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 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?

Yes  dog owners in the country spend more time walking their dogs than do dog owners in the city

Step-by-step explanation:

From the question we are told that

    The sample size from country is

     The  sample  size from city is  

     The  sample  mean for  country is  

      The  Sample  mean for  city is  

       The  standard deviation for country  is

        The standard deviation for  city is  

Let the level of significance is  

The null hypothesis is  

The  alternative hypothesis is  

 The  pooled standard deviation is  mathematically represented as

Here  df is the degree of freedom which is mathematically represented as

So

Generally the test statistics is mathematically represented

Generally the p-value  is obtained from the student t-distribution table   table , the value  is  

From the calculation we see that

So we reject the null hypothesis

Hence we can conclude that there is  sufficient evidence to support the claim that dog owners in the country spend more time walking their dogs than do dog owners in the city

i'm pretty sure it's 477,585,296, but perhaps my math is incorrect.

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation: