Lily, a statistician, decided to have a pet in her home, but she had a hard time making up her mind between cat and dog. A friend told her that the average lifespan of dogs is 5 years longer than that of cats. She was more of a cat person; however, if her friend's theory was true, she would get a dog. In order to verify her friend's theory, she investigated pets that had recently passed away and found the lifespan of 12 dogs and 15 cats.
$X$ (Sample of the lifespan of dog)


$Y$ (Sample of the lifespan of cat)


Dog

Lifespan(years)

Cat

Lifespan(years)

1

5.3

1

8.4

2

13.6

2

10.7

3

15

3

9.5

4

14

4

9.8

5

11

5

13

6

9.4

6

8.6

7

12.2

7

12

8

12.8

8

8

9

13.8

9

10.5

10

10

10

6.4

11

8.5

11

12.7

12

15.5

12

6.8



13

15



14

11



15

11.1



In addition to these data, she also learned from a doctor in an animal hospital that the lifespan of dogs is normally distributed with standard deviation equal to 4 years and that of cat is equal to 3 years. Then she went on to test her friend's theory using all the information she collected.
1.

Determine the null hypothesis:


Null Hypothesis: The average difference between the lifespan of dogs and cats is 5 years

2.

Substitute the information into the formula:


$\mathrm{Mean}\left(X\right)=11.6917$, $\mathrm{Mean}\left(Y\right)=10.2333$, $\mathrm{\beta}=5$, ${\mathrm{\sigma}}_{1}=4$, ${\mathrm{\sigma}}_{2}=3$, ${N}_{1}=10$, ${N}_{2}=15$


$z=\frac{\left(\left(11.6916710.23333\right)5\right)}{\sqrt{\left(\frac{{4}^{2}}{12}+\frac{{3}^{2}}{15}\right)}}=2.54715$


$p\mathrm{value}=\mathrm{Probability}\left(\rightZ>2.54715\left\right)=\mathrm{Probability}\left(Z<2.54715\right)+\mathrm{Probability}\left(Z>2.54715\right)=0.0108607$, $Z\u02dc\mathrm{Normal}\left(0,1\right)$


This statistical test provides evidence that the null hypothesis is false, so we reject the null hypothesis.
