Emeli Hagan

2021-02-09

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$6.80+0.76.$

1) Test for a significant difference between the variances of the two groups.

2) What is the appropriate procedure to test for a signifi- cant difference in means between the two groups?

3) Implement the procedure in Problem 8.3 using the critical-value method.

Similarly, the mean + 1 sd of In [calcium intake (mg)] among 40 females, 12 to 14 years of age, above the poverty level is

1) Test for a significant difference between the variances of the two groups.

2) What is the appropriate procedure to test for a signifi- cant difference in means between the two groups?

3) Implement the procedure in Problem 8.3 using the critical-value method.

Benedict

Skilled2021-02-10Added 108 answers

Given:

${n}_{1}=40,{s}_{1}=0.76$

${n}_{2}=25,{s}_{2}=0.64$

1) 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.

2) For this two sample t-test is used the reason is that population variances are equal.

3)

${H}_{0}:{\mu}_{1}={\mu}_{2}$

${H}_{1}:{\mu}_{1}\ne {\mu}_{2}$

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

${S}_{p}=\sqrt{\frac{({n}_{1}-1){s}_{1}^{2}+({n}_{2}-1){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}}$

$=\sqrt{\frac{(40-1){0.76}^{2}+(25-1)0.6462}{40+25-2}}$

$=0.71666$

And the t-statistics would be

$t=\frac{{\stackrel{\u2015}{x}}_{2}-{\stackrel{\u2015}{x}}_{2}}{{S}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}$

$=\frac{6.8-6.56}{0.71666\times \sqrt{\frac{1}{40}+\frac{1}{25}}}$

$=1.3136$

By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.

1) Hypothesis

Test statistics would be

Degree of freedom of numerator

Degree of freedom of denominator

P-value of the test

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.

2) For this two sample t-test is used the reason is that population variances are equal.

3)

Degree of freedom

For two tailed test critical values

If t is less than -1.998 or greater than 1.998, fall to reject the null hypothesis.

Pooled standard deviation

And the t-statistics would be

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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