A recent increase in sales of microchips has forced a computer company to buy a new processing machine to help keep up with demand. The builders of the new machine claim that it produces fewer defective microchips than the older machine. From a random sample of 90 microchips produced on the old machine, 5 were found to be defective. From a random sample of 83 microchips produced on the new machine, 3 were found to be defective. The quality control manager wants to construct a confidence interval to estimate the difference between the proportion of defective microchips from the older machine and the proportion of defective microchips from the new machine.

The final question is;

Why is it not appropriate to calculate a two-sample z-interval for a difference in proportions?

The normality of the sampling distribution of the difference in sample proportions cannot be established.

Step-by-step explanation:

To use the two-sample z-interval for a difference in proportions, some conditions must be met which help us to know the normality of the distribution.

The conditions are;

- The method of sampling for each population must be simple random sampling.

- The samples must be independent.

- Each sample must include at least 10 successes and 10 failures.

- Each population must be a minimum 20 times its sample.

In this case, we are not given the population size and as such we cannot ascertain the last condition I wrote above since the sample proportion we are given alone isn't sufficient. Also, they both don't include at least 10 failures. Thus the normality of the sample proportions cannot be established.

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.