Is the following paired differences (matched sample) hypothesis test and results correct? Please explain which steps or results are incorrect. Milk production promotes temporary loss of bone mass to provide adequate amounts of calcium. Data on total body bone mineral content (TBBMC) (g) for a sample both during lactation (L) and in postweaning period (P). (Let = true mean difference in TBBMC, postweaning minus lactation.) Assume normal distribution. Significance level .05 Subject 1 2 3 4 5 6 7 8 9 10 L 1928 2549 2825 1924 1628 2175 2114 2621 1843 2541 P 2126 2885 2895 1942 1750 2184 2164 2626 2006 2627 Does data suggest the true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g? I made a chart: Column 1 (subjects, numbered 1-10), Column 2: pop means for postweaning, Column 3: pop means for lactation, Column 4: difference (postweaning - lactation), Column 5: (diff-pop mean of differences)squared. State hypothesis: Ho: mean of differences = 0 Ha: mean of differences < 0 1st formula I used was mean of pop differences = sum of differences/sample size = 105.7 2nd formula I used was standard deviation of differences = 34.6150 Test statistic result = 7.3724 The p-value with a t-stat of 7.3724, and df = 9, is smaller than alpha (.05). In fact, lower than .005. (I do not know exact p-value, just looked at table.) Since p-value lower than alpha, we reject Ho and accept Ha with 95% confidence. The evidence shows that average post-weaning bone mineral contact exceeds that during lactation by more than 25g.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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Is the following paired differences (matched sample) hypothesis test and results correct? Please explain which steps or results are incorrect.

Milk production promotes temporary loss of bone mass to provide adequate amounts of calcium. Data on total body bone mineral content (TBBMC) (g) for a sample both during lactation (L) and in postweaning period (P). (Let = true mean difference in TBBMC, postweaning minus lactation.) Assume normal distribution. Significance level .05

Subject
	1	2	3	4	5	6	7	8	9	10
L	1928	2549	2825	1924	1628	2175	2114	2621	1843	2541
P	2126	2885	2895	1942	1750	2184	2164	2626	2006	2627

Does data suggest the true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g?

I made a chart: Column 1 (subjects, numbered 1-10), Column 2: pop means for postweaning, Column 3: pop means for lactation, Column 4: difference (postweaning - lactation), Column 5: (diff-pop mean of differences)squared.

State hypothesis:

Ho: mean of differences = 0

Ha: mean of differences < 0

1st formula I used was mean of pop differences = sum of differences/sample size = 105.7

2nd formula I used was standard deviation of differences = 34.6150

Test statistic result = 7.3724

The p-value with a t-stat of 7.3724, and df = 9, is smaller than alpha (.05). In fact, lower than .005. (I do not know exact p-value, just looked at table.)

Since p-value lower than alpha, we reject Ho and accept Ha with 95% confidence. The evidence shows that average post-weaning bone mineral contact exceeds that during lactation by more than 25g.

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.

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