Question

Denosumab and Osteoporosis. A clinical study was conducted to see whether an antibody called denosumab is effective in the treatment of osteoporosis in postmenopausal women, as reported in the article "Denosumab in Postmenopausal Women with Low Bone Mineral Density (New England Journal of Medicine, Vol. $354$. No. $8$, pp. $821-831$) by M. McClung et al. Postmenopausal women with osteoporosis were randomly assigned into groups that received either a placebo a six-month regimen of Denosumab at doses of $14$mg, $60$mg,$100$mg, or $210$mg. The following table provides summary statistics for the body-mass indexes (BMI) of the women in each treatment group.

At the $10\%$significance level, do the data provide sufficient evidence to conclude that a difference exists in mean BMI for women in the five different treatment groups? Note: For the degrees of freedom in this exercise:

Short answer

The data does not provide significant evidence that there is a difference in mean BMI for women in the five categories.

Step-by-step solution

Given information

The given data is

Explanation

The level of significance is $\alpha =0.10$

Let us do the test hypotheses

Null hypothesis

$H_0$: There is no difference exist in mean BMI for women in the five different groups.

Alternative hypothesis

$H_a$: There is difference exist in mean BMI for women in the five different groups.

The mean of all observations is

$\overline{x}=\frac{46(25.9)+54(25.8)+47(27.5)+42\left(26\right)+47(25.9)}{46+54+47+42+47}$

$=\frac{6186}{236}$

$=26.2136$

The treatment sum of the square is

$SSTR=\sum n_i{\left({\overline{x}}_i-\overline{x}\right)}^2$

$=46(25.9-26.2136{)}^2+54(25.8-26.2136{)}^2+47(27.5-26.2136{)}^2$

$+(42-1)(4.6{)}^2+(47-1\left)\right(4.3{)}^2$

role="math" localid="1652203360710" $=98.0766$

The error sum of squares is

$SSE=\sum \left(n_i-1\right){\left(s_i\right)}^2$

$=(46-1)(4.3{)}^2+(54-1\left)\right(5.3{)}^2+(47-1)(5.8{)}^2$

$+(42-1)(4.6{)}^2+(47-1\left)\right(4.3{)}^2$

$=5586.36$

Then, the total sum of squares is

$SST=SSTR+SSE$

$=98.0766+5586.36$

$=5684.4366$

The mean treatment of the sum of squares is

$MSTR=\frac{SSTR}{k-1}$

$=\frac{98.0766}{5-1}$

$=24.1834$

Then, the mean error of the sum of squares is

$MSE=\frac{SSE}{n-k}$

$=\frac{5586.36}{236-5}$

$=24.1834$

The F-static is

$F-static=\frac{MSTR}{MSE}$

$=\frac{24.5192}{24.1834}$

When $\alpha =0.10$and the critical value is $1.97$

F-static $(1.01)<$critical value $(1.97)$

As a result, the crucial value approach

At a $10\%$significant threshold, the null hypothesis is not rejected.

As a result, the results are not statistically significant at the $10\%$level.

As a result, the data does not provide significant evidence that there is a difference in mean BMI for women in the five categories.