Question

A veterinarian studying causes of enteroliths in horses suspects feeding too much alfalfa may be a culprit. The veterinarian hypothesizes that more than two thirds of all horses with enteroliths are fed mostly alfalfa. The veterinarian has access to horses that were examined at the veterinary clinic over the last five years. To find out about the diets of the horses it is necessary to contact the person responsible for feeding the horse prior to the diagnosis of enteroliths. This is time consuming and often it is not possible to contact the right person. It is more difficult the more time has elapsed. Therefore, the veterinarian decides to select the 62 horses diagnosed during the past five years of which 49 were fed mostly alfalfa. A z-test for H0: p = 0.67, Ha: p > 0.67. The P-value for this test is: 0.044. 0.066. 0.050. 0.022.

Answer 1

Answer: 0.022

Step-by-step explanation:

Let p be the population proportion of horses were fed mostly alfalfa.

We are given that :

$H_0: p=0.67\\ H_a:p>0.67$ , since alternative hypothesis ($H_a$) is right tailed , so the test is a right-tailed test.

Also, we have

Sample size= 62

Sample proportion= $\hat{p}=\dfrac{49}{62}\approx0.79$

Test statistic for population proportion:-

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Substitute all values ,

$z=\dfrac{0.79-0.67}{\sqrt{\dfrac{0.67(1-0.67)}{62}}}\\\\=\dfrac{0.12}{\sqrt{0.00356612903226}}\\\\=\dfrac{0.12}{0.059717}\\\\=2.00947551836\approx2.01$

P-value for right tailed test :

P(z>2.01)=1-P(z<2.01)              [∵P(Z>z=1-P(Z<z))]

=1-0.9777844                     [By using z-value table]

=0.0222156≈0.022

Hence, the P-value for this test is 0.022 .