Chapter 11.1, Problem 19E

To determine

(a)

To find:

The null and alternative hypotheses.

Answer to Problem 19E

Solution:

The null hypothesis is $H_0:{\mu }_1-{\mu }_2\le 0$ and the alternative hypothesis is $H_1:{\mu }_1-{\mu }_2>0$.

Explanation of Solution

Given information:

Lauren and Keri live in different states and disagree about who has higher electric bills. To settle their disagreement, the girls decided to sample electric bills in their area for the month of June and perform a hypothesis test that the mean monthly residential electric bill is higher for Lauren’s state than for Keri’s state. Is there evidence at the 0.01 Level to say that the mean monthly residential electric bill is higher for Lauren’s state than for Keri’s state.

The null hypothesis is a statement of no difference, that there is no significant difference between the two phenomena. It is considered to be true until it is nullified by statistical evidence for an alternative hypothesis.

An alternative hypothesis is a contradicting statement to the null hypothesis and states a significant difference between the two phenomena It is accepted when the null hypothesis is false.

Calculation:

From the given information:

Let ${\mu }_1$ and ${\mu }_2$ be the mean electric bill of the company in Lauren’s and Keri’s states respectively.

From the given information, the claim is to test that the residential electric bill is higher for Lauren’s state than for Keri’s state, i.e.

$\begin{array}{l}{\mu }_1>{\mu }_2\\ {\mu }_1-{\mu }_2>0\end{array}$

Therefore the null hypothesis is $H_0:{\mu }_1-{\mu }_2\le 0$ and the alternative hypothesis, $H_1:{\mu }_1-{\mu }_2>0$.

To determine

(b)

To find:

Which distribution to be used for test statistic, and the level of significance.

Answer to Problem 19E

Solution:

The distribution used is standard normal variate (one-tailed z test), and the level of significance is 1% or 0.01.

Explanation of Solution

Since both the sample sizes are greater than 30, normal distribution will be followed and the standard normal variate z-test will be applied.

According to the null hypothesis, one-tailed test is to be applied.

From the given information, the level of significance is 1% or 0.01.

To determine

(c)

To find:

The test statistic

Answer to Problem 19E

Solution:

The test statistic is 0.66

Explanation of Solution

Given information:

For Lauren’s state:

The sample size is 35, the mean monthly electric bill is $\$104.53$ and the population standard deviation is $\$17.81$.

For Keri’s state:

The sample size is 51, the mean monthly electric bill is $\$101.48$ and the population standard deviation is $\$25.30$.

Test statistics is a random variable which is calculated from the sample data and used in hypothesis testing.

Test statistic calculates the degrees of acceptance between sample data and null hypothesis.

Formula used:

The test statistic for a hypothesis test for two population means is:

$z=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

Where: ${\overline{x}}_1$ and ${\overline{x}}_2$ are the two sample means,

${\mu }_1-{\mu }_2$ is the hypothetical value of the difference between the two population means,

${\sigma }_1$ and ${\sigma }_2$ are the two population standard deviations,

and $n_1$ and $n_2$ are the two sample sizes.

Calculation:

The test statistic for a given hypothesis is,

$z=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

Substitute $104.53$ for ${\overline{x}}_1$, $17.81$ for ${\sigma }_1$, $35$ for $n_1$, $101.48$ for ${\overline{x}}_2$, $25.30$ for ${\sigma }_2$ and $51$ for $n_2$ in above equation.

$z=\frac{\left(104.53-101.48\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\left(17.81\right)}^2}{35}+\frac{{\left(25.30\right)}^2}{51}}}$

It is given that the null hypothesis for the given proportion is $H_0:{\mu }_1-{\mu }_2\le 0$. Substitute $0$ for ${\mu }_1-{\mu }_2$.

$\begin{array}{rll}z& =& \frac{3.05-0}{\sqrt{9.0627+12.5508}}\\ & =& \frac{3.05}{4.6490}\\ & =& 0.6561\\ & \approx & 0.66\end{array}$

Thus, the test statistic is $0.66$.

To determine

(d)

To find:

The conclusion by comparing the $p$-value to the level of significance.

Answer to Problem 19E

Solution:

The null hypothesis is accepted and it shows that there is no significant difference in the electric bills of the two companies.

There is insufficient evidence to say that the electric bill of the company in Lauren’s state is higher than by Keri’s state.

Explanation of Solution

$p$-value or probability value in a given statistical hypothesis is used to determine the significance of the results. A small $p$-value (mainly $\le 0.05$) indicates evidence against the null hypothesis.

Formula used:

The $p$- value is given by $\text{P}\left(z\ge Z\right)$, where Z is the value of the test statistic. $p$-value is equivalent to the area under the standard normal curve to the right of $z=Z$.

Calculation:

This is a right-tailed test, so p-value = P $\left(z\ge 0.66\right)$

$p$-value is equivalent to the area under the standard normal curve to the right of $z=0.66$

$p$-value $=0.2546$ at 0.01 level of significance.

Conclusion:

Test statistic is $0.66$ and the $p$-value is 0.2546.

Since the $p$-value is $>0.01$, the null hypothesis is accepted.

The null hypothesis is accepted and it shows that there is no significant difference in the electric bills of the two companies.