Chapter 9, Problem 138SE

a.

To determine

Fill the missing values in the table and to make conclusion about the test.

Answer to Problem 138SE

The completed table is,

variable	N	Mean	St dev	SE Mean	Z	P
x	14	26.541	2.032	0.401	1.3495	0.0886

Explanation of Solution

Calculation:

The SE Mean from the sample standard deviation is, $\frac{\sigma }{\sqrt{N}}$. Substitute $\sigma =1.5\text{ and SE Mean}=0.401$. Then,

$\begin{array}{c}SEMean=\frac{\sigma }{\sqrt{N}}\\ 0.401=\frac{1.5}{\sqrt{N}}\\ N={\left(\frac{1.5}{0.401}\right)}^2\\ =13.99\end{array}$

Thus, the value of $N=14$.

The value of $z_0$ is calculated as,

$z_0=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Substitute the values, $\overline{x}=26.541,\mu =26,$$\sigma =1.5,n=14$.

$\begin{array}{c}{z}_0=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ =\frac{26.541-26}{\frac{1.5}{\sqrt{14}}}\\ =\frac{0.541}{0.4008918}\\ =1.3495\end{array}$

Thus, the value of $z_0=1.3495$. The P-value for the test is calculated as,

$P\text{-value=}\left[1-\Phi \left(Z_0\right)\right]$

Substituting the value $z_0=1.3495$, the P-value is,

$\begin{array}{c}P\text{-value=}\left[1-\Phi \left(Z_0\right)\right]\\ =1-\Phi \left(1.3495\right)\end{array}$

Software procedure:

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

• Choose Calc > Probability Distribution> Normal Distribution.
• Choose Cumulative probability.
• In Input constant, enter 1.3495.
• Click OK.

Output using MINITAB software is given below:

Thus, the P-value is,

$\begin{array}{c}P\text{-value=}\left[1-\Phi \left(Z_0\right)\right]\\ =\left[1-\Phi \left(1.3495\right)\right]\\ =\left[1-0.9114\right]\\ =0.0886\end{array}$

Thus, the P-value is 0.0886.

b.

To determine

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

Answer to Problem 138SE

The test is one sided.

Explanation of Solution

Justification:

The hypothesis to be tested is $H_0:\mu =26\text{ vs }{H}_0:\mu >26$. Here, the alternative hypothesis contain the greater than symbol $(>)$.

Thus, the test is one sided test.

c.

To determine

Find the conclusion about the test if $\alpha =0.05$.

Answer to Problem 138SE

The null hypothesis is not rejected at 5% level of significance.

Explanation of Solution

Justification:

From the results of part (a), the P-value is 0.0886, which is greater than significance level $\alpha =0.05$. Thus, the null hypothesis is not rejected at 5% level of significance.

d.

To determine

Construct a 95% two sided confidence interval for the mean.

Answer to Problem 138SE

The 95% two sided confidence interval for the mean is (25.7553, 27.3268).

Explanation of Solution

Calculation:

The 95% confidence interval for the mean is, $\overline{x}-Z_{0.025}\frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+Z_{0.025}\frac{\sigma }{\sqrt{n}}$.

Substitute $\overline{x}=26.541,\sigma =1.5$ and $n=14$ in the above formula.

$26.541-Z_{0.025}\frac{1.5}{\sqrt{14}}<\mu <26.541+Z_{0.025}\frac{1.5}{\sqrt{14}}$

Software procedure:

Step by step procedure to obtain $z_{0.05}$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot > View Single, and then click OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability and both tails for the region of the curve to shade.
• Enter the value as 0.05.
• Click OK.

Output using MINITAB software is given below:

From the MINITAB output, the critical values are,  $z_{0.025}=1.96$.

Substituting the value, the 95% confidence interval for the mean is,

$\begin{array}{c}26.541-\left(1.96\right)\frac{1.5}{\sqrt{14}}<\mu <26.541+\left(1.96\right)\frac{1.5}{\sqrt{14}}\\ 26.541-0.7857<\mu <26.541+0.7857\\ 25.7553<\mu <27.3268\end{array}$

Thus, the 95% confidence interval for the mean is (25.7553, 27.3268).