1. NO ONE leaves the room during testing!!
If you have to use the restroom, do it NOW!!!
You may work on the exam review in teams, but you MUST remain relatively quiet.
Thanks for a GREAT year!! You all rock!!

2. 1) A random sample of 32 gas grills has a mean price of $630.90 and a standard deviation of $56.70.
Construct a 90% and a 95% confidence interval for the population mean (Page 318, #37)
Section 6.1 (Stat–Test–7)
Use 56.7 for σ, for 𝑥 , 32 for n
Use .90 for C-Level, and then run it again using .95 for C-Level
90% interval is
95% interval is
2) A survey shows that 51% of adults prefer to receive advertisements through the mail. The margin of error E is 5.2%. Construct a confidence interval for the proportion of adults who prefer to receive advertisements through the mail. (Page 339, #14.)
Section 6.3 – Simply add and subtract E from p.
.51±.052= .458−.562

3. 3) Your medical research team is investigating the mean cost of a 30-day supply of a certain heart medication. A pharmaceutical company thinks that the mean cost is less than $60. You want to support this claim. How would you write the null and alternate hypotheses? (Page 377, #47)
Section
Remember, if you want to support a claim, it MUST be the alternate hypothesis!!
𝐻 0 :𝜇≥60; 𝐻 𝑎 : 𝜇<60 (𝐶𝑙𝑎𝑖𝑚)

4. The money raised and spent (both in millions of U. S
The money raised and spent (both in millions of U.S. dollars) by all congressional campaigns for 8 recent 2-year periods is shown in the table. Use α = .05 to run test. (Pages 533 & 534; #15 & 23)
Section 9.3 Stat–Edit; put numbers into L1 (x) and L2 (y), run STAT–TEST–F
Remember, these are always two-tailed tests.
Press Alpha Trace to put equation into Y=.
𝐻 0 : 𝜌=0 𝐻 𝑎 : 𝜌≠0
A) Find the equation of the regression line.
𝑦=.9426𝑥
B) Find the values of r, 𝑟 2 and p.
𝑟=.997, 𝑟 2 =.994 p = 7.929E-8
C) Is the relationship between money raised and money spent a significant one?
YES – the p-value is < α, so we reject the null.
Since the null is that there is no relationship, and we rejected that, there is a significant relationship at the 5% level of significance.
Money Raised (x)
471.7
659.3
740.5
790.5
781.3
1047.3
969.5
1206.1
Money Spent (y)
446.3
680.2
725.2
765.3
740.4
1005.6
936.3
1156.8

5. Money Raised (x) Money Spent (y)
The money raised and spent (both in millions of U.S. dollars) by all congressional campaigns for 8 recent 2-year periods is shown in the table. Use α = .05 to run test. (Pages 533 & 534; #15 & 23)
D) Predict the amount of money spent when the amount of money raised is $775.8 million.
2nd Window, set table to start at 775.8
2nd Graph gives a y-value of
We predict that the money spent will be $ million.
E) Construct a 95% confidence interval for money spent when the amount of money raised is $775.8 million.
Stat–Test–F , Stat Calc 1 and 2nd VARS 4 (1 + c)/2 give you the necessary values.
Money Raised (x)
471.7
659.3
740.5
790.5
781.3
1047.3
969.5
1206.1
Money Spent (y)
446.3
680.2
725.2
765.3
740.4
1005.6
936.3
1156.8
STAT–TEST–F gives you:
𝑠 𝑒 =18.865
df = 6
STAT–Calc–1 on L1 gives you:
𝑥 =
𝑥 =6666.2
𝑥 2 =
𝑛=8
2nd–VARS–4 when
Area = (1+.95)/2 and df= 6
gives you 𝑡 𝑐 =2.447

6. The money raised and spent (both in millions of U. S
The money raised and spent (both in millions of U.S. dollars) by all congressional campaigns for 8 recent 2-year periods is shown in the table. Use α = .05 to run test. (Pages 533 & 534; #15 & 23)
𝐸= 𝑡 𝑐 𝑆 𝑒 𝑛 + 𝑛 𝑥 0 − 𝑥 2 𝑛 𝑥 2 − 𝑥 ;
𝐸= − −( ) 2 =49.153
752.83−49.153<𝑦< ;
<𝑦<801.98
We can be 95% certain that if $775.8 million is raised, then the amount spent will be between $703,677,000 and $801,983,000
Money Raised (x)
471.7
659.3
740.5
790.5
781.3
1047.3
969.5
1206.1
Money Spent (y)
446.3
680.2
725.2
765.3
740.4
1005.6
936.3
1156.8

7. 5) A company that makes cola drinks states that the mean caffeine content per one 12-ounce bottle of cola is 40 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 39.2 milligrams with a standard deviation of milligrams. At α = 0.01, can you reject the company’s claim? (Page 392, #39)
Section STAT–TEST–1
𝐻 0 :𝜇=40 (𝐶𝑙𝑎𝑖𝑚); 𝐻 𝑎 : 𝜇≠40
z = -.584
p = .559
Since p > α, fail to reject the null.
Because the claim and the null are the same thing, we cannot reject the claim, either.
At the 1% level of significance, there is not enough evidence to reject the claim that the mean caffeine content of a 12-ounce bottle of cola is 40 milligrams.

8. 6) You want to buy a microwave oven and will choose Model A if its repair costs are lower than Model B’s. You research the repair costs of 47 Model A ovens and 55 Model B ovens. Model A had a mean repair cost of $75, with a standard deviation of $ Model B had a mean repair cost of $80 with a standard deviation of $20. At α = 0.01, would you buy Model A? (Page 446, #21)
Section STAT–TEST–3
𝐻 0 : 𝜇 1 ≥ 𝜇 2 ; 𝐻 𝑎 : 𝜇 1 < 𝜇 2 (𝐶𝑙𝑎𝑖𝑚);
We are running Test 3 because our sample sizes are > 30.
z = p = .062
Since p > α, fail to reject the null.
Because we failed to reject the null, we also fail to support the claim.
No, at the 1% level of significance, there is insufficient evidence to prove that Model A has lower repair costs.

9. 𝐻 0 : 𝜇 𝑑 =0; 𝐻 𝑎 : 𝜇 𝑑 ≠0 (𝐶𝑙𝑎𝑖𝑚);
7) A manufacturing plant wants to determine whether its manager performance ratings (0-100) have changed from last month to this month. The table shows the plant’s performance ratings from the same six managers for last month and this month. At α = 0.05, is there enough evidence to conclude that the plant’s performance ratings have changed? (Page 469, #20)
Section 8.3
𝐻 0 : 𝜇 𝑑 =0; 𝐻 𝑎 : 𝜇 𝑑 ≠0 (𝐶𝑙𝑎𝑖𝑚);
STAT – Edit – Enter Last Month into L1 and This Month into L2
Highlight L3 and type in L1 – L2
Run STAT – TEST – 2 on L3 (use the Data option)
t = p = .147
Since p > α, fail to reject the null.
Because we failed to reject the null, we also fail to support the claim.
No, at the 5% level of significance, there is insufficient evidence to prove that performance ratings have changed.
Manager 1 2 3 4 5 6
Last Month 85 96 70 76 81 78
This Month 88 89 86 92

10. In a survey of 1150 adult males, 805 said they use the Internet
In a survey of 1150 adult males, 805 said they use the Internet. In a survey of 1050 females, 746 said they use the Internet. At α = 0.05, can you reject the claim that the proportions of Internet users are the same for both groups? (Page 477, #15)
Section 8.4 STAT – TEST – 6
𝐻 0 : 𝑝 1 = 𝑝 2 (𝐶𝑙𝑎𝑖𝑚); 𝐻 𝑎 : 𝑝 1 ≠ 𝑝 2
z = p = .590
Since p > α, fail to reject the null.
Because we failed to reject the null, we also fail to reject the claim.
At the 5% level of significance, there is insufficient evidence to reject the claim that the proportions of internet users are the same for both groups.

11. 9) A personnel director believes that the distribution of the reasons workers leave their jobs is different from the one listed below. The director randomly selects workers who recently left their jobs and asks each his or her reason for doing so. The results are shown in the table. At α = 0.01, are the distributions different? (Page 561, #10)
Claimed:
Limited Advancement Opportunity, 41%
Lack of Recognition, 25%
Low salary benefits, 15%
Unhappy with management, 10%
Bored or don’t know, 9%
Survey Results:
Limited Advancement Opportunity, 78
Lack of Recognition, 52
Low salary benefits, 30
Unhappy with management, 25
Bored or don’t know, 15
𝐻 0 : The distribution of reasons workers leave their jobs is: limited advancement opportunity, 41%; lack of recognition, 25%; low salary benefits, 15%; unhappy with management, 10%; bored or don’t know, 9%
𝐻 𝑎 : The distribution of reasons workers leave their jobs differs from the stated distribution.

12. Enter Observed values into L2 (78, 52, 30, 25, 15)
9) A personnel director believes that the distribution of the reasons workers leave their jobs is different from the one listed below. The director randomly selects workers who recently left their jobs and asks each his or her reason for doing so. The results are shown in the table. At α = 0.01, are the distributions different? (Page 561, #10)
Claimed:
Limited Advancement Opportunity, 41%
Lack of Recognition, 25%
Low salary benefits, 15%
Unhappy with management, 10%
Bored or don’t know, 9%
Survey Results:
Limited Advancement Opportunity, 78
Lack of Recognition, 52
Low salary benefits, 30
Unhappy with management, 25
Bored or don’t know, 15
Section 10.1 STAT – Edit –
Enter claimed percentages, in decimal form, into L1. (.41, .25, .15, .10, .09)
Enter Observed values into L2 (78, 52, 30, 25, 15)
Highlight L3 and type L1*n (200 in this case)
Double-check to be sure that all values in L3 are at least 5!

13. Run STAT–Test–D on L2 and L3 with 4 degrees of freedom (k – 1)
9) A personnel director believes that the distribution of the reasons workers leave their jobs is different from the one listed below. The director randomly selects workers who recently left their jobs and asks each his or her reason for doing so. The results are shown in the table. At α = 0.01, are the distributions different? (Page 561, #10)
Claimed:
Limited Advancement Opportunity, 41%
Lack of Recognition, 25%
Low salary benefits, 15%
Unhappy with management, 10%
Bored or don’t know, 9%
Survey Results:
Limited Advancement Opportunity, 78
Lack of Recognition, 52
Low salary benefits, 30
Unhappy with management, 25
Bored or don’t know, 15
Run STAT–Test–D on L2 and L3 with 4 degrees of freedom (k – 1)
p = .731.
Since p > α, fail to reject the null.
Because we failed to reject the null, we also fail to support the claim.
No, at the 1% level of significance, there is insufficient evidence to prove that the distribution of reasons workers leave their jobs differs from the claim.

14. H0: Grades are independent of the institution
10) The contingency table shows how a random sample of college freshmen graded the leaders of three types of institutions. At α = 0.05, can you conclude that the grades are related to the institution? (Page 573, #15)
GRADE
Institution A B C D E
Military 25 46 19 5 3
Religious 18 44 24 7
Media/Press 23 37 21 12
Section 10.2 STAT–TEST–C
H0: Grades are independent of the institution
Ha: Grades are dependent on the institution (Claim);
Enter contingency table given into Matrix A
2nd Matrix – Edit – 1
Matrix A should be a 3 x 5 matrix
Run Stat–Test–C on matrices A and B (this will ALWAYS be a right-tailed test).
Access Matrix B to be certain that all expected values are at least 5.

15. The null says that the variables are independent; we rejected that.
10) The contingency table shows how a random sample of college freshmen graded the leaders of three types of institutions. At α = 0.05, can you conclude that the grades are related to the institution? (Page 573, #15)
GRADE
Institution A B C D E
Military 25 46 19 5 3
Religious 18 44 24 7
Media/Press 23 37 21 12
𝛸 2 = ; p = E -8
Since p < α, reject 𝐻 0
The null says that the variables are independent; we rejected that.
Yes, there is enough evidence, at the 5% significance level, to conclude that grades are dependent on the institution.

16. 11) A realtor is comparing the prices of one-family houses in four cities. After randomly selecting one-family houses in the four cities and determining the price for each, the realtor creates the chart shown, with prices in thousands of dollars. At α = 0.10, can the realtor reject the claim that the mean price is the same for all four cities? (Page 597, #11)
City A
City B
City C
City D
207.0
257.9
253.9
200.9
179.5
215.1
138.9
215.3
224.6
199.3
234.2
285.9
220.3
201.5
198.1
210.7
282.7
221.9
226.7
321.3
240.4
248.8
173.1
202.1
120.8
267.2
190.0
217.4
238.6
158.7
188.9
328.7
208.5
177.9
157.3
257.1
254.0
187.3
237.4
284.0
128.7
201.3
239.4

17. 11) A realtor is comparing the prices of one-family houses in four cities. After randomly selecting one-family houses in the four cities and determining the price for each, the realtor creates the chart shown, with prices in thousands of dollars. At α = 0.10, can the realtor reject the claim that the mean price is the same for all four cities? (Page 597, #11)
Section 10.4 STAT–TEST–H
𝐻 0 : 𝜇 1 = 𝜇 2 = 𝜇 3 = 𝜇 4 (Claim)
𝐻 𝑎 : At least 1 mean is different from the others.
Stat – Edit
Enter City A into L1, City B into L2, City C into L3 and City D into L4
Run STAT–TEST–H on (L1,L2,L3,L4)
F = p = .055
Since < 0.10, reject the null
Yes, there is enough evidence, at the 10% significance level, to reject the claim that the mean price is the same for all four cities.