Question

A veterinarian wants to know if pit bulls or golden retrievers have a higher incidence of tooth decay at the age of three. The vet surveys 120 three-year old pit bulls and finds 30 of them have tooth decay. The vet then surveys 160 three-year old golden retrievers and finds 32 of them have tooth decay. Number the population of pit bulls and golden retrievers by 1 and 2, respectively. Refer to Exhibit 10.12. At the 10% significance level, can the vet conclude the proportion of pit bulls that have tooth decay is different than the proportion of golden retrievers that have tooth decay?

Answer 1

Answer:

$z=\frac{0.25-0.2}{\sqrt{0.221(1-0.221)(\frac{1}{120}+\frac{1}{160})}}=0.998$    

$p_v =2*P(Z>0.998)= 0.318$  

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the we have significant differences between the two proportions.  

Step-by-step explanation:

1) Data given and notation

$X_{1}=30$ represent the number of old pit bulls with tooth decay

$X_{2}=32$ represent the number of golden retrievers  with tooth decay

$n_{1}=120$ sample selected for 1

$n_{2}=160$ sample selected for 2  

$p_{1}=\frac{30}{120}=0.25$ represent the proportion of old pit bulls with tooth decay

$p_{2}=\frac{32}{160}=0.2$ represent the proportion of golden retrievers  with tooth decay

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)  

$\alpha=0.1$ significance level given

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:  

Null hypothesis:$p_{1} - p_{2}=0$  

Alternative hypothesis:$p_{1} - p_{2} \neq 0$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)  

Where $\hat p=\frac{X_{1}+X_{2}}{n_{2}+n_{2}}=\frac{30+32}{120+160}=0.221$  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.25-0.2}{\sqrt{0.221(1-0.221)(\frac{1}{120}+\frac{1}{160})}}=0.998$    

4) Statistical decision

Since is a two side test the p value would be:  

$p_v =2*P(Z>0.998)= 0.318$  

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the we have significant differences between the two proportions.