Question

The Ford Motor Company, as part of its quality control program, began returning to the supplier all shipments of steel that had defects or faulty chemistry. When Ford began this program, the defective rate in 100 shipments was 9%. A recent survey indicated that 2.2% of 136 shipments were defective. Does this represent a significant improvement in the quality of steel? Test at the 0.05 level. Show your work and upload a photo of your work. Write the null hypothesis, show the test statistic, the decision rule, and solve.

Answer 1

Ans:

sample proportion=0.022

sample size,n=136

Test statistic:

z=(0.022-0.09)/sqrt(0.09*(1-0.09)/136)

z=-2.77

p-value=P(z<-2.77)=0.0028

critical z value=-1.645

Rejection region: z<-1.645

Decision rule: Reject H0,if z<-1.645

As,p-value<0.05 or test statistic z is less than -1.645,we reject H0.

There is sufficient evidence to conclude that there is signficant improvement in quality of steel(as percentage of defective is decreased)