Chapter 8.4, Problem 7P

(a)

To determine

Identify whether it is appropriate to use a Student’s t distribution or not.

Find the degrees of freedom.

Answer to Problem 7P

Yes, it is appropriate to use a Student’s t distribution.

The degrees of freedom are 35.

Explanation of Solution

Calculation:

Conditions:

• When the d distribution considered in the study has the normal distribution or simply has a mound-shaped, symmetric distribution then the sampling distribution $\overline{d}$ has Student’s t distribution for any sample size n. The t statistic is used for testing.
• When the d distribution considered in the study is not normally distributed then the sampling distribution $\overline{d}$ has Student’s t distribution if the sample size n is greater than or equal 30. That is, $n\ge 30$.

Degrees of freedom:

The degrees of freedom for the t distribution is,

$d.f.=n-1$

In the formula n is the number of data pairs.

The sample size is 36 which greater than 30. Hence, the sampling distribution to be used is Student’s t distribution.

Degrees of freedom:

Substitute n as 36 in the degrees of freedom formula

$\begin{array}{c}d.f.=n-1\\ =36-1\\ =35\end{array}$

Hence, the degrees of freedom are 35.

(b)

To determine

State the hypotheses.

Answer to Problem 7P

The hypotheses is,

Null hypothesis: $H_0:{\mu }_d=0$.

Alternative hypothesis: $H_1:{\mu }_d\ne 0$.

Explanation of Solution

Calculation:

Let $\begin{array}{cccccc}Name& SHIFT& TARGETPRODUCTIVITY& PRODUCTIVITYUNIT& ACTUALPRODUCTIVITY& \#ofHoursConnected\\ NIVAS& I& 100 & pages & 75 & 8hours \end{array}$ denotes the population mean of the differences.

From the given information the value of $\alpha$ is 0.05, and the claim that the population mean of the differences is different from 0.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\mu }_d=0$

Alternative hypothesis:

$H_1:{\mu }_d\ne 0$

(c)

To determine

Find the t-value of the sample test statistic.

Answer to Problem 7P

The t-value of the sample test statistic is 2.4.

Explanation of Solution

Calculation:

Test statistic for t:

The t statistic value for sample test statistic $\overline{d}$ is,

$t=\frac{\overline{d}-{\mu }_d}{\left(\frac{s_d}{\sqrt{n}}\right)}$

In the formula $\overline{d}$ is sample test statistic, $\begin{array}{cccccc}Name& SHIFT& TARGETPRODUCTIVITY& PRODUCTIVITYUNIT& ACTUALPRODUCTIVITY& \#ofHoursConnected\\ NIVAS& I& 100 & pages & 75 & 8hours \end{array}$ is specified in $H_0$, $s_{\sigma }$ is sample standard deviation of the differences d, and n is the number of data pairs.

T-statistic:

Substitute $\overline{d}$ as 0.8, ${\mu }_d$ as 0, $s_{\sigma }$ as 2, and n as 36 in the test statistic formula

$\begin{array}{c}t=\frac{0.8-0}{\left(\frac{2}{\sqrt{36}}\right)}\\ =\frac{0.8}{0.3333}\\ =2.40\end{array}$

Hence, the t-value of the sample test statistic is 2.40.

(d)

To determine

Find the P-value.

Answer to Problem 7P

The P-value is 0.0218.

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 35.
• Click the Shaded Area tab.
• Choose X Value and Both Tails, for the region of the curve to shade.
• Enter the X value as 2.40.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0109 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.0109\\ =0.0218\end{array}$

Hence, the P-value is 0.0218.

(e)

To determine

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

Answer to Problem 7P

The null hypothesis is rejected.

Explanation of Solution

Calculation:

From part (d), the P-value is 0.0218.

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 0.0218 and the level of significance is 0.05.

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

That is, $0.0218\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.05.

(f)

To determine

Interpret the results.

Explanation of Solution

Calculation:

From part (e), the null hypothesis is rejected. This shows that, there is sufficient evidence that the population mean of the differences is different from 0 at 0.05 level of significance.