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 1928 2549 2825 1924 1628 2175 2114 2621 1843 2541
P 2126 2885 2895 1942 1750 2184 2164 2626 2006 2627

Required:
a. Does 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.]

b. Calculate a lower confidence bound using a 95% confidence level for the true average difference between TBBMC during postweaning and lactation.

Answer 1

Answer:

- 179.981

Step-by-step explanation:

The hypothesis :

H0 : μL - μP ≥ 25

H0 : μL - μP < 25

The sample mean difference ;Xd

d = L - P

Xd = Σd/n

d = -198,-336,-70,-18,-122,-9,-50,-5,-163,-86

Xd = - 1057 / 10

Xd = - 105.7

Using calculator ;

Standard deviation of difference, Sd = 103.845

The test statistic :

T = Xd ÷ (Sd/√n)

T = -105.7 ÷ (103.845/√10)

T = - 3.219

Decision region :

Reject H0 ; If Pvalue < α

The Pvalue : df = n - 1 ; 10 - 1 = 9

Pvalue(-3.219, 9) ; two-tailed = 0.00525

Hence, reject H0

B.) The confidence interval for difference in mean :

Xd ± Tcritical[Sd/√n]

Tcritical at 95%, df = 9

Tcritical = 2.262

C.I = -105.7 ± 2.262[103.845/√10]

C.I = -105.7 ± 74.281076

Lower boundary: - 105.7 - 74.281076 = - 179.9810