Question

Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk production. A paper gave the following data on total body bone mineral content (TBBMC) (g) for a sample both during lactation (L) and in the postweaning period (P).

Subject
	1	2	3	4	5	6	7	8	9	10
L	1927	2547	2825	1923	1628	2175	2111	2621	1843	2543
P	2127	2885	2895	1945	1750	2183	2164	2626	2006	2628

(a) Does the data suggest that true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g? State and test the appropriate hypotheses using a significance level of 0.05. [Note: The appropriate normal probability plot shows some curvature but not enough to cast substantial doubt on a normality assumption.] (Use μ_D = μ_P − μ_L.)

H₀: μ_D = 25
H_a: μ_D > 25 H₀: μ_D = 25
H_a: μ_D < 25     H₀: μ_D = 25
H_a: μ_D ≠ 25 H₀: μ_D = 25
H_a: μ_D ≤ 25

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t	=	2
P-value	=	3

State the conclusion in the problem context.

Fail to reject H₀. The data suggests that the true average total body bone mineral content during postweaning does not exceed that during lactation by more than 25 g. Reject H₀. The data suggests that the true average total body bone mineral content during postweaning does not exceed that during lactation by more than 25 g.     Fail to reject H₀. The data suggests that the true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g. Reject H₀. The data suggests that the true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g.

(b) Calculate an upper confidence bound using a 95% confidence level for the true average difference between TBBMC during postweaning and during lactation. (Round your answer to two decimal places.)
5 g

Answer 1

a)

H₀: μ_D = 25
H_a: μ_D > 25

t =2.48

p value =0.018

since p value <0.05

Reject H₀. The data suggests that the true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g.

b)

for 95% upper CI; and 9 degree of freedom, value of t=			1.833
therefore confidence interval=sample mean -/+ t*std error
margin of errror          =t*std error=		60.356

upper confidence limit                    =

166.95