Week 5 LectureGM 533Applied Managerial Statistics

B. Heard(Do not copy, post or distribute without my permission. Students are free to download a copy for personal use)

GM 533 Week 5

Checkpoint Examples

GM 533 Week 5

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GM 533 Week 5

•A new computer company keeps up with its customers’ satisfaction on a 10 point scale. The company thinks their customer satisfaction rating is at least 8.2. Suppose that the company wishes to use the random sample of 33 satisfaction ratings to provide evidence supporting the claim that the mean customers’ satisfaction rating exceeds 8.2.

GM 533 Week 5• a: Letting μ represent the mean customer satisfaction

rating for the company, set up the null hypothesis H0 and the alternative hypothesis Ha needed if we wish to attempt to provide evidence supporting the claim that μ exceeds 8.2.

• b: The random sample of 33 satisfaction ratings yields a sample mean of x- = 8.295 . Assuming that s equals 0.28, use critical values to test H0 versus Ha at each of α = .10, .05, .01, and .001.

• c: Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of α = .10, .05, .01, and .001.

• d: How much evidence is there that the mean customer satisfaction rating exceeds 8.2?

GM 533 Week 5

•a: Letting μ represent the mean customer satisfaction rating for the company, set up the null hypothesis H0 and the alternative hypothesis Ha needed if we wish to attempt to provide evidence supporting the claim that μ exceeds 8.2.

GM 533 Week 5

a: Since the claim is that the mean exceeds 8.2, it will be our alternative. If they noted it was “at least 8.2” it would be the null. The null always contains equality (it will be either =, ≤ or ≥ ). So we have

Ho: μ ≤ 8.2Ha: μ > 8.2 (claim)

GM 533 Week 5

•b: The random sample of 33 satisfaction ratings yields a sample mean of x- = 8.295 . Assuming that s equals 0.28, use critical values to test H0 versus Ha at each of α = .10, .05, .01, and .001.

GM 533 Week 5Calculate z, I used a template I made.

Compare to values at .10, .05, .01, and .001.

I calculate z to be 1.949

Sample MeanPopulation Mean Sample Std Dev. Sample Size

8.295 8.2 0.28 33

Calculated z to compare to critical values

1.949

GM 533 Week 5 I then compared this value to the z’s calculated

for the various alpha’s (on my template also).I compared to the “Right Tailed” column since we

are dealing with “greater than.”

z Right Tailed

0.1 1.28

0.05 1.64

0.01 2.33

0.001 3.09

As you can see z’s at 0.1 and 0.05 are less than 1.949, but z’s at 0.01 and 0.001 are greater than 1.949. So we would reject Ho at 0.1 and 0.05, but not at 0.01 and 0.001.

GM 533 Week 5

c: Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of α = .10, .05, .01, and .001.

GM 533 Week 5Using your calculated z, find the p value (I used template).Compare to alpha’s of 0.1, 0.05, 0.01 and 0.001. The

calculated value (right tailed value) of 0.0256 again falls in between 0.05 and 0.01. So we would reject Ho at 0.1 and 0.05, but not at 0.01 and 0.001.

P-Value Method

z Left Tailed Right Tailed Two Tailed

1.949 0.9744 0.0256 0.0513

GM 533 Week 5

•d: How much evidence is there that the mean customer satisfaction rating exceeds 8.2?

GM 533 Week 5

How much evidence is there?Well, at alpha’s of 0.1 and 0.05, we would

accept the claim that the mean customer satisfaction rating is more than 8.2.

At alpha’s of 0.01 and 0.001, we would not be able to reject the null hypothesis than the satisfaction score is 8.2 or less.

So there is pretty strong evidence (in the 95% confidence range, think 1- alpha)

GM 533 Week 5

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GM 533 Week 5• The Energy Shotz Energy Drink Company has just

installed a new bottling process that will fill 8-ounce bottles of the popular Super Shotz Energy Drink. Both overfilling and underfilling bottles are undesirable: Underfilling leads to customer complaints and overfilling costs the company considerable money and loss of product. In order to verify that the filler is set up correctly, the company wishes to see whether the mean bottle fill, μ, is close to the target fill of 8 ounces. To this end, a random sample of 64 filled bottles is selected from the output of a test filler run. If the sample results cast a substantial amount of doubt on the hypothesis that the mean bottle fill is the desired 8 ounces, then the filler’s initial setup will be readjusted.

GM 533 Week 5

•a) The energy drink company wants to set up a hypothesis test so that the filler will be readjusted if the null hypothesis is rejected. Set up the null and alternative hypotheses for this hypothesis test.

GM 533 Week 5

We want to know if it’s 8 or not.

Ho: μ = 8 (claim)Ha: μ ≠ 8

GM 533 Week 5•b) Suppose that Company has just installed

a new bottling process that will fill 8-ounce decides to use a level of significance of α = .01, and suppose a random sample of 64 bottle fills is obtained from a test run of the filler. For each of the following sample mean, x- = 8.02 − determine whether the filler’s initial setup should be readjusted. Use a critical value, a p-value, and a confidence interval. Assume that s equals .1.

GM 533 Week 5• The calculated z of 1.60 is within the bounds of +/- 2.58, so

we can not reject Ho.

z (alpha) Left Tailed Right TailedTwo Tailed

(+/-)

0.1 -1.28 1.28 1.64

0.05 -1.64 1.64 1.96

0.01 -2.33 2.33 2.58

0.001 -3.09 3.09 3.29

GM 533 Week 5•The p value of 0.1096 (Two tailed because

of = sign) is greater than 0.01, so we can not reject Ho.

P-Value Method

z Left Tailed Right Tailed Two Tailed1.600 0.9452 0.0548 0.1096

Compare to0.1

0.050.01

0.001

GM 533 Week 5• The confidence interval is [7.99, 8.05],

we are good to go. No readjustment needed!

Confidence Interval

C (1 - alpha) s n (sample size) x bar Left Right

0.99 0.1 64 8.02 7.9878 8.0522Margin of

Error

0.0322

0.995

GM 533 Week 5

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GM 533 Week 5• An excellent score on the Kindergarten Aptitude

Test (KAT) is a 9.1 out of ten points. • a: Letting μ represent the mean score on the

KAT, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that μ exceeds 9.1.

• b: The mean and the standard deviation of a sample of n = 49 Kindergarten Aptitude Test Takers ratings are x- = 9.192 and s = 0.19. Use a critical value to test the hypotheses you set up in your hypothesis by setting α equal to .01.

GM 533 Week 5

•Letting μ represent the mean score on the KAT, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that μ exceeds 9.1.

Ho: μ ≤ 9.1Ha: μ > 9.1 (claim)

GM 533 Week 5•The mean and the standard deviation of a

sample of n = 49 Kindergarten Aptitude Test Takers ratings are x- = 9.192 and s = 0.19. Use a critical value to test your hypothesis by setting α equal to .01.

GM 533 Week 5Sample Mean Population Mean Sample Std Dev. Sample Size

9.192 9.1 0.19 49Calculated z to compare to critical values

3.389

Compare toz (alpha) Left Tailed Right Tailed Two Tailed (+/-)

0.1 -1.28 1.28 1.640.05 -1.64 1.64 1.960.01 -2.33 2.33 2.58

0.001 -3.09 3.09 3.29

P-Value Methodz Left Tailed Right Tailed Two Tailed

3.389 0.9996 0.0004 0.0007

Compare to0.1

0.050.01

0.001

GM 533 Week 5

•At a z of .01 we get 2.33 on the right tailed (greater than). Since 3.389 is greater than 2.33 we reject Ho and say that we accept the claim that the mean score is greater than 9.1.

•On the p-value method, 0.1,0.05,0.01 and 0.001 are all greater than the 0.0004 so we reject Ho and accept that our mean is greater than 9.1. There would be very strong evidence.

GM 533 Week 5

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GM 533 Week 5

•Consider a medical company that wishes to determine whether a new ingredient, catalyst ST-109, changes the mean hourly yield of its process from the historical process mean of 200 pounds per hour. When six trial runs are made using the new catalyst, the following yields (in pounds per hour) are recorded: 190,195,201,209,219 and 237.

GM 533 Week 5

•a: Letting μ be the mean of all possible yields using the new ingredient, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence that μ differs from 200 pounds.

GM 533 Week 5

Ho: μ = 200Ha: μ≠ 200 (claim)

GM 533 Week 5•b: The mean and the standard deviation of the sample

of 6 catalyst yields are x- = 208.5 and s = 17.3407 . Using a critical value and assuming approximate normality, test the hypotheses you set up in part α by setting α equal to .01. The p-value for the hypothesis test is given in the Excel output below. Interpret this p-value.

t-statistic1.201

p-value*0.135 •p-value made up, not calculated

in the case of this example.

GM 533 Week 5190 Sample Mean Population Mean Sample Std Dev. Sample Size

195 208.5 200 17.3407 6

201 Calculated t to compare to critical values

209 1.201

219

237 Compare to

t (alpha) t alpha/2 (2 tailed)

Mean 0.1 2.015

208.5 0.05 2.571

Std Dev 0.01 4.032

17.3407 0.001 6.869

GM 533 Week 5

•Calculated t of 1.201 is less than t(.01/2) of 4.302, so I would not reject Ho.

•p-value of 0.135 is larger than 0.1, 0.5, 0.01 and 0.001 so again there is strong evidence that I can not reject that the mean is 200 pounds thus providing no evidence to support the claim that it is different from 200 pounds.

GM 533 Week 5

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GM 533 Week 5

•The manufacturer of the ACME Apple Slicer claims that 98 percent of its slicers last at least one year without breaking. In order to test this claim, a consumer group randomly selects 250 consumers who have owned and used an ACME Apple Slicer for at least one year. Of these 250 consumers, 246 say that their slicer is still slicing away, while 4 say that their slicers just don’t cut it anymore.

GM 533 Week 5•a.: Letting p be the proportion of slicers

that last one year without a problem, set up the null and alternative hypotheses that the consumer group should use to attempt to show that the manufacturer’s claim is false or that it is less than 98%.

GM 533 Week 5

Ho: p ≥ 0.98Ha: p < 0.98 (claim)

GM 533 Week 5

•b.: Use critical values and the previously given sample information to test the hypotheses you set up in part a by setting α equal to .10, .05, .01, and .001. How much evidence is there that the manufacturer’s claim is false?

GM 533 Week 5Total # of the Total p hat mu n sigma (calc) Calculated z test statistic

250 246 0.984 0.98 250 0.008854377 0.452

* p hat is a p with a rooftop

Compare toz (alpha) Left Tailed Right Tailed Two Tailed (+/-)

0.1 -1.28 1.28 1.640.05 -1.64 1.64 1.960.01 -2.33 2.33 2.58

0.001 -3.09 3.09 3.29

P-Value Methodz Left Tailed Right Tailed Two Tailed

0.452 0.6743 0.3257 0.6514

Compare to0.1

0.050.01

0.001

GM 533 Week 5• The calculated z test statistic of 0.452 is to the

right of the left tailed values of z for every alpha given, thus I can NOT reject the null hypothesis that p = 0.98

• There is strong evidence! (Thus I can’t agree with the consumer group’s claim that it is less than 98% or better yet I can not reject the null that it is greater than or equal to 98%)

• By the p-value method, 0.6743 is greater than 0.1, 0.05, 0.01 and 0.001 thus again I can NOT reject the null hypothesis.

GM 533 Week 5•c.: Do you think the results of the

consumer group’s survey have practical importance? Explain your opinion.

•Yes, I do, but not what they expected. “p hat” was calculated to be 0.984 which is above the company’s claimed 98%, it seems there is strong evidence that we CAN NOT reject the company’s claim and definitely can’t agree with the consumer group that it is less than that.

GM 533 Week 5

•I will post these charts in my Statcave at

www.facebook.com/statcave

I will also make my Excel template available at the Statcave by Wednesday at the latest. (I needed to change a couple of things so it handles negative z-scores correctly).