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.

20 points   

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.

20 points   

QUESTION 3

1. Bryce works for Big Cheese Wheel, a cheese company as a quality control manager. Usually, 3% of cheese is bad enough and has to be thrown out. Bryce's new hire, Rachel, found just 1% of cheese is bad (based on testing 205 lbs of cheese). Bryce wonders if Rachel is making mistakes (claiming bad cheese is actually good) or if this is just random and uses a hypothesis test to answer this question. To three decimal places, what is the calculated value of Bryce's hypothesis test?

10 points   

QUESTION 4

1. Using your answer from the previous question, is this statistically significant (note this is a two-tailed test)?

   		Yes, at the 95% level
   		Yes, at the 99% level
   		Yes, at the 99.9% level
   		It is not statistically significant

10 points   

QUESTION 5

1. The average 60-watt (a unit of power) bulb emits about 800 lumens (a measure of brightness) of light. Suppose the population standard deviation is 30 lumens. Also suppose you're testing a new watt bulb to see if it emits more light with the same 60 watts of power. With a sample of 100 bulbs and 99% confidence, what is the minimum average your sample can have and still be counted as statistically significant?

20 points   

QUESTION 6

1. Cassidy wants to know how little excersice someone can do and still lose weight. Suppose she gets 24 volunteers to do 20 minutes of walking a day for a week. She weighs them before and after the week in question and discovers they lost an average of ten pounds with a sample standard deviation of 18. At 99% significance, what was the result of the test?

   		Not statistically significant because the calculated score was less than the critical score.
   		Statistically significant because the calculated score was greater than the critical score.
   		Statistically significant because the calculated score was less than the critical score.
   		Not statistically significant because the calculated score was greater than the critical score.

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