Question

QUESTION 5

1. For each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts. A large shipment has just arrived. A quality control manager randomly selects 50 of the parts from the shipment and finds that 9 parts are defective. Is this sufficient evidence to reject the entire shipment? Use a .05 level of significance to conduct the appropriate hypothesis test.

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H0 : p = 0.10

Ha : p <0.10

n =50

x = 9

= x / n = 9 / 50 = 0.18

P0 = 0.10

1 - P0 = 1 - 0.10 =0.90

Test statistic = z

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

= 0.18 - 0.10/ [(0.10*0.90) / 50]

= 1.88

Test statistic = z =1.88

P(z < 1.88 ) = 0.9699

P-value = 0.9699

= 0.05

P-value >

0.9699 > 0.05

Fail to reject the null hypothesis .

There is insufficient evidence to suggest tha