Chapter 9.4, Problem 13E

(a)

To determine

The appropriate null and alternate hypotheses.

Answer to Problem 13E

Null hypothesis:

$H_0:p=0.80$

That is, the percentage of emails that are spam is not differs from $80\%$.

Alternative hypothesis:

$H_1:p\ne 0.80$

That is. the percentage of emails that are spam differs from $80\%$.

Explanation of Solution

Given information:

According to Secure List, $71.8\%$ of all email sent in July $2012$ was spam.

A system manager suspects that the spam email percentage at his company may be $80\%$.

In an inspection he finds that $382$ out of $500$ emails are spam.

Calculation:

A system manager suspects that the spam email percentage at his company may be $80\%$.

So null and alternate hypotheses would be given as-

Null hypothesis:

$H_0:p=0.80$

That is, the percentage of emails that are spam is not differs from $80\%$.

Alternative hypothesis:

$H_1:p\ne 0.80$

That is. the percentage of emails that are spam differs from $80\%$.

(b)

To determine

The test satistic $Zz$.

Answer to Problem 13E

$z=-2.01$

Explanation of Solution

Given information:

Same as part $a$.

Formula used:

$\begin{array}{l}\widehat{p}=\frac{x}{n}\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\end{array}$

Calculation:

A system manager suspects that the spam email percentage at his company may be $80\%$.

i.e. $p_0=0.80$.

In an inspection it is found that $382$ out of $500$ emails are spam.

The value of $\widehat{p}$ is obtained below.

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{382}{500}\\ =0.764\end{array}$

Calculating the test statistic

$\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.764-0.80}{\sqrt{\frac{0.80\left(1-0.80\right)}{500}}}\\ =\frac{-0.036}{0.0179}\\ =-2.01\end{array}$

Thus, the value of test statistic $z$ is $-2.01$.

(c)

To determine

Whether the percentage of emails that are spam differs from $80\%$ for $\alpha =0.05$.

Answer to Problem 13E

It can be concluded that there is evidence at level of significance $\alpha =0.05$ for the percentage of emails that are spam differs from $80\%$.

Explanation of Solution

Given information:

According to Secure List, $71.8\%$ of all email sent in July $2012$ was spam.

A system manager suspects that the spam email percentage at his company may be $80\%$.

In an inspection he finds that $382$ out of $500$ emails are spam.

$\alpha =0.05$.

Concepts used:

From the two tailed test.

If $z\ge z_{\frac{\alpha }{2}}\left(=1.96\right)$, then reject the null hypothesis or

If $z\le -z_{\frac{\alpha }{2}}\left(=-1.96\right)$, then reject the null hypothesis.

Calculation:

From part $b$ of this exercise, the value of test statistic is $z=-2.01$ and

From Table $9.1$, the critical value for the level of significance $\alpha =0.05\text{ is }{z}_{\frac{\alpha }{2}}=\pm 1.96$.

From the two tailed test.

If $z\ge z_{\frac{\alpha }{2}}\left(=1.96\right)$, then reject the null hypothesis or

If $z\le -z_{\frac{\alpha }{2}}\left(=-1.96\right)$, then reject the null hypothesis.

Since $z\le -z_{\frac{\alpha }{2}}\left(=-1.96\right)$.

Therefore the null hypothesis is rejected.

Hence, it can be concluded that there is evidence at level of significance $\alpha =0.05$ for the percentage of emails that are spam differs from $80\%$.

(d)

To determine

Whether the percentage of emails that are spam differs from $80\%$ for $\alpha =0.01$.

Answer to Problem 13E

It can be concluded that there is no evidence at level of significance $\alpha =0.01$ for the percentage of emails that are spam differs from $80\%$.

Explanation of Solution

Given information:

According to Secure List, $71.8\%$ of all email sent in July $2012$ was spam.

A system manager suspects that the spam email percentage at his company may be $80\%$.

In an inspection he finds that $382$ out of $500$ emails are spam.

$\alpha =0.05$.

Calculation:

From part $b$ of this exercise, the value of test statistic is $z=-2.01$

Since the test is two-tailed and level of significance $\alpha =0.01$.

From Table $9.1$, the critical value for the level of significance $\alpha =0.01\text{ is }{z}_{\frac{\alpha }{2}}=\pm 2.576$.

Rejection rule:

From the two-tailed test,

If $z\ge z_{\frac{\alpha }{2}}\left(=2.576\right)$, then reject the null hypothesis or

If $z\le -z_{\frac{\alpha }{2}}\left(=-2.576\right)$, then reject the null hypothesis.

Conclusion for $\alpha =0.01$

The value of test statistic is $-2.01$ and the critical value is $z_{\frac{\alpha }{2}}=\pm 2.576$

Here, the value test statistic lies between lower and upper critical value. Therefore, the null hypothesis is not rejected.

Hence, it can be concluded that there is no evidence at level of significance $\alpha =0.01$ for the percentage of emails that are spam differs from $80\%$.