The mean + 1 sd of In [calcium intake (mg)] among 25 females, 12 to 14 years of age, below the poverty level is 6.56 + 0.64. Similarly, the mean + 1 s

Step 1
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Given:
$n_1=40,s_1=0.76$
$n_2=25,s_2=0.64$
Step 2
8.2
Hypothesis
$H_0:{\sigma }_1={\sigma }_2$
$H_1:{\sigma }_1\ne {\sigma }_2$
Test statistics would be
$F=\frac{S_1^2}{S_2^2}$
$=\frac{{0.76}^2}{{0.64}^2}$
$=1.41$
Degree of freedom of numerator $40-1=39$
Degree of freedom of denominator $25-1=24$
P-value of the test $=0.3759$
Therefore P-value is greater than 0.05, therefore it is fail to reject the null hypothesis.
Conclusion is that there is no significant difference between the populations.
Step 3
8.3
For this two sample t-test is used the reason is that population variances are equal.
Step 4
8.4
$H_0:{\mu }_1={\mu }_2$
$H_1:{\mu }_1\ne {\mu }_2$
Degree of freedom
$=n_1+n_2-2$
$=40+25-2$
$=63$
For two tailed test critical values $\pm 1.998$
If t is less than -1.998 or greater than 1.998, fall to reject the null hypothesis.
Pooled standard deviation
$S_p=\sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}}$
$=\sqrt{\frac{(40-1){0.76}^2+(25-1){0.64}^2}{40+25-2}}$
$=0.71666$
And the t-statistics would be
$t=\frac{{\bar{x}}_1-{\bar{x}}_2}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$
$=\frac{6.8-6.56}{0.71666\times \sqrt{\frac{1}{40}+\frac{1}{25}}}$
$=1.3136$
By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.