Chapter 11.2, Problem 8E

To determine

(a)

To state:

Null hypothesis and alternative hypothesis.

Answer to Problem 8E

Solution:

Null Hypothesis is

$H_0:{\mu }_1-{\mu }_2\ge 0$

Alternative Hypothesis is

$H_1:{\mu }_1-{\mu }_2\text{< }0$

Explanation of Solution

Given:

A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 11 days following no advertisements, the mean was 17.8 purchasing customers with a standard deviation of 3.5 customers. On 6 days following advertising, the mean was 22.8 purchasing customers with a standard deviation of 2.8 customers. Assume that the population variances are equal.

Let the days when no advertisements are played on radio corresponds to poupulation1 and the days when advertisements are played corresponds to population 2. Let ${\mu }_1$ be the mean number of customers of population 1 and ${\mu }_2$ mean number of customers. When written mathematically, this is ${\mu }_1<{\mu }_2$ and hence the alternative hypothesis. By subtracting ${\mu }_2$ from both sides of the above inequality result is ${\mu }_1-{\mu }_2<0$.

The null hypothesis is that the mean number of customers is not lower on days following no advertisements than the mean number of customers on days following advertisements. The null hypothesis is stated as follows-

$H_0:{\mu }_1-{\mu }_2\ge 0$

Alternative hypothesis-.

$H_1:{\mu }_1-{\mu }_2\text{< }0$

Thus, the hypotheses are stated as follows:

$H_0:{\mu }_1-{\mu }_2\ge 0$

$H_1:{\mu }_1-{\mu }_2\text{< }0$

To determine

(b)

The type of distribution to use for the test statistics and state the level of significance.

Answer to Problem 8E

Solution:

The t-test statistic for equal variances is appropriate and the level of significance for this test is.

$\alpha =0.01$

Explanation of Solution

Given:

A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 11 days following no advertisements, the mean was 17.8 purchasing customers with a standard deviation of 3.5 customers. On 6 days following advertising, the mean was 22.8 purchasing customers with a standard deviation of 2.8 customers. Assume that the population variances are equal.

Given that the difference between two population means when both population variances are unknown but assumed to be equal and random samples drawn are independent. Both the population distributions are approximately normal. The t-test statistic for equal variances is appropriate. The level of significance for this test is $\alpha =0.01$.

Therefore, the t-test statistic for equal variances is appropriate and the level of significance for this test is $\alpha =0.01$.

To determine

(c)

To calculate:

The test statistic.

Answer to Problem 8E

Solution:

The test statistic is -3.0006.

Explanation of Solution

Given:

A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 11 days following no advertisements, the mean was 17.8 purchasing customers with a standard deviation of 3.5 customers. On 6 days following advertising, the mean was 22.8 purchasing customers with a standard deviation of 2.8 customers. Assume that the population variances are equal.

Formula used:

The test statistic for a hypothesis test for two population means is given by

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{(n_1-1)s_1{}^2+(n_2-1)s_2{}^2}{n_1+n_2-2}}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Such that both population standard deviations are unknown and assumed to be equal, the samples taken are independent, simple random samples, and either both sample sizes are at least 30 or both population distributions are approximately normal. The respective sample means are ${\overline{x}}_1$ and ${\overline{x}}_2$, ${\mu }_1-{\mu }_2$ is the presumed value of the difference between the two population means from the null hypothesis, also the respective sample standard deviation are $s_1$, $s_2$ and the respective sample sizes are $n_1$ and $n_2$.

The number of degrees of freedom for the t-distribution of the test statistic is $n_1+n_2-2$.

Calculation:

Given information, the sample means are ${\overline{x}}_1=17.8$ and ${\overline{x}}_2=22.8$. The sample standard deviations are $s_1=3.5$ and $s_2=2.8$, the sample sizes are $n_1=11$ and $n_2=6$. The null hypothesis is taken as $H_0:{\mu }_1-{\mu }_2\ge 0$.

The standard deviation for the first sample is $s_1=3.5$. The variance for the first sample is calculated as,

$\begin{array}{rll}{s}_1{}^2& =& 3.5\times 3.5\\ & =& 12.25\end{array}$... $(1)$

The standard deviation for the second sample is $s_2=2.8$.The variance for the second sample is calculated as,

$\begin{array}{rll}{s}_2{}^2& =& 2.8\times 2.8\\ & =& 7.84\end{array}$... $(2)$

The test statistic for a given hypothesis test is

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{(n_1-1)s_1{}^2+(n_2-1)s_2{}^2}{n_1+n_2-2}}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

From equation $(1)$ and $(2)$, substitute $17.8$ for ${\overline{x}}_1$, $22.8$ for ${\overline{x}}_2$, $12.25$ for ${\text{s}}_{\text{1}}{}^2$, $54.76$ for ${\text{s}}_2{}^2$, $12$ for $n_1$, and $15$ for $n_2$ in above equation

$t=\frac{\left(17.8-22.8\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{(11-1)\times 12.25+(6-1)\times 7.84}{11+6-2}}\sqrt{\frac{1}{11}+\frac{1}{6}}}$

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

$\begin{array}{rll}t& =& \frac{\left(17.8-22.8\right)-0}{\sqrt{\frac{(11-1)\times 12.25+(6-1)\times 7.84}{11+6-2}}\sqrt{\frac{1}{11}+\frac{1}{6}}}\\ & =& -3.0006\end{array}$

Thus, the test statistic is -3.0006.

To determine

(d)

To draw:

The conclusion and interpret the decision.

Answer to Problem 8E

Solution:

The null hypothesis is rejected and it is concluded that there is sufficient evidence at the 0.01 level of significance to support the claim that the mean number of customers is not lower on days following no advertisements than the mean number of customers on days following advertisements

Explanation of Solution

Given:

A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 11 days following no advertisements, the mean was 17.8 purchasing customers with a standard deviation of 3.5 customers. On 6 days following advertising, the mean was 22.8 purchasing customers with a standard deviation of 2.8 customers. Assume that the population variances are equal.

Calculation:

Rejection Regions for hypothesis tests for two population means reject the null hypothesis, $H_0$ if:

$t\le -t_{\alpha }$ for a left tailed test

$t\ge t_{\alpha }$ for a right tailed test

$|t|\le t_{\frac{\alpha }{2}}$ for a two tailed test

The alternative hypothesis contains “$<$ “this is a left tailed test. Since the population variances are equal, the number of degrees of freedom for this test is

$\begin{array}{rll}{n}_1+n_2-2& =& 11+6-2\\ & =& 15\end{array}$.

It is given that $\alpha =0.01$ The critical t-value for a left tailed test with 15 degrees of freedom and 0.01 level of significance is

$t_{\alpha }=2.602$

The calculated value of the test statistic is -3.0006 less than the critical value; it does falls in the rejection region. The null hypothesis is rejected.

Thus, it is concluded that there is sufficient evidence at the 0.01 level of significance to support the claim the mean number of customers is not lower on days following no advertisements than the mean number of customers on days following advertisements