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

Tyra 2021-01-10 Answered
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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Expert Answer

Alix Ortiz
Answered 2021-01-11 Author has 9558 answers

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.

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