Week 5 Charts for GM 533

1. GM 533 Week 5
Yes, it’s tough…
Just do it!

2. GM 533 Week 5
Checkpoint Examples

3. GM 533 Week 5

4. 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 36 satisfaction ratings to provide
evidence supporting the claim that the mean
customers’ satisfaction rating exceeds 8.2.

5. 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 x̄ = 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?

6. 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.

7. 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 ≤ 8.2
Ha: μ > 8.2 (claim)

8. GM 533 Week 5
• b: The random sample of 33 satisfaction
ratings yields a sample mean of x̄ = 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.

9. GM 533 Week 5
Calculate z, I used a template I made. Compare
to values at .10, .05, .01, and .001.
I calculate z to be 1.949
Sample Mean
Population
Mean Sample Std Dev. Sample Size
8.295 8.2 0.28 33
Calculated z to compare to critical values
1.949

10. 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.

11. 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.

12. GM 533 Week 5
Using 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

13. GM 533 Week 5
• d: How much evidence is there that the mean
customer satisfaction rating exceeds 8.2?

14. 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)

15. GM 533 Week 5

16. 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.

17. 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.

18. GM 533 Week 5
We want to know if it’s 8 or not.
Ho: μ = 8 (claim)
Ha: μ ≠ 8

19. 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, x̄
= 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.

20. 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 Tailed
Two 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

21. 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 Tailed
1.600 0.9452 0.0548 0.1096
Compare to
0.1
0.05
0.01
0.001

22. 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 8.0522
Margin of
Error
0.0322
0.995

23. GM 533 Week 5

24. 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 x̄ = 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.

25. 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 ≤ 9.1
Ha: μ > 9.1 (claim)

26. GM 533 Week 5
• The mean and the standard deviation of a
sample of n = 49 Kindergarten Aptitude Test
Takers ratings are x̄ = 9.192 and s = 0.19. Use
a critical value to test your hypothesis by
setting α equal to .01.

27. GM 533 Week 5
Sample Mean Population Mean Sample Std Dev. Sample Size
9.192 9.1 0.19 49
Calculated z to compare to critical values
3.389
Compare to
z (alpha) Left Tailed Right Tailed Two 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
P-Value Method
z Left Tailed Right Tailed Two Tailed
3.389 0.9996 0.0004 0.0007
Compare to
0.1
0.05
0.01
0.001

28. 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.

29. GM 533 Week 5

30. 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.

31. 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.

32. GM 533 Week 5
Ho: μ = 200
Ha: μ≠ 200 (claim)

33. GM 533 Week 5
• b: The mean and the standard deviation of the sample
of 6 catalyst yields are x̄ = 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-statistic
1.201
p-value*
0.135
•p-value made up, not calculated
in the case of this example.

34. GM 533 Week 5
190 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

35. 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.

36. GM 533 Week 5

37. 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.

38. 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%.

39. GM 533 Week 5
Ho: p ≥ 0.98
Ha: p < 0.98 (claim)

40. 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?

41. GM 533 Week 5
Total # 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 to
z (alpha) Left Tailed Right Tailed Two 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
P-Value Method
z Left Tailed Right Tailed Two Tailed
0.452 0.6743 0.3257 0.6514
Compare to
0.1
0.05
0.01
0.001

42. 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.

43. 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.

44. 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 as soon as I can.