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 = 30 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 = 26.2 with a standard deviation of s = 2.97. In the general population of Alzheimer’s patients, the standardized test is known to have a population mean of μ = 28.1. 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 ≥ 28.1; H₁: MSAM-eSAM-e < 28.1 H₀: MSAM-eSAM-e ≥ 28.1; H₁: MSAM-eSAM-e > 28.1 H₀: μSAM-eSAM-e ≥ 28.1; H₁: μSAM-eSAM-e < 28.1 H₀: μSAM-eSAM-e = 28.1; H₁: μSAM-eSAM-e ≠ 28.1 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. 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 Alzheimer’s patients.

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 = 30 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 = 26.2 with a standard deviation of s = 2.97. In the general population of Alzheimer’s patients, the standardized test is known to have a population mean of μ = 28.1. 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 ≥ 28.1; H₁: MSAM-eSAM-e < 28.1 H₀: MSAM-eSAM-e ≥ 28.1; H₁: MSAM-eSAM-e > 28.1 H₀: μSAM-eSAM-e ≥ 28.1; H₁: μSAM-eSAM-e < 28.1 H₀: μSAM-eSAM-e = 28.1; H₁: μSAM-eSAM-e ≠ 28.1 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. 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 Alzheimer’s patients.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

The scores from the sample produced a mean of M = 26.2 with a standard deviation of s = 2.97. In the general population of Alzheimer’s patients, the standardized test is known to have a population mean of μ = 28.1. 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 ≥ 28.1; H₁: MSAM-eSAM-e < 28.1

H₀: MSAM-eSAM-e ≥ 28.1; H₁: MSAM-eSAM-e > 28.1

H₀: μSAM-eSAM-e ≥ 28.1; H₁: μSAM-eSAM-e < 28.1

H₀: μSAM-eSAM-e = 28.1; H₁: μSAM-eSAM-e ≠ 28.1

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.

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 Alzheimer’s patients.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

Given information:

Sample size $(n) = 30$

Sample mean $(M) = 26.2$

Sample standard deviation $(s) = 2.97$

Population mean $(μ) = 28.1$

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