91Ó°ÊÓ

The given data is

The level of significance is $\mathrm{Î\pm }=0.10$

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

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

$=\frac{6186}{236}$

$=26.2136$

The treatment sum of the square is

$SSTR=∑n_i{\left({\stackrel{¯}{x}}_i-\stackrel{¯}{x}\right)}^2$

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

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

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

The error sum of squares is

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

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

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

$=5586.36$

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 $\mathrm{Î\pm }=0.10$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.