Question

2. For the following experiment/question, pick the most appropriate statistical test. You have the following statistical tests as choices: some may be used more than once, others not at all. Assume homogeneity of variance (where applicable) and the validity of parametric tests (where applicable), unless something is directly stated (e.g., “the data are not at all normal”) or otherwise indicated (viz., by the inspection of the data) which would indicate a strong and obvious violation of an assumption. This means you must inspect the data for violations of all assumptions. Unless it is stated that the population parameter is known, assume it isn’t. Please do not concern yourself with any intervening variable that you may perceive. Finally, please don’t think about what the data would look like in reality. Assume that the question represents reality. Please simply write the letter for the test as your answer. [You can, but need not, add an explanation.] If none of the tests are appropriate, the answer is P) none of the above. Here are the tests: A: one sample z-test B: one-sample t-test C: t-test for the difference between means for two related samples D: t-test for the difference between means for two independent samples with homogeneity of variance E: t-test for the difference between means for two independent samples with heterogeneity of variance F: a one sample z-test for proportions (or a chi-square goodness of fit) G: chi-square goodness of fit only (where a one sample z-test of proportions isn’t appropriate) H: a two-sample z-test for the difference between proportions (or a chi-square test of independence) I: chi-square test of independence only (where a two-sample z-test for the difference between proportions isn’t appropriate.) J: simple regression K: multiple regression L: one-way ANOVA M: two-way ANOVA N: Mann-Whitney O: Wilcoxon P: none of the above. Does watching Barney increase altruism in children? Amanda put nine kids in a room where they watched Barney and another nine kids is a quiet room without television. Then the kids filled out a questionnaire that determined their altruism score (which we can assume to be interval). The scores are: Barney: 43, 55, 63, 57, 77, 65, 84, 73, 66 Quiet: 67, 55, 74, 63, 56, 33, 45, 53, 47

Answer 1

Tests:

A: one sample z-test

B: one-sample t-test

C: t-test for the difference between means for two related samples

D: t-test for the difference between means for two independent samples with homogeneity of variance

E: t-test for the difference between means for two independent samples with heterogeneity of variance

F: a one sample z-test for proportions (or a chi-square goodness of fit)

G: chi-square goodness of fit only (where a one sample z-test of proportions isn’t appropriate)

H: a two-sample z-test for the difference between proportions (or a chi-square test of independence)

I: chi-square test of independence only (where a two-sample z-test for the difference between proportions isn’t appropriate.)

J: simple regression K: multiple regression

L: one-way ANOVA

M: two-way ANOVA

N: Mann-Whitney

O: Wilcoxon

P: none of the above.

Problem:

Amanda put nine kids in a room where they watched Barney and another nine kids is a quiet room without television. Then the kids filled out a questionnaire that determined their altruism score (which we can assume to be interval).

We have to see if watching Barney increase altruism in children.

We are also given the scores of two scenario.

The scores are:

Barney: 43, 55, 63, 57, 77, 65, 84, 73, 66

Quiet: 67, 55, 74, 63, 56, 33, 45, 53, 47

Sample size = 9 < 30

We will be doing comparison between two indepedent groups using mean score.

Sample of Children represent the behavior of all children and therefore can be assumed to be normally distributed

As, children as from same population they can be said of having equal variances in scores.

So,given all these conditions we can use t-test for independent samples with equal variances.

D: t-test for the difference between means for two independent samples with homogeneity of variance

Test:

	Barney	Quiet
	43	67
	55	55
	63	74
	57	63
	77	56
	65	33
	84	45
	73	53
	66	47
Count	9	9
Mean	64.8	54.8
Sample Stdev	12.4	12.4

From calculations also we can see they have equal variances.

The provided sample means are shown below:

Xˉ1​=64.8 and Xˉ2​=54.8

Also, the provided sample standard deviations are:

s1​=12.4, s2​=12.4

and the sample sizes are n1​=9 and n2​=9.

(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 right-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=16. 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 right-tailed test is tc​=1.746, for α=0.05 and df=16.

The rejection region for this right-tailed test is R={t:t>1.746}.

(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=1.711≤tc​=1.746, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0532, and since p=0.0532≥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 greater than μ2​, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is −2.392<μ1​−μ2​<22.392.

Graphically

For your understanding I am also posting the test,

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