Question

Please give a detailed explanation, not just the response. If possible, a way to work out the problem in Excel is helpful! Thanks in advance!

According to the Carnegie unit system, the recommended number of hours students should study per unit is 2. Are statistics students' study hours less than the recommended number of hours per unit? The data show the results of a survey of 16 statistics students who were asked how many hours per unit they studied. Assume a normal distribution for the population.

0, 3.2, 0, 3.2, 2, 1.3, 0.1, 0, 3.1, 2.1, 2.4, 1.5, 1.5, 2.4, 0.8, 2.6

What can be concluded at the αα = 0.01 level of significance?

1. For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
2. The null and alternative hypotheses would be:

H0:H0:  ? μ p  Select an answer ≠ < = >       

H1:H1:  ? p μ  Select an answer ≠ > < =    

3. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? ≤ >  αα
6. Based on this, we should Select an answer accept reject fail to reject  the null hypothesis.
7. Thus, the final conclusion is that ...
   • The data suggest the population mean is not significantly less than 2 at αα = 0.01, so there is sufficient evidence to conclude that the population mean study time per unit for statistics students is equal to 2.
   • The data suggest the populaton mean is significantly less than 2 at αα = 0.01, so there is sufficient evidence to conclude that the population mean study time per unit for statistics students is less than 2.
   • The data suggest that the population mean study time per unit for statistics students is not significantly less than 2 at αα = 0.01, so there is insufficient evidence to conclude that the population mean study time per unit for statistics students is less than 2.
8. Interpret the p-value in the context of the study.
   • If the population mean study time per unit for statistics students is 2 and if you survey another 16 statistics students, then there would be a 11.81302557% chance that the sample mean for these 16 statistics students would be less than 1.64.
   • There is a 11.81302557% chance of a Type I error.
   • If the population mean study time per unit for statistics students is 2 and if you survey another 16 statistics students, then there would be a 11.81302557% chance that the population mean study time per unit for statistics students would be less than 2
   • • There is a 11.81302557% chance that the population mean study time per unit for statistics students is less than 2.
   • Interpret the level of significance in the context of the study.
     • There is a 1% chance that the population mean study time per unit for statistics students is less than 2.
     • There is a 1% chance that students just don't study at all so there is no point to this survey.
     • If the population mean study time per unit for statistics students is less than 2 and if you survey another 16 statistics students, then there would be a 1% chance that we would end up falsely concluding that the population mean study time per unit for statistics students is equal to 2.
     • If the population mean study time per unit for statistics students is 2 and if you survey another 16 statistics students, then there would be a 1% chance that we would end up falsely concluding that the population mean study time per unit for statistics students is less than 2.

Answer 1

The sample size is n = 16.

The provided sample data along with the data required to compute the sample mean Xˉ and sample variance s^2 are shown in the table below:

	hours study	hours study2
	0	0
	3.2	10.24
	0	0
	3.2	10.24
	2	4
	1.3	1.69
	0.1	0.01
	0	0
	3.1	9.61
	2.1	4.41
	2.4	5.76
	1.5	2.25
	1.5	2.25
	2.4	5.76
	0.8	0.64
	2.6	6.76
Sum =	26.2	63.62

The sample mean Xˉ is computed as follows:

Also, the sample variance s^2 is

Therefore, the sample standard deviation s is

The provided sample mean is Xˉ=1.638

and the sample standard deviation is s = 1.175,

and the sample size is n = 16.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 22

Ha: μ < 22

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

Based on the information provided, the significance level is α=0.01,

and the critical value for a left-tailed test is

t_c = -2.602.

The rejection region for this left-tailed test is

R=t:t<−2.602

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

t = −1.232 ≥ tc​ = −2.602,

it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p = 0.1184,

and since p = 0.1184 ≥ 0.01,

it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ is less than 2, at the 0.01 significance level

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