Question

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

Comparing two groups
ANSWERED
asked 2021-03-12
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 \(95\%\) Cl for the difference in means between the two groups.

Answers (1)

2021-03-13
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
\(H_{0} : \sigma_{1} = \sigma_{2}\)
\(H_{1} : \sigma_{1} \neq \sigma_{2}\)
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
\(H_{0} : \mu_{1} = \mu_{2}\)
\(H_{1} : \mu_{1} \neq \mu_{2}\)
Degree of freedom
\(= n_{1} + n_{2} - 2\)
\(= 40 + 25 - 2\)
\(= 63\)
For two tailed test critical values \(\pm1.998\)
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
\(t=\frac{\overline{x}_{1}-\overline{x}_{2}}{S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\)
\(= \frac{6.8-6.56}{0.71666\times \sqrt{\frac{1}{40}+\frac{1}{25}}}\)
\(= 1.3136\)
By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.
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