Question

6. Two-tailed hypothesis testing - Step by step

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 Alzheimer’s. She obtains a sample of n = 20 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 = 24.4 with a standard deviation of s = 6.02. In the general population of Alzheimer’s patients, the standardized test is known to have a population mean of μ = 27.2. 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: μSAM-eSAM-e ≥ 27.2; H₁: μSAM-eSAM-e < 27.2

H₀: μSAM-eSAM-e ≤ 27.2; H₁: μSAM-eSAM-e > 27.2

H₀: μSAM-eSAM-e ≠ 27.2; H₁: μSAM-eSAM-e = 27.2

H₀: μSAM-eSAM-e = 27.2; H₁: μSAM-eSAM-e ≠ 27.2

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:

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 Alzheimer’s patients.

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

11.0003.0786.31412.70631.82163.65720.8161.8862.9204.3036.9659.92530.7651.6382.3533.1824.5415.84140.7411.5332.1322.7763.7474.60450.7271.4762.0152.5713.3654.03260.7181.4401.9432.4473.1433.70770.7111.4151.8952.3652.9983.49980.7061.3971.8602.3062.8963.35590.7031.3831.8332.2622.8213.250100.7001.3721.8122.2282.7643.169110.6971.3631.7962.2012.7183.106120.6951.3561.7822.1792.6813.055130.6941.3501.7712.1602.6503.012140.6921.3451.7612.1452.6242.977150.6911.3411.7532.1312.6022.947160.6901.3371.7462.1202.5832.921170.6891.3331.7402.1102.5672.898180.6881.3301.7342.1012.5522.878190.6881.3281.7292.0932.5392.861200.6871.3251.7252.0862.5282.845210.6861.3231.7212.0802.5182.831220.6861.3211.7172.0742.5082.819230.6851.3191.7142.0692.5002.807240.6851.3181.7112.0642.4922.797250.6841.3161.7082.0602.4852.787260.6841.3151.7062.0562.4792.779270.6841.3141.7032.0522.4732.771280.6831.3131.7012.0482.4672.763290.6831.3111.6992.0452.4622.756300.6831.3101.6972.0422.4572.750400.6811.3031.6842.0212.4232.704600.6791.2961.6712.0002.3902.6601200.6771.2891.6581.9802.3582.617∞0.6741.2821.6451.9602.3262.576

Answer 1

H₀: μSAM-e = 27.2; H₁: μSAM-e ≠ 27.2

The t Distribution:

The critical t scores (the values that define the borders of the critical region) are -2.861, 2.861.

The estimated standard error is 1.346 .

The t statistic is -2.080.

The t statistic not in the critical region. Therefore, the null hypothesis is not rejected.

Therefore, the researcher cannot conclude that SAM-e has a significant effect on the moods of Alzheimer’s patients.

27.200	hypothesized value
24.400	mean 1
6.020	std. dev.
1.346	std. error
20	n
19	df
-2.080	t
.0513	p-value (two-tailed)