GM 533 Week 6 - Help with Part B of the Project

1. GM 533
Week 6
Part B of the Project Help
B Heard
Do not post these charts without my
permission. Students may download
one copy for personal use.

2. Part B of the Project Example

3. Part B of the Project Example

4. Part B of the Project Example
• Your manager has speculated the following:
a) the average (mean) annual income was less than $50,000
(Changed all data),
b) the true population proportion of customers who live in an City C
area exceeds 40%,
c) the average (mean) number of years lived in the current home is
less than 12 years,
d) the average (mean) credit balance for suburban customers is more
than $4400.
1) Using the sample data, perform the hypothesis test for each of the
above situations in order to see if there is evidence to support your
manager’s belief in each case a.-d. In each case use the Seven
Elements of a Test of Hypothesis, in Section 6.2 of your text book
with α = .05, and explain your conclusion in simple terms. Also be
sure to compute the p-value and interpret.

5. Part B of the Project Example
a) the average (mean) annual income was less
than $50,000 (Changed all data)
▫ I found the average for annual incomes to be
46.06 or $46,060 and the standard deviation to
be 13.712 or $13,712
▫ Set up Hypothesis Test
 Ho: µ = 50 (I’m keeping in mind that’s
thousands)
 H1: µ < 50

6. Part B of the Project Example
• For α = 0.05 and “<“ in the Ha, I found
z = -1.645, so the “Rejection Region” would be
z < -1.645
• Now I calculate the test statistic
• z = (46.06-50)/1.939 = -2.032 because σx bar =
13.712/sqrt(50) = 1.93917 (I rounded a little)

7. Part B of the Project Example
• My calculated test statistic of -2.032 does fall in
the rejection region of z < -1.645 therefore I
would reject the null hypothesis and say there “is
sufficient evidence to indicate µ < 50 or
$50,000.”

8. Part B of the Project Example
b) the true population proportion of customers
who live in City C area exceeds 40%,
• 13 of the 50 surveyed live in City C, which is 26%
or 0.260, this is my point estimate for p.
• Thus I would have
▫ Ho: p = 0.40 vs. Ha: p > 0.40

9. Part B of the Project Example
• To conduct the large sample z-test,we must first
verify the sample size is large enough.
▫ nPo = 50(0.40) = 20 and 50(1-0.40) = 30, both
are larger than 15, so we conclude the sample size
is large enough to apply the large sample z test.

10. Part B of the Project Example
• z = (0.26 – 0.400)/0.69282 = -2.02 where s p hat =
sqrt(((0.40)(0.60))/50) = 0.069282
• This is a one tailed (upper or right since Ha has “>”).
Our rejection region would be z > 1.645
• -2.02 is definitely not greater than + 1.645 (not in
the rejection region) so WE WOULD NOT Reject
Ho.
• By not rejecting the Ho, we are saying there is
insufficient evidence to conclude the true population
of customers who live in City C is greater than 40%

11. Part B of the Project Example
c) the average (mean) number of years lived in the
current home is less than 12 years,
▫ I found the average number of years in the
current home for my survey data to be exactly
12.0 (remember I changed the data) and the
standard deviation to be 5.135
▫ Set up Hypothesis Test
 Ho: µ = 12
 H1: µ < 12

12. Part B of the Project Example
• For α = 0.05 and “<“ in the Ha, I found
z = -1.645, so the “Rejection Region” would be
z < -1.645
• Now I calculate the test statistic
• z = (12.0-12)/0.7262 = 0 because σx bar =
5.135/sqrt(50) = 0.7262 (I rounded a little)

13. Part B of the Project Example
• My calculated test statistic of 0 does NOT fall in
the rejection region of z < -1.645 therefore I
would NOT reject the null hypothesis and say
there “is insufficient evidence to indicate µ < 12.”

14. Part B of the Project Example
d) the average (mean) credit balance for everyone
(I CHANGED) customers is more than $4400.
▫ I found the average credit balance for those
surveyed $3961.08 (remember I changed the
data) and the standard deviation to be 927.7
▫ Set up Hypothesis Test
 Ho: µ = 4400
 H1: µ > 4400

15. Part B of the Project Example
• For α = 0.05 and “>“ in the Ha, I found z =
+1.645, so the “Rejection Region” would be z > 1.645
• Now I calculate the test statistic
• z = (3961.08-4400)/131.2 = -3.346 because σx bar =
927.7/sqrt(50) = 131.2 (I rounded a little)
• The calculated z being negative on this (I changed
the data), it should be obvious it is not in the
rejection region.

16. Part B of the Project Example
• My calculated test statistic of -3.346 does NOT
fall in the rejection region of z > 1.645 therefore
I would NOT reject the null hypothesis and say
there “is insufficient evidence to indicate µ
>4400.”

17. Part B of the Project Example
• Don’t forget that on number 1, parts a-d, you
were also asked to be sure to compute the p-
value and interpret.

18. Part B of the Project Example
2) Follow this up with computing 95% confidence
intervals for each of the variables described in
a.-d., and again interpreting these intervals.

19. Part B of the Project Example

20. Part B of the Project Example
You would say that you
are 95% confident that
the true mean income
lies between $42,260
and $49,860.

21. Part B of the Project Example
This would be
similar for c and
d.
I will cover
calculating the
confidence limits
for part b on the
next chart.
Please DON’T
FORGET THAT
YOUR PART D IS
FOR SUBURBAN
ONLY. I DID ALL
OF MY DATA.

22. Part B of the Project Example
• For the confidence interval on part b, I have “p
hat” which is 13/50 = 0.26 (Mine was for City C)

23. Part B of the Project Example
3) Write a report to your manager about the
results, distilling down the results in a way that
would be understandable to someone who does
not know statistics. Clear explanations and
interpretations are critical.
 Write a good report using correct grammar. Try
to find the correct symbols to use, etc. This is not
a problem set, it is a report! Cover all of the
questions asked, in the order they were asked, but
in the FORM of a Report (not like a numbered
Homework assignment).