Question

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 21 country owners and 20 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 3.0. The mean number of hours country owners spent walking theirs dogs per week was 15.0. The standard deviation of the number of hours spent walking the dog by owners in the country was 4.0. Do dog owners in the country spend more time walking their dogs than do dog owners in the city?




Using an alpha level of .05 (t= 2.32), what is the conclusion you are entitled to draw as a result of this test?

Answer 1

Answer:

Yes, there is enough evidence to support the claim that dog owners in the country spend more time walking their dogs than do dog owners in the city (P-value=0.0000263).

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that dog owners in the country (sample 2) spend more time walking their dogs than do dog owners in the city (sample 1).

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0$

The significance level is 0.05.

The sample 1, of size n1=20 has a mean of 10 and a standard deviation of 3.

The sample 2, of size n2=21 has a mean of 15 and a standard deviation of 4.

The difference between sample means is Md=-5.

$M_d=M_1-M_2=10-15=-5$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{20}+\dfrac{4^2}{21}}\\\\\\s_{M_d}=\sqrt{0.45+0.762}=\sqrt{1.212}=1.101$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-5-0}{1.101}=\dfrac{-5}{1.101}=-4.54$

The degrees of freedom for this test are:

$df=n_1+n_2-1=20+21-2=39$

This test is a left-tailed test, with 39 degrees of freedom and t=-4.54, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t<-4.54)=0.0000263$

As the P-value (0.0000263) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that dog owners in the country spend more time walking their dogs than do dog owners in the city.

Answer 2

Using the t-distribution, it is found that since the test statistic is t = 4.54 > 2.32, it can be concluded that dog owners in the country spend more time walking their dogs than do dog owners in the city.

At the null hypothesis, we test if dog owners in the country and in the city spend the same amount of time walking their dogs, that is:

$H_0: \mu_{Co} - \mu_{Ci} = 0$

At the alternative hypothesis, we test if dog owners in the country spend more time, that is:

$H_1: \mu_{Co} - \mu_{Ci} > 0$

The standard errors are:

$s_{Co} = \frac{4}{\sqrt{21}} = 0.8729$

$s_{Ci} = \frac{3}{\sqrt{20}} = 0.6708$

The distribution of the differences has:

$\overline{x} = \mu_{Co} - \mu_{Ci} = 15 - 10 = 5$

$s = \sqrt{s_{Co}^2 + s_{Ci}^2} = \sqrt{0.8729^2 + 0.6708^2} = 1.1009$

We have the standard deviation for the samples, hence, the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu$ is the value tested at the null hypothesis, for this problem $\mu = 0$, hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{5 - 0}{1.1009}$

$t = 4.54$

Since the test statistic is t = 4.54 > 2.32, it can be concluded that dog owners in the country spend more time walking their dogs than do dog owners in the city.

A similar problem is given at https://brainly.com/question/17192140