Question

For each exercise, answer the following along with any additional questions. Assume group variances are equal (unless the problem is ran via statistical software). |  Provide the null and alternative hypotheses in formal and plain language for appropriate two-tailed test (viz., dependent or independent) at the 0.05 significance level  Do the math and reject/accept null at a=.05. State your critical t value.  Explain the results in plain language.  Calculate the 95% confidence interval for the difference in means and state both formally and in plain language if appropriate.

The U.S. Department of Veterans Affairs mandates two day classes on sexual harassment for new employees. As a potential cost reduction, the personnel office is considering switching to an online course. There is concern that performance will be lower with the computer, self-paced based training. The Department conducts a trial with new employees where a random group are assigned the online system. After either the online (6 subjects) or in-class training (8 subjects), employees are assessed two weeks after training with a standardized test concerning knowledge of sexual harassment regulations and behavior with scores ranging 0-100 (100 is the best score). Data is as follows

In-class: 70, 70, 60, 60, 65, 55, 65, 80

On-line: 75, 65, 80, 90, 70, 75

Answer 1

first we calculate mean and standard deviation for both the samples:

	In class	In class2
	70	4900
	70	4900
	60	3600
	60	3600
	65	4225
	55	3025
	65	4225
	80	6400
Sum =	525	34875

	On-line	On-line2
	75	5625
	65	4225
	80	6400
	90	8100
	70	4900
	75	5625
Sum =	455	34875

The provided sample means are shown below:

Also, the provided sample standard deviations are:

and the sample sizes are and

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

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=12. 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​=2.179, for α=0.05 and df = 12.

The rejection region for this two-tailed test is R={t:∣t∣>2.179}.

(3) Test Statistics

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

t= -2.326

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=2.326>tc​=2.179, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0384, and since p=0.0384<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 population mean μ1​ is different than μ2​, at the 0.05 significance level.

95% confidence interval:

We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known. The following information has been provided about each of the samples:

Sample Mean 1	65.625
Sample Standard Deviation 1	7.763
Sample Size 1	8
Sample Mean 2	75.833
Sample Standard Deviation 2	8.612
Sample Size 2	6

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are.

The critical value for \alpha = 0.05α=0.05 and df=12 degrees of freedom is .

The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

Since we assume that the population variances are equal, the standard error is computed as follows:

Now, we finally compute the confidence interval:

​

Therefore, based on the data provided, the 95\%95% confidence interval for the difference between the population means μ1​−μ2​ is −19.772<μ1​−μ2​<−0.64

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