Question

A machine, when working properly, produces only 2% defective items. Periodically a sample is taken and tested to make sure that the machine is working properly. Suppose a random sample of 121 items produced by this machine was tested and was found to have 5 defective items at α=0.05, what is the critical proportion to compare to the sample proportion when deciding on whether the proportion of the population must be > 0.02? Answer in 4 decimal places.

Please show work.

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H0 : p = 0.02

Ha : p > 0.02

n = 121

x = 5

= x / n = 5 / 121 = 0.04

P0 = 0.02

1 - P0 = 1 - 0.02 = 0.98

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.04 - 0.02/ [(0.02*0.98) /121 ]

= 1.57

Test statistic = z = 1.57

P(z >1.57 ) = 1 - P(z < 1.57 ) = 1 -0.9418

P-value = 0.0582

= 0.05

P-value >

0.0582 > 0.05

Fail tom reject the null hypothesis .

There is not sufficient evidence to conclude that the population propration is greater than 2%, at the 0.05 significance level.