Chapter 9, Problem 81CE

a.

To determine

State the hypotheses.

Answer to Problem 81CE

Null hypothesis:

$H_0:\mu \ge 90$

Alternative hypothesis:

$H_1:\mu <90$

Explanation of Solution

The given information is that, the data represents the scores for Friday quizzes.

Bob’s claim that he is really a 90+ performer.

The test hypotheses are given below:

Null hypothesis:

$H_0:\mu \ge 90$

Alternative hypothesis:

$H_1:\mu <90$

b.

To determine

State the formula for the test statistic and also describe the decision rule at 1% level of significance.

Answer to Problem 81CE

The formula for test statistic is,

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

Decision rule is, reject $H_0$ if $t_{\text{calc}}<-2.998$.

Explanation of Solution

Test statistic:

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.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =8-1\\ =7\end{array}$

Critical value:

Software procedure:

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

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV(0.01,7)”
• Output using Excel software is given below:

From the output, the critical value of Student’s t is $t_{0.01}=-2.998$.

Decision rule for left-tailed test at $\alpha =0.01$:

If $t_{\text{calc}}<-2.998$, then reject the null hypothesis.

c.

To determine

Find the test statistic and state the conclusion.

Answer to Problem 81CE

There is evidence to infer the Bob’s claim that he is really a 90+ performer.

Explanation of Solution

Calculation:

• Sample mean and variance:
• The formula for mean is,
• $\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$
• The formula for standard deviation is,
• $s=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$
• The value of ${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$ and mean is calculated as follows:

Score	$\left(x_i-\overline{x}\right)$	${\left(x_i-\overline{x}\right)}^2$
80	–8.375	70.14063
85	–3.375	11.39063
95	6.625	43.89063
92	3.625	13.14063
89	0.625	0.390625
84	–4.375	19.14063
90	1.625	2.640625
92	3.625	13.14063
${\displaystyle \sum x_i}=707$		${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}=173.875$
$\overline{x}=88.375$

• The standard deviation is,
• $\begin{array}{c}s=\sqrt{\frac{173.875}{8-1}}\\ =\sqrt{24.8393}\\ =4.984\end{array}$
• Thus, the standard deviation is 4.984.

Substitute $\overline{x}=88.375$, ${\mu }_0=90$, $s=4.984$ and $n=8$ in the test statistic formula.

$\begin{array}{c}{t}_{\text{calc}}=\frac{88.375-90}{\frac{4.984}{\sqrt{8}}}\\ =\frac{-1.625}{\left(\frac{4.984}{2.8284}\right)}\\ =\frac{-1.625}{1.7621}\\ =-0.92\end{array}$

Thus, the test statistic is –0.92.

Conclusion for critical value method:

Here, the test statistic is greater than the critical value.

That is, $t_{\text{calc}}\left(=-0.92\right)>-2.998$

Therefore, the null hypothesis is not rejected.

Hence, there is evidence to infer the Bob’s claim that he is really a 90+ performer.

d.

To determine

Describe the assumptions for given hypotheses tests.

Explanation of Solution

The assumption is necessary to validate the given test is the population is normally distributed.

e.

To determine

Find the p-value by using Excel and interpret it.

Answer to Problem 81CE

The p-value is 0.1941.

There is evidence to infer the Bob’s claim that he is really a 90+ performer.

Explanation of Solution

Calculation:

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =8-1\\ =7\end{array}$

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(–0.92,7)”.
• Output using Excel software is given below:

Thus, the p-value is 0.1941.

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, $p\text{-value}\left(=0.1941\right)>\alpha \left(=0.01\right)$.

Therefore, the null hypothesis is not rejected.

Hence, there is evidence to infer the Bob’s claim that he is really a 90+ performer.