Question

S-adenosyl methionine (SAM-e) is a naturally occurring compound in human cells that is thought to have an effect on depression symptoms. Suppose that a researcher is interested in testing SAM-e on patients who are struggling with cancer. She obtains a sample of n = 25 patients and asks each person to take the suggested dosage each day for 4 weeks. At the end of the 4-week period, each individual takes the Beck Depression Inventory (BDI), which is a 21-item, multiple-choice self-report inventory for measuring the severity of depression.

The scores from the sample produced a mean of M = 20.8 with a standard deviation of s = 5.18. In the general population of cancer patients, the standardized test is known to have a population mean of μ = 23.2. Because there are no previous studies using SAM-e with this population, the researcher doesn’t know how it will affect these patients; therefore, she uses a two-tailed single-sample t test to test the hypothesis.

From the following, select the correct null and alternative hypotheses for this study:

H₀: MSAM-eSAM-e ≥ 23.2; H₁: MSAM-eSAM-e < 23.2

H₀: μSAM-eSAM-e ≤ 23.2; H₁: μSAM-eSAM-e > 23.2

H₀: MSAM-eSAM-e ≥ 23.2; H₁: MSAM-eSAM-e > 23.2

H₀: μSAM-eSAM-e = 23.2; H₁: μSAM-eSAM-e ≠ 23.2

Assume that the depression scores among patients taking SAM-e are normally distributed. You will first need to determine the degrees of freedom. There are _______________ degrees of freedom.

Use the t distribution table to find the critical region for α = 0.01.

The t Distribution:

df	Proportion in One Tail
	0.25	0.10	0.05	0.025	0.01	0.005
	Proportion in Two Tails Combined
	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

The critical t scores (the values that define the borders of the critical region) are_________ .

The estimated standard error is______- .

The t statistic is________ .

The t statistic______ in the critical region. Therefore, the null hypothesis___________ rejected.

Therefore, the researcher   conclude that SAM-e has a significant effect on the moods of cancer patients.

Answer 1

Correct null and alternative hypothesis: H₀: μSAM-eSAM-e = 23.2; H₁: μSAM-eSAM-e ≠ 23.2.
Degrees of freedom = = 25 - 1 = 24.

The t statistic is not in the critical region. Therefore, the null hypothesis is not rejected. Therefore, the researcher   conclude that SAM-e has a significant effect on the moods of cancer patients.