Chapter 11.1, Problem 18E

To determine

(a)

To find:

The null and alternative hypotheses.

Answer to Problem 18E

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:

Joe wants to get the best possible price on a used luxury car. He lives near the border of two states, and he believes that prices are better across the state line, i.e., the mean selling price for used luxury cars is lower in the other 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:

${\mu }_1>{\mu }_2$

Let ${\mu }_1$ and ${\mu }_2$ be the mean selling prices of used luxury cars at Joe’s state and the other state respectively. From the given information, the claim made is that the mean selling prices of used luxury cars are better in the other 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 18E

Solution:

The distribution used is standard normal variate (one-tailed z test), and the level of significance is 5% or 0.05.

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 5% or 0.05.

To determine

(c)

To find:

The test statistic

Answer to Problem 18E

Solution: The test statistic is 2.05

Explanation of Solution

Given information:

For Joe’s state:

The sample size is 31, the mean selling price is $\$62,065$ and the population standard deviation is $\$1,625$.

For the other state:

The sample size is 40, the mean selling price is $\$61,300$ and the population standard deviation is $\$1,475$.

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 $62065$ for ${\overline{x}}_1$, $1625$ for ${\sigma }_1$, $31$ for $n_1$, $61300$ for ${\overline{x}}_2$, $1475$ for ${\sigma }_2$ and $40$ for $n_2$ in above equation.

$z=\frac{\left(62065-61300\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\left(1625\right)}^2}{31}+\frac{{\left(1475\right)}^2}{40}}}$

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

$\begin{array}{rll}z& =& \frac{765-0}{\sqrt{85181.4516+54390.6250}}\\ & =& \frac{765}{373.5935}\\ & =& 2.0477\\ & \approx & 2.05\end{array}$

Thus, the test statistic is $2.05$.

To determine

(d)

To find:

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

Answer to Problem 18E

Solution:

The null hypothesis is and it shows the selling prices of used luxury cars are better in the other state.

Yes, there is sufficient evidence to show that the mean selling prices of used luxury cars are lower in the other 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 $\left(z\ge 2.05\right)$

$p$-value is equivalent to the area under the standard normal curve to the right of $z=2.05$

$p$-value $=0.0202$ at 0.05 level of significance.

Conclusion:

Test statistic is $-1.67$ and the $p$-value is 0.0202.

Since the $p$-value is $<0.05$, the null hypothesis is rejected.

This shows the selling prices of used luxury cars are better in the other state.