Chapter 9.5, Problem 35SE

a.

To determine

Find the test statistic and p-value.

State the conclusion for given level of significance.

Answer to Problem 35SE

The test statistic is 1.5 and the p-value is 0.1544.

The conclusion is that null hypothesis is not rejected.

Explanation of Solution

Calculation:

The given information is that, the sample mean is $\overline{x}=203$, the sample standard deviation is $s=8$, the sample size is $n=16$ and the level of significance is $\alpha =0.025$.

The test hypotheses are given below:

Null hypothesis: $H_0:\mu =200$

Alternative hypothesis: $H_1:\mu \ne 200$

The formula for test statistic is,

$t_{\text{calc}}=\frac{\overline{x}-{\mu }_0}{\frac{s}{\sqrt{n}}}$

Where $\overline{x}$ is the sample mean, ${\mu }_0$ is the population mean, s is the sample standard deviation and n is the sample size.

Substitute $\overline{x}=203$, ${\mu }_0=200$, $s=8$ and $n=16$ in the test statistic formula.

$\begin{array}{c}{t}_{\text{calc}}=\frac{203-200}{\frac{8}{\sqrt{16}}}\\ =\frac{3}{\left(\frac{8}{4}\right)}\\ =\frac{3}{2}\\ =1.5\end{array}$

Thus, the test statistic is 1.5.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =16-1\\ =15\end{array}$

p-value:

Software procedure:

Step-by-step software procedure to obtain p-value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.DIST.2T(1.5, 15)”
• Output using Excel software is given below:

Thus, the p-value for two-tailed test is 0.1544.

Decision rule:

If $p\text{-value}<\alpha$, then reject the null hypothesis.

Conclusion for p-value method:

Here, the p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.1544\right)>\alpha \left(=0.025\right)$.

Therefore, the null hypothesis is not rejected.

b.

To determine

Find the test statistic and p-value.

State the conclusion for given level of significance.

Answer to Problem 35SE

The test statistic is –2 and the p-value is 0.0285.

The conclusion is that null hypothesis is rejected.

Explanation of Solution

Calculation:

The given information is that, the sample mean is $\overline{x}=198$, the sample standard deviation is $s=5$, the sample size is $n=25$ and the level of significance is $\alpha =0.05$.

The test hypotheses are given below:

Null hypothesis: $H_0:\mu \ge 200$

Alternative hypothesis: $H_1:\mu <200$

Substitute $\overline{x}=198$, ${\mu }_0=200$, $s=5$ and $n=25$ in the test statistic formula.

$\begin{array}{c}{t}_{\text{calc}}=\frac{198-200}{\frac{5}{\sqrt{25}}}\\ =\frac{-2}{\left(\frac{5}{5}\right)}\\ =-2\end{array}$

Thus, the test statistic is –2.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =25-1\\ =24\end{array}$

p-value:

Software procedure:

Step-by-step software procedure to obtain p-value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.DIST(–2, 24,1)”
• Output using Excel software is given below:

Thus, the p-value for left-tailed test is 0.0285.

Conclusion for p-value method:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.0285\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

c.

To determine

Find the test statistic and p-value.

State the conclusion for given level of significance.

Answer to Problem 35SE

The test statistic is 3.75 and the p-value is 0.0003.

The conclusion is that null hypothesis is rejected.

Explanation of Solution

Calculation:

The given information is that, the sample mean is $\overline{x}=205$, the sample standard deviation is $s=8$, the sample size is $n=36$ and the level of significance is $\alpha =0.05$.

The test hypotheses are given below:

Null hypothesis: $H_0:\mu \le 200$

Alternative hypothesis: $H_1:\mu >200$

Substitute $\overline{x}=205$, ${\mu }_0=200$, $s=8$ and $n=36$ in the test statistic formula.

$\begin{array}{c}{t}_{\text{calc}}=\frac{205-200}{\frac{8}{\sqrt{36}}}\\ =\frac{5}{\left(\frac{8}{6}\right)}\\ =\frac{5}{1.3333}\\ =3.75\end{array}$

Thus, the test statistic is 3.75.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =36-1\\ =35\end{array}$

p-value:

Software procedure:

Step-by-step software procedure to obtain p-value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.DIST.RT(3.75, 35)”
• Output using Excel software is given below:

Thus, the p-value for right-tailed test is 0.0003.

Conclusion for p-value method:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.0003\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.