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Bone mineral densities (mg/cm 2 ) were measured on a sample of n = 9 women (age

ID: 3172152 • Letter: B

Question

Bone mineral densities (mg/cm2) were measured on a sample of n = 9 women (age 45 to 65) with low bone mineral densities (BMD) who had been randomly selected and assigned to a treatment group that took the drug conjugated equine estrogen (CEE) for 3 years. CEE is used as a treatment for low bone density and investigators wanted to examine its effectiveness. The responses, rounded to the nearest integer, are reported in Table 1.


Assuming that this sample is representative of the entire population of women in this age group with low bone density, research clinicians would like to test the hypothesis that, on average, women on CEE therapy will have density levels that are greater than 1305. Let = 0.025


Note: you should carry at least 5 decimal precision for any intermediate calculations then round your answer as indicated in the problem


E) The point estimate for the true standard deviation for BMD for women (age 45 - 65) taking CEE is:
      ??? mg/cm2
     Note: round your answer to the nearest hundredth

F) The test statistic for this data set is: ????
     Note: Round your answer to the nearest hundredth.

Table 1: Bone Mineral Densities for women between 45 - 65 years old 1317 1314 1301 1315 1322 1306 1333 1320 1286

Explanation / Answer

E) The point estimate for the true standard deviation for BMD for women (age 45 - 65) taking CEE is:

Formula for Standard deviation is given as below:

SD = Sqrt[(X - mean)^2 / (n – 1)]

Calculation table for computation of SD is given as below:

X

(X - mean)^2

1317

18.77774889

1314

1.777768889

1301

136.1111889

1315

5.444428889

1322

87.11104889

1306

44.44448889

1333

413.4443089

1320

53.77772889

1286

711.1112889

Total

11814

1472

Mean

1312.66667

SD

13.56466

(X - mean)^2 = 1472

(X - mean)^2 / (n – 1) = 1472/8 = 184

SD = sqrt(184) = 13.56

F) The test statistic for this data set is:

The formula for test statistic is given as below:

Test statistic = t = (Xbar - µ) / [SD/sqrt(n)]

We are given,

Xbar = 1312.66667

µ = 1305

SD = 13.56465997

Sample size = n = 9

Test statistic = t = (1312.66667 – 1305) / [13.56465997/sqrt(9)]

Test statistic = t = 7.66667/4.521553322

Test statistic = t = 1.695582496

Test statistic = 1.70

X

(X - mean)^2

1317

18.77774889

1314

1.777768889

1301

136.1111889

1315

5.444428889

1322

87.11104889

1306

44.44448889

1333

413.4443089

1320

53.77772889

1286

711.1112889

Total

11814

1472

Mean

1312.66667

SD

13.56466