Answer to Question #255330 in Statistics and Probability for Zee

Question

The park officials of Kruger Park believe that 50% of the buffaloes in the southern part have tuberculosis.

The characteristic being studied is the occurrence of tuberculosis. A random sample of

100 buffalo is obtained from the southern part of the Kruger Park for which 65 are tested positively

for tuberculosis.

(a) Do the data indicate that the proportion of buffalo in the southern part that have tuberculosis

is greater than assumed? Perform a proper hypothesis test and test at the 5% significance

level. Clearly show how you draw a conclusion when you test this hypothesis, using both

methods, i.e.

critical value approach, and p-value approach

Answer 1

The null and alternate hypothesis

$H_o : P=0.50\\ H_1 : P>0.50\\$

Test statistic

Where n = 100, x= 65

$p= \frac{x}{n}= \frac{65}{100}=0.65\\ p_o=0.50\\ z= \frac{p-p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}= \frac{0.65-0.50}{\sqrt{\frac{0.50(1-0.50)}{100}}}=3$

Critical value:

$z_{\alpha}=z_{0.05}=1.64\\ z= 3> 1.64$

Reject the null hypothesis.

P-value $P[z>3]=0.0013$

P-value $=0.0013< \alpha = 0.03$

Therefore reject the null hypothesis.

Conclusion

There is sufficient evidence to support that the proportion of buffalo in the southern part that has tuberculosis is greater than assumed.

95% lower one-sided confidence bound

$=p-z_{\alpha /2} * \sqrt{\frac{p(1-p)}{n}}=0.65-1.96 * \sqrt{\frac{0.65(1-0.65)}{100}}=0.56 <0.60$

Conclusion

There is evidence that buffalo in the southern part that has tuberculosis is greater if the occurrence of tuberculosis is more than an additional 10 % of the existing proportion.