Question

Test whether there is a difference in the means of the two lists of numbers. Pay attention to the variances. Answer the following questions. Use alpha = 0.05.

Set 1: 44, 11, 27, 30, 27, 30, 17, 12 , 17, 9, 14, 24, 13, 9, 13, 15, 13, 15, 15, 23, 53, 33, 10, 6, 17, 14, 5, 25, 18, 23, 13, 12, 8, 17, 12, 36, 10, 4, 6, 32

Set 2: 20, 12, 23, 12, 25, 11, 21, 13, 25, 24, 36, 22, 2, 24, 15, 12, 22, 24, 17, 8, 13, 21, 18, 24, 27, 14, 12, 18, 20, 8, 10, 4, 14, 7, 26, 27, 12, 22, 10, 10

Is this a one or two-tailed test?

T=?

P=?

Decision (fail to reject Ho or Reject)?

Thank you so much for helping!

Answer 1

The data provided is:

	Set 1	Set 2
	44	20
	11	12
	27	23
	30	12
	27	25
	30	11
	17	21
	12	13
	17	25
	9	24
	14	36
	24	22
	13	2
	9	24
	13	15
	15	12
	13	22
	15	24
	15	17
	23	8
	53	13
	33	21
	10	18
	6	24
	17	27
	14	14
	5	12
	25	18
	18	20
	23	8
	13	10
	12	4
	8	14
	17	7
	12	26
	36	27
	10	12
	4	22
	6	10
	32	10
Count	40	40
Average	18.3	17.125
St. Dev	10.84908	7.418575
Variance	117.7026	55.03526

Testing for equality of variance

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: ​

Ha: ​

This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used.

Rejection Region

Based on the information provided, the significance level is α=0.05, and the rejection region for this two-tailed test is R={F:F<0.529 or F>1.891}.

Test Statistics

The F-statistic is computed as follows:

(4) The decision about the null hypothesis

Since from the sample information we get that F=2.139>FU​=1.891, it is then concluded that the null hypothesis is rejected. Hence variances are not equal i.e. there is enough evidence to claim that the population variance ​ is different than the population variance ​, at the α=0.05 significance level.

T-test for two Means – Unknown Population Standard Deviations

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The p-value is

Since it is observed that ∣t∣=0.565≤tc​=1.995, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.5736, and since p=0.5736≥0.05, it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

Graphically

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