Question

1.An investigator predicts that dog owners in the country spend more time walking their dogs than do dog owners in the city. The investigator gets a sample of 12 country owners and 15 city owners. The mean number of hours per week that city owners spend walking their dogs is 10.0. The standard deviation of hours spent walking the dog by city owners is 4.0. The mean number of hours country owners spent walking their dogs per week was 12.0. The standard deviation of the number of hours spent walking the dog by owners in the country was 5.0. Do dog owners in the country spend more time walking their dogs than do dog owners in the city?

Accomplish the table below:

Dog Owners

Mean

Standard Deviation

Sample Size

Country

City

Answer 1

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{5^2}{4^2}=1.5625$$

The critical values for $\alpha=0.05, df_1=n_1=1=11$degrees of freedom, $df_2=n_2=1=14$ degrees of freedom, are $F_L = 0.2977$ and $F_U = 3.0946,$ and since $F = 1.563,$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\le\mu_2$

$H_a:\mu_1>\mu_2$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=n_1+n_2-2=12+15-2=25$$

Based on the information provided, the significance level is $\alpha = 0.05,$

and the degrees of freedom are $df = 25$ degrees of freedom.

Hence, it is found that the critical value for this right-tailed test is $t_c = 1.708141,$ for $\alpha = 0.05$ and $df = 25.$

The rejection region for this right-tailed test is $R = \{t: t > 1.708141\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{12-10}{\sqrt{\dfrac{(12-1)(5)^2+(15-1)(4)^2}{12+15-2}(\dfrac{1}{12}+\dfrac{1}{15})}}$$

$$\approx1.1559$$

Since it is observed that $t = 1.1559 \le t_c = 1.708141,$

it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for right-tailed test, $df=25$ degrees of freedom, $t=1.1559,$ is $p=0.129325,$ and since $p = 0.129325 \ge 0.05=\alpha,$ it is concluded that the null hypothes is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$

is greater than $\mu_2,$ at the $\alpha = 0.05$ significance level.