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on ses aplia.com/af/servlet/quiz?cto thomas.myers1-0003 &quiz; action takeQuizBquiz probGuid ONAPCOA8010100000036c7fa30080000E e 6. Thettest for two independent samples One-tailed example using tables Aa Aa Most engaged couples expect or at least hope that they will have high levels of marital satisfaction. However, because 54% of first marriages end in divorce, social scientists have begun investigating influences on marital satisfaction. [Data source: This data was obtained from National Center for Health Statistics.) suppose a counseling psychologist sets out to look at the role of sexual orientation in relationship longevity. He decides to measure marital satisfaction in a group of homosexual couples and a group of heterosexual couples. He chooses the Marital satisfaction Inventory, because it refers to partner and relationship rather than "spouse and "marriage," which makes it useful for research with both and nontraditional couples. the Marital satisfaction Inventory indicate greater There is one score per couple Assume that these scores are satisfaction. normally distributed and that the variances of the scores are the same among homosexual couples as among heterosexual couples. The psychologist thinks that homosexual couples will have less reationship satisfaction than heterosexual couples. He identifies the null and alternative hypotheses as This is a The psychologist collects the data. A group of 31 homosexual couples scored an average of 21 with a sample isfaction Inv stan da evation of e here to searchExplanation / Answer
Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 < 2
Alternative hypothesis: 1 > 2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s1^2/n1) + (s2^2/n2)]
SE = sqrt[(9^2/31) + (12^2/30] = 2.72266
DF = (s1^2/n1 + s2^2/n2)^2 / { [ (s1^2 / n1)^2 / (n1 - 1) ] + [ (s2^2 / n2)^2 / (n2 - 1) ] }
DF = (9^2/31 + 12^2/30)^2 / { [ (12^2 / 31)^2 / (30) ] + [ (12^2 / 30)^2 / (29) ] }
DF = 54.95 / { 0.71925 + 0.79448 } = 54.9 / 1.51373 = 36.268
t = [ (x1 - x2) - d ] / SE = [ (21.7 - 25.5) - 0 ] / 2.72266 = -1.3956939
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is
The P-Value is 0.08575.
The result is not significant at p < 0.05.
Interpret results. Since the P-value (0.08575) is more than the significance level (0.05), we can accept the null hypothesis.
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