Question

(1 point) The table below lists the weights of some colleage students in September and later in February of their freshman year. September weight 62 54 71 65 53 58 74 49 66 56 69 74 61 56 71 51 70 52 70 78 59 60 73 60 February weight 67 53 67 71 56 56 77 54 67 55 64 70 56 59 74 56 69 53 65 72 59 57 77 54 At the 0.05 significance level, test the claim of no difference between September weights and February weights. The test statistic is

The critical value is

Answer 1

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	62	67	-5
	54	53	1
	71	67	4
	65	71	-6
	53	56	-3
	58	56	2
	74	77	-3
	49	54	-5
	66	67	-1
	56	55	1
	69	64	5
	74	70	4
	61	56	5
	56	59	-3
	71	74	-3
	51	56	-5
	70	69	1
	52	53	-1
	70	65	5
	78	72	6
	59	59	0
	60	57	3
	73	77	-4
	60	54	6
Average	63	62.833	0.167
St. Dev.	8.46	8.036	3.953
n	24	24	24

For the score differences we have

(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 paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=23.

Hence, it is found that the critical value for this two-tailed test is , for α=0.05 and df = 23

The rejection region for this two-tailed test is R={t:∣t∣>2.069}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=0.207≤tc​=2.069, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.8382, and since p=0.8382≥0.05, 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 μ1​ is different than μ2​, at the 0.05 significance level.

Using excel

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