means that the students study less than the 1961 national average of 24 hours per week}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean hours = 19 hours

             s = sample standard deviation = 10 hours

             n = sample of students = 25

2. So, Standard error (SE) of $\bar X$ is given by =  $\frac{s}{\sqrt{n} }$  = $\frac{10}{\sqrt{25} }$  = 2.

3. The test statistics  =   $\frac{19-24}{\frac{10}{\sqrt{25} } }$  ~ $t_2_4$

                                    =  -2.50

Hence, the value of test statistics is -2.50.

4. Now, P-value of the test statistics is given by;

      P-value = P( $t_2_4$ < -2.50) = 0.0098 or 0.01

• If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

• If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

5. Now, here the P-value is 0.01 or 1% which is clearly smaller than the level of significance of 0.05, so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students study less than the 1961 national average of 24 hours per week.