Question

Use the following distribution of sample data

X=study hours

12

12

11

8

8

3

and calculate the mean, SS, the variance and the standard deviation.

Mean=? SS=? variance=? standard deviation=?

The above data (in question #2)represents data collected from a small random sample of students who reported how many hours they study per week.  Use the mean, standard deviation and n from the data above and alpha =.05 in a hypothesis test to determine whether students spend the recommended 12 hours per week or whether they spend less than 12 hours a week on average studying--use the space below to write the 1)  null and alternative hypothesis, 2) the critical value from the table, 3)your own calculated test statistic, and your 4) decision.

Answer 1

(A)

The sample size is n = 6. The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	Study hours	Study hours2
	12	144
	12	144
	11	121
	8	64
	8	64
	3	9
Sum =	54	546

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

(B)

The sample mean is and the sample standard deviation is s = 3.464, and the sample size is n = 6.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 12

Ha: μ < 12

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

The significance level is α = 0.05, and the critical value for a left-tailed test is tc ​= −2.015.

The rejection region for this left-tailed test is R={t : t < − 2.015}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = −2.121 < tc ​= −2.015, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0437, and since p = 0.0437 < 0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 12, at the 0.05 significance level.

Therefore, there is enough evidence to claim that the students spend less than 12 hours a week on average studying at the 0.05 significance level.