On a measure of well-being, the 49 children of alcoholics had a mean of \(\displaystyle{26.1}\ {\left(={7.2}\right)}\) and the 449 subjects in the control group had a mean of 28.8 \(\displaystyle{\left({5}={6.4}\right)}\). The difference scores between the matched subjects from the two groups has a mean of \(\displaystyle{2.7}\ {\left({s}={9.7}\right)}\)

a) Since the groups were matched according to age and gender. The groups to be compared are dependent samples.

b) All staps for two sided significance test are as follows:

1)Assumptions: The assumptions on which this analysis is based are:

i) Data must be quantitative.

ii) The sample of difference scores must be a random sample from a population of such difference scores.

ii) The difference scores have a population distribution that is approximately normal.

2) Hypotheses: Let \(\displaystyle\mu_{{1}}{\quad\text{and}\quad}\mu_{{2}}\) are the population means of two groups (alcoholic and non.alcoholic groups) and \(\displaystyle\mu_{{1}}\ —\ \mu_{{2}}\ =\ \mu_{{d}}\).We want to test is there any difference between these two population means. So our hypotheses are

Null: \(H_0:\ \mu_d\ =\ 0 (\text{that is}\ \mu_1\ =\ \mu_2)\)

Altemative: \(H_a:\ \mu_d\ \neq\ 0 (\text{that is}\ \mu_1\ \neq\ \mu_2)\)

3) Test Statistic: We are given that the sample mean difference, \(\displaystyle\overline{{x}}_{{d}}\ =\ {2.7}\), deviation of the difference scores, \(\displaystyle{s}_{{d}}={9.7}\). The standard error: \(\displaystyle{s}{e}={\frac{{{s}_{{d}}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={\frac{{{9.7}}}{{\sqrt{{{49}}}}}}\)

\(\displaystyle=\ {1.3857}\)

The t-statistic for the significance test of \(\displaystyle{H}_{{0}}:\ \mu_{{d}}\ =\ {0}\) against \(H_a :\ \mu_d\ \neq\ 0 is: t = \frac{\overline x_{d}}{se}\)

\(\displaystyle=\ {\frac{{{2.7}}}{{{1.3857}}}}\)

\(\displaystyle=\ {1.9485}\)

4) P-Value: The P-value is the two-tail probability rom a t distribution. From calculator, we get P-value \(\displaystyle=\ {0.057}\)

5) Conclusion: Since, P-Value \(\displaystyle=\ {0.057}\) is greater than 0.05, we accept \(\displaystyle{H}_{{0}}\). We can conclude that there is no difference between the population means at significance level of 0.05

c) We assume that the difference scores have a population distribution that is approximately normal and our sample is a random sample from this distribution.