Question

You are the manager of a manufacturing company that produces both plumbuses and dinglebops. Each plumbus is defective with probability .15 and each dinglebop is defective with probability .1. In a case like this, the variance of the number of defective items of each type is as follows. If the defective probability is p, the variance σ^2 is n*p*(1-p).

A) If you produce 500 of each type of item what is the expected number and standard deviation of the TOTAL number of defective items?
Expected number:______
Standard deviation:______

B) Of the 500 plumbuses produced, you will be fired if at least 90 are defective. Assuming the number of defective plumbuses is normally distributed, what is the probability of you being fired?

C) You are now interested in computing the number of plumbuses your company can produce in a day. You collect data over the next 100 days and find the sample mean to be 47 plumbuses and the sample standard deviation to be 9. Compute a 56% confidence interval for the number of plumbuses produced in a day.
( _____ , _____ )

D) A rival manufacturer reports that over 20 days they produced an average of 45 plumbuses per days with a sample standard deviation of 5. You wish to show that your company does better using the information from part C. Carry out a hypothesis test with α = .01 to try to statistically make your case. What is the standard error we should use, what is your p-value, and what is your conclusion?
SE: ____
p-value: ____

Pick one of these conclusions:

Fail to reject the null hypothesis because the p-value is greater than .01

Reject the null hypothesis because the p-value is greater than .01

Fail to reject the null hypothesis because the p-value is less than .05

Reject the null hypothesis because the p-value is greater than .05

Reject the null hypothesis because the p-value is less than .05

Reject the null hypothesis because the p-value is less than .01

Fail to reject the null hypothesis because the p-value is less than .01

Fail to reject the null hypothesis because the p-value is greater than .05

Answer 1

A)

Let X shows the number of defective plumbus and Y shows the number of defective dinglebop.

Here X has binomial distribution with parameters as follow:

n = 500 and p = 0.15

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And Y has binomial distribution with parameters as follow:

n = 500 and p = 0.1

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Now let T shows the total number of defective. So,

T = X+Y

T will be have approximate normal distribution with following parameters

Expected number: 122
Standard deviation:10.4283

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B)

Using continuity correction factor we need to find the probability

P(X >= 90) = P(X > 89.5)

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c)

d)

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The p-value is

p-value = 0.0851

Fail to reject the null hypothesis because the p-value is greater than .01

Note: Option "Fail to reject the null hypothesis because the p-value is greater than .05" is also correct.