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

Tyra

Tyra

Answered question

2021-01-10

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 26.1(s=7.2) and the 49 subjects in the control group had a mean of [28.8(s=6.4)]. The difference scores between the matched subjects from the two groups had a mean of 2.7(s=9.7).
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?

Answer & Explanation

Alix Ortiz

Alix Ortiz

Skilled2021-01-11Added 109 answers

On a measure of well-being, the 49 children of alcoholics had a mean of 26.1 (=7.2) and the 449 subjects in the control group had a mean of 28.8 (5=6.4). The difference scores between the matched subjects from the two groups has a mean of 2.7 (s=9.7)
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 μ1andμ2 are the population means of two groups (alcoholic and non.alcoholic groups) and μ1  μ2 = μd.We want to test is there any difference between these two population means. So our hypotheses are
Null: H0: μd = 0(that is μ1 = μ2)
Altemative: Ha: μd  0(that is μ1  μ2)
3) Test Statistic: We are given that the sample mean difference, xd = 2.7, deviation of the difference scores, sd=9.7. The standard error: se=sdn
=9.749
= 1.3857
The t-statistic for the significance test of H0: μd = 0 against Ha: μd  0is:t=xdse
= 2.71.3857
= 1.9485
4) P-Value: The P-value is the two-tail probability rom a t distribution. From calculator, we get P-value = 0.057
5) Conclusion: Since, P-Value = 0.057 is greater than 0.05, we accept H0. 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.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?