Chapter 8.3, Problem 32E

a.

To determine

State the null and alternative hypothesis.

Answer to Problem 32E

Null hypothesis:

$\underset{\_}{H_0:\mu =16}$.

Alternative hypothesis:

$\underset{\_}{H_1:\mu \ne 16}$.

Explanation of Solution

Calculation:

The result of t-test is given.

Here, the direction of the test is two-tailed.

Null hypothesis:

$\underset{\_}{H_0:\mu =16}$.

Here, the mean value of the population is equal to 16.

Alternative hypothesis:

$\underset{\_}{H_1:\mu \ne 16}$.

Here, the mean value of the population is not equal to 16.

b.

To determine

Find the value of $\overline{x}$.

Answer to Problem 32E

The $\overline{x}$-value is 13.2874.

Explanation of Solution

Calculation:

From the given output, the value of $\overline{x}$ is 13.2874.

Thus, the $\overline{x}$-value is 13.2874.

c.

To determine

Find the value of s.

Answer to Problem 32E

The s-value is 6.0989.

Explanation of Solution

Calculation:

From the given output, the value of s is 6.0989.

Thus, the s-value is 6.0989.

d.

To determine

Find the value of degrees of freedom.

Answer to Problem 32E

The value of degrees of freedom is 10.

Explanation of Solution

Calculation:

The formula for degrees of freedom in t-test is n – 1.

From the output, the value of n is 11.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =11-1\\ =10\end{array}$

Thus, the degrees of freedom is 10.

e.

To determine

Check whether the null hypothesis is reject or not at 0.05 level of significance.

Answer to Problem 32E

The null hypothesis is not rejected at 0.05 level of significance.

Explanation of Solution

Calculation:

Decision rule:

• If $P\text{-value}\le \alpha$, then reject the null hypothesis.
• Otherwise, do not reject the null hypothesis.

From the output, the value of P-value is 0.171 and the level of significance is 0.05.

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

Thus, the null hypothesis is not rejected.

f.

To determine

Find the t statistic.

Answer to Problem 32E

The value of t statistic is 2.332.

Explanation of Solution

Calculation:

The given hypotheses is that,

Null hypothesis:

$H_0:\mu =9$.

Alternative hypothesis:

$H_1:\mu >9$.

Here, the direction of the test is right-tailed.

The formula for t test statistic is,

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Here $\overline{x}$ is the sample mean, $\mu$ is the population mean, s is the sample standard deviation and n is the sample size.

Substitute the value $\overline{x}$ as 13.2874, $\mu$ as 9, s as 6.0989 and n as 11.

$\begin{array}{c}t=\frac{13.2874-9}{\frac{6.0989}{\sqrt{11}}}\\ =\frac{4.2874}{1.8389}\\ =2.332\end{array}$

Thus, the t statistic is 2.332.

g.

To determine

Find the P-value.

Answer to Problem 32E

The P-value is 0.02095.

Explanation of Solution

Calculation:

P-value:

Software procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

• Choose Graph > Probability Distribution Plot > choose View Probability> OK.
• From Distribution, choose t.
• Enter df as 10.
• Choose X value and Right Tail for the region of the curve to shade.
• Enter the value as 2.332.
• Click OK.
• Output using MINITAB software is given below:

From the output, the P-value is 0.02095.

Thus, the P-value is 0.02095.

h.

To determine

Check whether the null hypothesis is reject or not at 0.05 level of significance.

Answer to Problem 32E

The null hypothesis is rejected at 0.05 level of significance.

Explanation of Solution

Calculation:

Decision rule:

• If $P\text{-value}\le \alpha$, then reject the null hypothesis.
• Otherwise, do not reject the null hypothesis.

From the output, the value of P-value is 0.02095 and the level of significance is 0.05.

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

Thus, the null hypothesis is rejected.