Easy Talk phones have an electronic component that is produced by WiredUp Inc. Under normal operations 8% of these components are defective. Every week, the production process is tested by inspecting 10 of the new components produced. If 2 (or 20%) or more are defective they shut down the production process for further inspection. The null and alternative hypotheses of the test are:

(null) H0 : π = 0.08

(alternative) Ha : π > 0.08

Where π is the proportion of components being produced that are defective at the time of the test.

(a) What is the test statistic that is being used in the test?

(b) What is the rejection region?

(c) Use an appropriate applet to determine the probability they make a Type I Error and shut down the production process unnecessarily.

(d) Suppose the machines producing the electronic component get out of adjustment and 15% of the components being produced are defective. Use an appropriate applet to find the probability the test will reveal that the machine is out of adjustment.

(e) Now use a theory-based method (and not an applet) to calculate the probability in (d) again.

(f) Both the calculations in (d) and (e) are approximate. Which do you expect to be more accurate? Explain.

(g) Suppose the change from an 8% to 15% defective rate is important for the company to detect. Briefly describe how the company might modify the test, without increasing the probability of a Type I Error, to increase the probability of detecting that the alternative hypothesis is true when the defective rate is 15%.

Answer and Step-by-step explanation:

(a)

The test statistic used in the test is number of new components produced which are defective.

(b)

We reject the null hypothesis H0 if the number of new components produced which are defective is greater than or equal to 2.

(c)

Probability of Type I Error = p(Reject H0 | π = 0.08)

= p(X \ge 2 | π = 0.08) = p(X > 1 | π = 0.08)

= 0.1878825

(Since X ~ Binomial(n = 10, π = 0.08) we use R code pbinom(1, 10, 0.08, lower.tail = FALSE) to find the above value)

(d)

Probability of Type I| Error = p(Fail to Reject H0 | π = 0.15)

= p(X < 2 | π = 0.15) = p(X \le 1 | π = 0.15)

= 0.5442998

(Since X ~ Binomial(n = 10, π = 0.15) we use R code pbinom(1, 10, 0.15) to find the above value)

(e)

When π = 0.15,

Standard error of proportion, SE = \sqrt{\pi(1-\pi)/n} = \sqrt{0.15*(1-0.15)/10} = 0.1129159

Sample proportion for X = 2 is p = 0.2

Probability of Type I| Error = p(Fail to Reject H0 | π = 0.15)

= p(X < 2 | π = 0.15) = p(\hat{p} < 0.2 | π = 0.15)

= p[Z < (0.2 - 0.15)/0.1129159]

= p[Z < 0.4428]

= 0.6710448

(f)

The applet calculations will be more accurate as in theory-based we have appoximated the binomial distribution by Normal distribution.

(g)

To increase the probability of detecting that the alternative hypothesis is true when the defective rate is 15%, we need to increase the sample size. The sample size reduces the standard error and the probability of Type II error.

in mathematics, a constraint is a condition of an optimization problem that the solution must satisfy. there are several types of constraints—primarily equality constraints, inequality constraints, and integer constraints. the set of candidate solutions that satisfy all constraints is called the feasible set.

hoped i'm right lol