Question

QUESTION 1

1. Joyce works in a toy factory maintaining machines. Machines should inject exactly 500 grams of plastic into a mold to make a toy plane. Joyce wonders if these toy machines might need cleaning, but it's a lot of work taking apart the machines to do a proper job. She grabs planes from 24 different machines at random and weighs them. Her average is 495 grams. She knows there's a standard devation of 10 grams for all machines (Joyce has been doing this for a while). What is Joyce's conclusion as a result of this test? (Check all that apply.) Note the critical z value for 90% confidence is 1.645.

   		Fail to reject the null hypothesis.
   		Reject the null hypothesis at 90% confidence.
   		Reject the null hypothesis at 95% confidence.
   		Reject the null hypothesis at 99% confidence.

QUESTION 2

1. Now suppose Joyce doesn't have population standard deviation (perhaps because these are new machines); she only has her sample standard deviation which is 12. Using the same information, (500 grams average, sample mean of 495, sample size of 24), what are the results of Joyce's analysis? (Check all that apply.) You will need to use one of the tables available on Blackboard (knowing which one is part of the question).

   		Fail to reject the null hypothesis.
   		Reject the null hypothesis at 90% confidence.
   		Reject the null hypothesis at 95% confidence.
   		Reject the null hypothesis at 99% confidence.

Answer 1

Question 1)

Answer)

As the population s.d is known here we can use standard normal z table to estimate the interval

Null hypothesis Ho : u = 500

Alternate hypothesis Ha : u not equal to 500

Critical value z for 90% confidence level from z table is 1.645

Margin of error (MOE) = Z*S.D/√N = 1.645*10/√24 = 3.36

Interval is given by

(Mean - MOE, Mean + MOE)

(495-3.36, 495+3.36)

(491.64, 498.36)

As the interval does not have 500(null hypothesised value in it) we reject the null hypothesis Ho

Reject the null hypothesis at 90% confidence

Question 2)

Now here we have sample s.d so we will use t distribution

Degrees of freedom is = n-1 = 23

For 23 dof and 90% confidence level, critical value t from t table is 1.71

Margin of error (MOE) = t*s.d/√n = 1.71*12/√24 = 4.2

Interval is given by

(Mean - MOE, Mean + MOE)

[490.8, 499.2].

You can be 90% confident that the population mean (μ) falls between 490.8 and 499.2.

This interval also does not contain 500

Reject the null hypothesis at 90% confidence