Question

Two sample hypothesis testing: use the data analysis add in to test the claim that the average scores for all statistics students taking test 1 and 2 are the same using a 5% level of significance. Fill each of the cells with required information.

test 1 Scores	test 2 Scores
91	85
83	76
75	68
85	52
90	99
80	64
67	96
82	92
88	81
87	45
95	74
91	62
73	83
80	79
83	87
92	91
94	90
68	48
75	88
91	70
79	58
95	88
87	79
76	93
91	89
85	90
59	86
70	75
69	77
78	94

Two Sample Hypothesis Test			Identify the claim below
Hypothesis Statements	Ho				Insert Data Analysis test here
	Ha
Standardized Test Statistic
P-value
Reject or Fail to Reject H0?
Conclusion about the claim

Answer 1

First we calculate mean and standard deviation of both the samples:

	test 1 scores	test 1 scores2
	91	8281
	83	6889
	75	5625
	85	7225
	90	8100
	80	6400
	67	4489
	82	6724
	88	7744
	87	7569
	95	9025
	91	8281
	73	5329
	80	6400
	83	6889
	92	8464
	94	8836
	68	4624
	75	5625
	91	8281
	79	6241
	95	9025
	87	7569
	76	5776
	91	8281
	85	7225
	59	3481
	70	4900
	69	4761
	78	6084
Sum =	2459	204143

The sample mean Xˉ is computed as follows:

Also, the sample variance is

Therefore, the sample sandartd deviation is

	test 2 scores	test 2 scores2
	85	7225
	76	5776
	68	4624
	52	2704
	99	9801
	64	4096
	96	9216
	92	8464
	81	6561
	45	2025
	74	5476
	62	3844
	83	6889
	79	6241
	87	7569
	91	8281
	90	8100
	48	2304
	88	7744
	70	4900
	58	3364
	88	7744
	79	6241
	93	8649
	89	7921
	90	8100
	86	7396
	75	5625
	77	5929
	94	8836
Sum =	2359	191645

The sample mean Xˉ is computed as follows:

Also, the sample variance is

Therefore, the sample sandartd deviation is

Now we perform the required test:

The provided sample means are shown below:

Xˉ1​=81.967

Xˉ2​=78.633

Also, the provided sample standard deviations are:

s1​=9.445

s2​=14.561

and the sample sizes are

n1​=30 and

n2​=30.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​ or scores of both the tests are same.

Ha:μ1​ ≠μ2​ or scores of both the tests are different.

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.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=58. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.002, for α=0.05 and df=58.

The rejection region for this two-tailed test is R={t:∣t∣>2.002}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=1.052≤tc​=2.002, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.2971, and since p=0.2971≥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 or scores of both the tests are same.

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