 # 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 Kyran Hudson 2021-03-12 Answered
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 sd of In [calcium intake (mg)] among 40 females, 12 to 14 years of age, above the poverty level is $6.80+0.76.$
8.2 Test for a significant difference between the variances of the two groups.
8.3 What is the appropriate procedure to test for a signifi- cant difference in means between the two groups?
8.4 Implement the procedure in Problem 8.3 using the critical-value method.
8.5 What is the p-value corresponding to your answer to Problem 8.4?
8.6 Compute a $95\mathrm{%}$ Cl for the difference in means between the two groups.
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Step 1
We are authorised to answer three subpart at a time, since you have not mentioned which part you are looking for, so we are answering the first three subpart, please re post your question separately for the remaining subpart.
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 $±1.998$
If t is less than -1.998 or greater than 1.998, fall to reject the null hypothesis.
Pooled standard deviation
${S}_{p}=\sqrt{\frac{\left({n}_{1}-1\right){S}_{1}^{2}+\left({n}_{2}-1\right){S}_{2}^{2}}{{n}_{1}+{n}_{2}-2}}$
$=\sqrt{\frac{\left(40-1\right){0.76}^{2}+\left(25-1\right){0.64}^{2}}{40+25-2}}$
$=0.71666$
And the t-statistics would be
$t=\frac{{\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}}{{S}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}$
$=\frac{6.8-6.56}{0.71666×\sqrt{\frac{1}{40}+\frac{1}{25}}}$
$=1.3136$
By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.