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 23 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 their 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 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:

$H_{0}: \mu_{C o}-\mu_{C i}=0$

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

$H_{1}: \mu_{C o}-\mu_{C i}>0$

The standard errors are:

$\begin{aligned} &s_{C o}=\frac{4}{\sqrt{21}}=0.8729 \\ &s_{C i}=\frac{3}{\sqrt{20}}=0.6708 \end{aligned}$

The distribution of the differences has:

$\begin{aligned} &\bar{x}=\mu_{C o}-\mu_{C i}=15-10=5 \\ &s=\sqrt{s_{C o}^{2}+s_{C i}^{2}}=\sqrt{0.8729^{2}+0.6708^{2}}=1.1009 \end{aligned}$

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

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

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

$\begin{aligned} t &=\frac{\bar{x}-\mu}{s} \\ t &=\frac{5-0}{1.1009} \\ t &=4.54 \end{aligned}$

Since the test statistic is $\mathbf{t}=\mathbf{4} . \mathbf{5 4}>\mathbf{2} .32,$ it can be concluded that dog owners in the country spend more time walking their dogs than do dog owners in the city.