Question

A researcher is interested in determining whether difficulties experienced during giving birth have any influence on the eventual development of mother/child relationship. By examining hospital records, the researcher identifies women who had easy and difficult deliveries and gives them a questionnaire measuring the quality of relationship between mother and child. An independent samples t-test is performed and the following summary of the results is obtained:

Descriptive Statistics

Variable	Mean	SD	n
Difficult Delivery	82	9.62	5
Easy Delivery	68	8.37	5

t-test

t		df	Mean Difference	Standard error of the mean difference	Sig.	95% CI of the difference
						Lower	Upper
2.46		8	14	5.7	0.04	0.86	27.14

a. Using the degrees of freedom, what sample size was used in this test? Just report the number.

b. Reproduce the standard error of the mean difference using information from the tables. Show your work.

c. Reproduce the test statistic using information from the second table only. Show your work.

d. State your decision about the null based on the test statistic (i.e., do not use p-value or CI). Limit your answer to one full sentence.

e. State your decision about the null based on the p-value (i.e., do not use test statistic or CI). Assume the test was conducted as the alpha level of .05. Limit your answer to one full sentence.

f. State your decision about the null based on the confidence interval (i.e., do not use test statistic or p-value). Limit your answer to one full sentence.

g. Writing in full sentences, interpret the result (Step 5). Make sure to include the CI in your interpretation.   

h. Would your decision change if the sample size was n=6 in each group? Explain how you would know this without doing any calculations

i. If an ANOVA test were conducted instead, what would be the F-ratio? Round to 2 decimals.

Answer 1

a.

Sample size, n= df + 2 = 8+2 = 10

b.

Standard error of the mean difference = $\dpi{100} \small \sqrt{(\sigma1^2/n1)+(\sigma2^2/n2)}$ =$\dpi{100} \small \sqrt{(9.62^2/5)+(8.37^2/5)}$ =5.7

c.

The test statistic is: t = mean difference/Standard error of the mean difference = 14/5.7 = 2.46

d.

Decision about the null based on the test statistic:

Since the test statistic of 2.46 > t_critical of 2.31, we reject the null hypothesis at 0.05 significance level.

e.

Decision about the null based on the p-value:

p-value for the test statistic is =0.04

Since p-value of 0.04 < significance level of 0.05, we reject the null hypothesis at 0.05 significance level.

f.

Decision about the null based on the confidence interval:

Since the 95% CI of the difference of means: [0.86,27.14] does not contain 0, we reject the null hypothesis at 0.05 significance level.

g.

Interpretation:

Since the null hypothesis is rejected at 0.05 significance level, there is a sufficient evidence to claim that the difference in means is significant.

The interval [0.86,27.14] contains the true difference between means 95% of the time.

Hence, difficulties experienced during giving birth have innfluence on the eventual development of mother/child relationship.

h.

The decision would not change if the sample size was n=6 in each group

p-value will decrease when the sample size increases, i.e.,as the degrees of freedom, df increases. So, if n=6 in each group instead of 5, p-value will be lower than the previous one and so, less than 0.05 significance level since previous p-value is also less than 0.05.

(because - as the df decreases, the t-distribution has thicker tails which means p-value increases and so, if df increases then the p-value will decrease. So, df = 10 means lower p-value than that of df=8)

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i.

F = t² = (2.46)² = 6.05

Thus, F-ratio would be 6.05