In: Statistics and Probability
A company that packages peanuts states that at a maximum 6% of the peanut shells contain no nuts. At random, 300 peanuts were selected and 21 of them were empty.
With a significance level of 1%, can the statement made by the company be accepted? What is the P-Value?
Solution:
Point estimate = sample proportion = 
 = x / n = 0.07
This a right (One) tailed test.
The null and alternative hypothesis is,
Ho: p = 0.06
Ha: p 
 0.06
Test statistics
z = (
- 
 ) / 
*(1-
)
/ n
= ( 0.07 - 0.06) / 
 (0.06*0.94) /300
= 0.729
P-value = P(Z > z )
= 1 - P(Z < 0.729 )
= 1 - 0.7670
= 0.2330