Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics as$\frac{\text{Larger }{s}^2}{\text{smaller }{s}^2}>3$

From the output by substituting the values we get

  $\frac{5.067}{2.238}=2.264$

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

Answer to Problem 10.24E

The observed value for the test statistics is $t_{STAT}=0.365$ . The P-value associated with the test statistics for two tailed test is $P-value=0.722$ .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is $t_{STAT}=0.365$ .

The P-value associated with the test statistics for two tailed test is $P-value=0.722$ .

The observed value for the test statistics is $t_{STAT}=0.365$ . The P-value associated with the test statistics for two tailed test is $P-value=0.722$ .

Conclusion:

Therefore, observed value for the test statistics is $t_{STAT}=0.365$ and the P-value associated with the test statistics for two tailed test is $P-value=0.722$ .

(c)

To determine

To find: the pooled estimate $s^2$ of the population variance

Answer to Problem 10.24E

The pooled estimate $s^2$ of the population variance is $3.524$ .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate $s^2$ of the population varianceis $3.524$

Conclusion: Thus, the pooled estimate $s^2$ of the population variance is $3.524$ .

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

  $(-1.890,2.698)$ . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

  $\left((\overline{x_1}-\overline{x_2})\pm t_{\alpha }\sqrt{s^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\right)$

From the above output by substituting the value, we get

  $\left((\overline{x_1}-\overline{x_2})\pm t_{\alpha }\sqrt{s^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\right)$

  $\left(\left(28.667-28.268\right)\pm 2.201\sqrt{3.524\left(\frac{1}{6}+\frac{1}{7}\right)}\right)$

  $(-1.890,2.698)$

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

  $(-1.890,2.698)$ . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.