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

`z=\frac{0.25-0.2}{\sqrt{0.221(1-0.221)((1)/(120)+(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.

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)=(30)/(120)=0.25` represent the proportion of old pit bulls with tooth decay

`p_(2)=(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) \\eq 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)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(2)+n_(2))=(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)((1)/(120)+(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.