Chapter 10, Problem 93SE

a.

To determine

Fill the missing values.

Answer to Problem 93SE

The missing values are as follows,

Explanation of Solution

Given info:

A computer output is provided for two sample T-test.

Calculation:

Test statistic:

$T_0=\frac{{\overline{X}}_1-{\overline{X}}_2-{\Delta }_0}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_1}}}$

Software procedure:

Step-by-step software procedure to obtain the test statistic and P-value using MINITAB software is as follows,

• Choose Stat > Basic Statistics > 2-Sample t.
• Select Summarized data.
• In Sample 1, enter Sample size as 20, Sample Mean as 11.87 and standard deviation as 2.23.
• In Sample 2, enter Sample size as 20, Sample Mean as 12.73 and standard deviation as 3.19.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Select Assume equal variances.
• Click OK in all the dialogue boxes

Thus, the test statistic is ­–0.99 and the P-value is 0.329.

b.

To determine

Check whether it is a two-sided test or one-sided test.

Answer to Problem 93SE

The test is two-sided test.

Explanation of Solution

Justification:

From the MINTAB output in part (a), it is observed that the alternative hypothesis is, $H_1:{\mu }_1-{\mu }_2\ne 0$.

Thus, this is a two-sided test.

c.

To determine

Explain the conclusion when $\alpha =0.05$ and $\alpha =0.10$.

Answer to Problem 93SE

The null hypothesis is not rejected in both the cases.

Explanation of Solution

Justification:

For $\alpha =0.05$:

Conclusion:

From the MINITAB output in part (a), the P-value 0.329 is greater than the significance level 0.05.

That is, $P-\text{value}\left(=0.329\right)>\alpha \left(=0.05\right)$

Hence the null hypothesis is not rejected.

For $\alpha =0.10$:

Software procedure:

Step-by-step software procedure to obtain the test statistic and P-value using MINITAB software is as follows,

• Choose Stat > Basic Statistics > 2-Sample t.
• Select Summarized data.
• In Sample 1, enter Sample size as 20, Sample Mean as 11.87 and standard deviation as 2.23.
• In Sample 2, enter Sample size as 20, Sample Mean as 12.73 and standard deviation as 3.19.
• Choose Options.
• In Confidence level, enter 90.
• In Alternative, select not equal.
• Select Assume equal variances.
• Click OK in all the dialogue boxes

Output using the MINITAB software is as follows,

Conclusion:

Here the P-value 0.329 is greater than the significance level 0.10.

That is, $P-\text{value}\left(=0.329\right)>\alpha \left(=0.10\right)$

Hence the null hypothesis is not rejected