Question

7) A study examined language acquisition in learning-impaired children. The reseateno designed two computer modules that taught the same content but instructional approaches. The study refers to the two computer module used different s as Module 1" and Module 2. " Children at each module, students took a quiz that asked the children to identify 50 of correct identifications by the 22 children in the Module identifications by the 22 children in the Module 2 group were summarized in the table below: en were assigned at random to either Module 1 or Module 2. At the end of words. The numbers of group wwer group and trds. The recorded. Key data are 1 group and the numbers of correct were recpnd Number of correct identifications 23.05 19.74 Module 1 4.25 3.36 Module 2 The study wants to conduct a hypothesis test to determine whether there is a difference between the mean number of correct identifications for students using Module 1 and the mean number of correct identifications for students using Module 2. They do not have arny information suggesting that either module should result in a higher (or lower) number of correct identifications State which hypothesis test would be appropriate in this situation. The formula sheet for hypothesis testing includes all the hypothesis tests we will cover in STAT100. Provide the a) name of the hypothesis test as listed in the first column of the formula sheet State the null and alternative hypotheses for this hypothesis test. Be sure to use appr notation and symbols. b) opriate Is this a two-sided test or a one-sided test? Explain why. c) You can assume that the conditions for this test have been met. Calculate the test statistic for this test. Use the appropriate formula from the formula sheet for hypothesis testing Show all work and round any values to three decimal places. d) Draw a diagram showing where the test statistic you calculated in part d) would be located in a distribution assuming that the null hypothesis is true. Shade in the region under the curve which represents the p-value for this test e) f) Use the appropriate distribution (either the z-distribution (Standard Normal) or the t- distribution) and determine the p-value for this test. Explain how you determined the p-valu Draw a conclusion for this test and explain why you arrived at your conclusion. 9)

Answer 1

This question is related to ascertaining if there exists difference between two samples /groups on the basis of the hypothesis test of significance.

Now before getting into the details of the solution, let us spend some time exploring the nature of the samples.

There are two samples of size 22 each and both of them comprise of elements (kids) chosen at random.

Hence we can say that the samples are random and independent of each other .

Now the size of the samples is 20 ( <30).

this is less than the normal permissible sample size . Hence we shall take help of t test for hypothesis validation since t test is applicable for small sample sizes.

Hence for the given problem we shall use the hypothesis testing via t test for two independent samples.

Now if we look at the problem it wants to validate if there exists a difference between the two sample means.

So we shall design our hypothesis model accordingly assuming that for null hypothesis there is no difference in the sample means and for alternate hypothesis that the two samples are not equal to each other. (i.e, have difference,hence)

We shall use the 95% level for the test of confidence level.

with these understanding and data .lets start

The provided sample means are shown below: 23.05 X2 19.74 Also, the provided sample standard deviations are: 14.25 82-3.36 and the sample sizes are n22 and n222. The following null and alternative hypotheses need to be tested: на: 1 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. A F-test is used to test for the equality of variances. The following F-ratio is obtained: 4. AXǐz: 1.6 F-Si_ 4.252 F-362 it.uP The critical values are Fi0.415 and Fu 2.409, and since F1.6, then the null hypothesis of equal variances is not rejected.

Re Based on the information provided, the significance level is α-0.05, and the degrees of freedom are df-42. 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-2.018, for α-0.05 and df 42. The rejection region for this two-tailed test is R-It: t > 2.018 Since it is assumed that the population variances are equal, the t-statistic is computed as follows 23.05 19.74 0.748 22-1)4.25+(22-1)3.3621 1 222:2 22+22-2 4 Since it is observed that 비ー0.748-t-2.018, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p 0.4586, and since p0.45860.05, 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 μι is different than μ2, at the 0.05 significance level. Confidence Interval The 95% confidence interval is-5.62 < μι-μ2 < 12.24. Graphically T-Test t = 0.748, p = 0.4586 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 4.0 -3.5 -3.0 -2.5 2.0 1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

WE conclude that the the mean sample of the two cases is not different significantly at 95% confidence .