Chapter 8, Problem 10CRP

(a)

To determine

Categorize type of test used.

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 10CRP

The type of the test used is difference of means.

The level of significance is 0.05.

The null hypothesis is $H_0:{\mu }_1={\mu }_2$.

The alternative hypothesis is $H_1:{\mu }_1\ne {\mu }_2$.

Explanation of Solution

Calculation:

Let ${\mu }_1$ denotes the average off-schedule time of bus line A, ${\mu }_2$ denotes the average off-schedule time of bus line B.

From the given information the value of $\alpha$ is 0.05, and a significant difference in average off-schedule times.

The averages of two populations are compared. Since two samples are used to test difference between means with unknown standard deviations, the appropriate test would be difference of mean.

Hence, the level of significance is 0.05.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

(b)

To determine

Identify the sampling distribution to be used.

Mention the assumption to test.

Find the sample test statistic.

Find the corresponding distribution value.

Answer to Problem 10CRP

The sampling distribution to be used is Student’s t distribution.

The sample test statistic is –2.735.

The t value is $\pm 2.009$.

Explanation of Solution

Calculation:

Conditions:

• When the distribution of $x_1$ (population 1) and $x_2$ (population 2) has the normal distribution or mounded shaped with unknown population standard deviations ${\sigma }_1$, ${\sigma }_2$ then sampling distribution of ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution for any sample sizes of $n_1,n_2$. The t statistic is used for testing.
• When the distribution of $x_1$ (population 1) or $x_2$ (population 2) are not normally distributed with unknown population standard deviations ${\sigma }_1$, ${\sigma }_2$ then sampling distribution of ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution for sample sizes of $n_1\ge 30,n_2\ge 30$.

Degrees of freedom:

The degrees of freedom for the t distribution is,

$d.f.=\text{smaller}\left\{n_1-1,n_2-1\right\}$

In the formula $n_1$ is the sample size from population 1 and $n_2$ is the sample size from population 2.

The two sample t statistic is,

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

In the formula, ${\mu }_1$ is mean of first population, ${\mu }_2$ is mean of second population, ${\overline{x}}_1$ is the sample mean from population 1, ${\overline{x}}_2$ is the sample mean from population 2, $s_1^2$ sample variance of population 1, $s_2^2$ is sample variance of population 2, $n_1$ is the sample size from population 1 and $n_2$ is the sample size from population 2.

The population distributions are mound-shaped and symmetric for problems with small samples that involve testing a mean or difference of means is assumed.

It is clear that the distribution of off-schedule times is approximately normal and population standard deviation is unknown. Hence, the sampling distribution to be used is Student’s t distribution.

Degrees of freedom:

Substitute $n_1$ as 51, $n_2$ as 60 in the degrees of freedom formula

$\begin{array}{c}d.f.=\text{smaller}\left\{51-1,60-1\right\}\\ =\text{smaller}\left\{50,59\right\}\\ =50\end{array}$

The degrees of freedom are 50.

The sample value is,

$\begin{array}{c}{\overline{x}}_1-{\overline{x}}_2=53-62\\ =-9\end{array}$

The value of ${\overline{x}}_1-{\overline{x}}_2$ is –9.

T-statistic:

Substitute ${\overline{x}}_1$ as 53, ${\overline{x}}_2$ as 62, ${\mu }_1-{\mu }_2$ as 0, $s_1$ as 19, $s_2$ as 15, $n_1$ as 51, and $n_2$ as 60 in the test statistic formula

$\begin{array}{c}t=\frac{\left(53-62\right)-0}{\sqrt{\frac{{19}^2}{51}+\frac{{15}^2}{60}}}\\ =\frac{-9}{3.2907}\\ =-2.735\end{array}$

Hence, the sample test statistic is –2.735.

Critical value:

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 50.
• In the row of two-tail area locate the level of significance $\alpha =0.05$.
• The intersecting value of row and columns is 2.009.

Hence, the t value for the two tailed is $\pm 2.009$.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Answer to Problem 10CRP

The P-value is 0.0086.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 50.
• Click the Shaded Area tab.
• Choose X Value and Both Tails, for the region of the curve to shade.
• Enter the X value as –2.735.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0043 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.0043\\ =0.0086\end{array}$

Hence, the P-value of the test statistic is 0.0086.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.05 or not.

Answer to Problem 10CRP

The null hypothesis is rejected.

The data is statistically significant at level 0.05.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0086.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0086 and the level of significance is 0.05.

The P-value is less than the level of significance.

That is, $0.0086\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.05.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that, there is sufficient evidence that the difference in average off-schedule times at 0.05 level of significance.