Answer:
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 [tex]n_1 = 21[/tex]
The sample size from city is [tex]n_2 = 23[/tex]
The sample mean for country is [tex]\= x_1 = 15 [/tex]
The Sample mean for city is [tex]\= x_2 = 10[/tex]
The standard deviation for country is [tex]\sigma _1 = 4[/tex]
The standard deviation for city is [tex]\sigma _2 = 3[/tex]
Let the level of significance is [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o : \mu_1 = \mu_2[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 > \mu_2[/tex]
The pooled standard deviation is mathematically represented as
[tex]s = \sqrt{ \frac{s_1 ^2 * (n_1 - 1 ) + s_2 ^2 * (n_2 - 1 )}{ df} }[/tex]
Here df is the degree of freedom which is mathematically represented as
[tex]df = n_1 + n_2 - 2[/tex]
[tex]df = 21 + 23 -2 [/tex]
[tex]df = 42[/tex]
So
[tex]s = \sqrt{ \frac{4 ^2 * (15.0 - 1 ) + 3 ^2 * (10 - 1 )}{ 42} }[/tex]
[tex]s = 3.5[/tex]
Generally the test statistics is mathematically represented
[tex]t = \frac{\= x_1 - \= x_2 }{ s * \sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex]
[tex]t = \frac{ 15 -10 }{ 3.5 * \sqrt{\frac{1}{ 21 } + \frac{1}{23} } }[/tex]
[tex]t = 4.733[/tex]
Generally the p-value is obtained from the student t-distribution table table , the value is
[tex]P(T > 4.733)= t_{4.733, 42 } = 0.000013 [/tex]
From the calculation we see that
[tex]p-value < \alpha[/tex]
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
