Question

10. The t test for two independent samples - One-tailed example using tables

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 the National Center for Health Statistics.]

Suppose a clinical 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 traditional and nontraditional couples. Higher scores on the Marital Satisfaction Inventory indicate greater satisfaction. There is one score per couple. Assume that these scores are 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 relationship satisfaction than heterosexual couples. He identifies the null and alternative hypotheses as:

H₀: μhomosexual couples

   μheterosexual couples

H₁: μhomosexual couples

   μheterosexual couples

This is a    tailed test.

The psychologist collects the data. A group of 31 homosexual couples scored an average of 21.7 with a sample standard deviation of 9 on the Marital Satisfaction Inventory. A group of 30 heterosexual couples scored an average of 25.5 with a sample standard deviation of 12. Use the t distribution table. To use the table, you will first need to calculate the degrees of freedom.

The degrees of freedom are   .

The t distribution

	Proportion in One Tail
	0.25	0.10	0.05	0.025	0.01	0.005
	Proportion in Two Tails Combined
df	0.50	0.20	0.10	0.05	0.02	0.01
1	1.000	3.078	6.314	12.706	31.821	63.657
2	0.816	1.886	2.920	4.303	6.965	9.925
3	0.765	1.638	2.353	3.182	4.541	5.841
4	0.741	1.533	2.132	2.776	3.747	4.604
5	0.727	1.476	2.015	2.571	3.365	4.032
6	0.718	1.440	1.943	2.447	3.143	3.707
7	0.711	1.415	1.895	2.365	2.998	3.499
8	0.706	1.397	1.860	2.306	2.896	3.355
9	0.703	1.383	1.833	2.262	2.821	3.250
10	0.700	1.372	1.812	2.228	2.764	3.169
11	0.697	1.363	1.796	2.201	2.718	3.106
12	0.695	1.356	1.782	2.179	2.681	3.055
13	0.694	1.350	1.771	2.160	2.650	3.012
14	0.692	1.345	1.761	2.145	2.624	2.977
15	0.691	1.341	1.753	2.131	2.602	2.947
16	0.690	1.337	1.746	2.120	2.583	2.921
17	0.689	1.333	1.740	2.110	2.567	2.898
18	0.688	1.330	1.734	2.101	2.552	2.878
19	0.688	1.328	1.729	2.093	2.539	2.861
20	0.687	1.325	1.725	2.086	2.528	2.845
21	0.686	1.323	1.721	2.080	2.518	2.831
22	0.686	1.321	1.717	2.074	2.508	2.819
23	0.685	1.319	1.714	2.069	2.500	2.807
24	0.685	1.318	1.711	2.064	2.492	2.797
25	0.684	1.316	1.708	2.060	2.485	2.787
26	0.684	1.315	1.706	2.056	2.479	2.779
27	0.684	1.314	1.703	2.052	2.473	2.771
28	0.683	1.313	1.701	2.048	2.467	2.763
29	0.683	1.311	1.699	2.045	2.462	2.756
30	0.683	1.310	1.697	2.042	2.457	2.750
40	0.681	1.303	1.684	2.021	2.423	2.704
60	0.679	1.296	1.671	2.000	2.390	2.660
120	0.677	1.289	1.658	1.980	2.358	2.617
∞	0.674	1.282	1.645	1.960	2.326	2.576
	0.50	0.20	0.10	0.05	0.02	0.01

With α = 0.05, the critical t-score (the value for a t-score that separates the tail from the main body of the distribution, forming the critical region) is   . (Note: If your df value is not included in this table, look up the critical values for both surrounding df values and select the larger t value to use as your critical t-score. If you fail to reject the null hypothesis, you can later check the smaller t value to decide whether to interpolate. However, for the purposes of this problem, you can just assume that if your t statistic is not more extreme than the larger t value, you will not reject the null hypothesis. Also, the table includes only positive t values. Since the t distribution is symmetrical, for a one-tailed test where the alternative hypothesis is less than, simply negate the t value provided in the table.)

To calculate the t statistic, you first need to calculate the estimated standard error of the difference in means. To calculate this estimated standard error, you first need to calculate the pooled variance. The pooled variance is   . The estimated standard error of the difference in means is   . (Hint: For the most precise results, retain four decimal places from your calculation of the pooled variance to calculate the standard error.)

Calculate the t statistic. The t statistic is   . (Hint: For the most precise results, retain four decimal places from your previous calculation to calculate the t statistic.)

The t statistic   lie in the critical region for a one-tailed hypothesis test. Therefore, the null hypothesis is   . The psychologist   conclude that homosexual couples have less relationship satisfaction than heterosexual couples.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Homosexual = u_Heterosexual
Alternative hypothesis: u_Homosexual < u_Heterosexual

This 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).

DF = n₁ + n₂ - 2

DF = 31 + 30 -2

DF = 59

t_critical = 2.000

Rejection region is t > 2.00

With α = 0.05, the critical t-score is 2.000.

The pooled variance is 114.407.

The estimated standard error of the difference in means is 2.7394.
t = [ (x₁ - x₂) - d ] / SE

t = - 1.387

The t statistic is - 1.387

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 population means, and SE is the standard error.

The t statistic (1.387) does not lie in the critical region for a one-tailed hypothesis test. Therefore, the null hypothesis is not rejected. The psychologist cannot conclude that homosexual couples have less relationship satisfaction than heterosexual couples.