# Effect of alcoholic parents A study1 compared personality characteristics between 49 children of alcoholics and a control group of 49 children of nona

Effect of alcoholic parents A study1 compared personality characteristics between 49 children of alcoholics and a control group of 49 children of nonalcoholics who were matched on age and gender. On a measure of well being, the 49 children of alcoholics had a mean of $$\displaystyle{26.1}{\left({s}={7.2}\right)}$$ and the 49 subjects in the control group had a mean of $$\displaystyle{\left[{28.8}{\left({s}={6.4}\right)}\right]}$$. The difference scores between the matched subjects from the two groups had a mean of $$\displaystyle{2.7}{\left({s}={9.7}\right)}$$.
a. Are the groups to be compared independent samples or depentend samples? Why?
b. Show all speps of a test equality of the two population means for a two-sided alternative hypotesis. Report the P-value and interpret.
c. What assumptions must you make the inference in part b to be valid?

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Alix Ortiz

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.