In: Statistics and Probability
An exterminator claims that no more than 10% of the homes he treats have termite problems within 1 year after treatment. In a sample of 100 homes, local officials find that 14 had termites less than 1 year after being treated. At the 0.05 level of significance, evaluate the credibility of the exterminator’s statement. (P-Value Approach)
Solution :
Given that,
= 0.10
1 - = 0.90
n = 100
x = 14
Level of significance = = 0.05
Point estimate = sample proportion = = x / n = 0.14
This a right (One) tailed test.
The null and alternative hypothesis is,
Ho: p = 0.10
Ha: p 0.10
Test statistics
z = ( - ) / *(1-) / n
= ( 0. - 0.10) / (0.10*0.90) /100
= 1.333
P-value = P(Z > z )
= 1 - P(Z < 1.333 )
= 1 - 0.9087
= 0.0913
The p-value is p = 0.0913, and since p = 0.0913 > 0.05, it is concluded that fail to reject the null hypothesis .