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# 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
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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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