AP Stat
Chapter 26-27 Review
Chi Square & Linear-regression t-test
1. You wonder if a bag of M & M’s is made up of 1/6 orange, brown, yellow, green, blue & red.
Your friend weights each color in the bag and finds 10 grams of orange, 11 grams of brown,
9 grams of yellow, 7 grams of green, 13 grams of blue & 10 grams of red.
What could you conclude from running a Chi Squared test of the expected colors
of M & M’s?
Answer: Can’t do a Chi Squared because M & M’s were not counted.
2. How many df are there for a chi-square test of homogeneity based on a table that has
5 rows & 6 columns?
Answer:
df = 20
3. Find the number of times you expect each face of a dice to come up if you roll it 90 times.
Answer:
15
20-29
30-39
40-49
50-59
60 & over
10
12
17
20
21
Did not succeed in
20
lowering blood pressure
18
13
10
9
Succeeded in lowering
blood pressure
A new blood pressure medication is tested on 30 volunteers from 5 different age groups. The
Results are shown on the table above.
4. If a Chi-square test were done, what kind of test would be conducted?
Answer: Chi-square test of homogeneity
5. Find the expected number of successful cases of 40-49 year olds.
Answer: (30/150) x 70 = 14
6. Suppose we are testing an SAT prep program. At the end we want to see if the amount
of points gained depends on the number of hours a student spent preparing for the SAT.
What kind of test would we do?
Answer: Linear Regression t-test
Flavor
Frequency
Grape
530
Lemon
Lime
470
420
Orange Strawberry
610
585
Trix cereal comes in five fruit flavors, and each flavor has a different shape. A curious
Student methodically sorted an entire box of the cereal & found the distribution of flavors
For the pieces of cereal in the box as shown above:
Is there evidence that the flavors are evenly distributed?
P: The distribution of the different flavors in a box of Trix
H:
Ho: The flavors are evenly distributed
Ha: The flavors in the box are not evenly distributed.
A:
Expected Counts: 2615/5 = 523 for each flavor
ECF’s >5, data is counted data, sample is representative of a typical box of Trix
N:
Chi-square GOF test
T:
ECF’s shown above, df = 4
O:
(530 523) 2
(585 523) 2
 
 ... 
 47.5717
523
523
2
M:
Reject Ho at the .05 level
S:
There is strong evidence that the distribution of Trix is not the same by flavor.
Orange & strawberry have more than expected & lemon & lime have less than
expected.
A sample of the non-academic interests of students at a high school is shown below. The
observed counts are shown versus the expected counts
Observed/Expected
10th Graders
11th Graders
12th Graders
No Extracurriculars
8/17.96
15/14.85
25/15.19
Sports Team
15/11.60
10/9.59
6/9.81
Club
12/10.10
9/8.35
6/8.55
Band
8/5.61
4/4.64
3/4.75
Choir
5/3.74
3/3.09
2/3.17
Other
4/2.99
2/2.47
2/2.53
a. Show how this distribution would be changed to meet the ECF condition
Observed/Expected
10th Graders
11th Graders
12th Graders
No Extracurriculars
8/17.96
15/14.85
25/15.19
Sports
15/11.60
10/9.59
6/9.81
Club
12/10.10
9/8.35
6/8.55
Band/Choir/Other
17/12.34
9/10.20
7/10.45
A hypothesis test is run where the null hypothesis is the distribution of non-academic
activities is the same across grade levels against the alternative hypothesis that the
distribution of non-academic activities is not the same across grade levels.
b. What kind of test is this?
Answer: χ2 Homogeneity Test
c. The results of the test are shown below. State the conclusion.
Answer: The distribution of non-academic activities is not the same across grade levels. It
appears that students do less non-academic activities as they progress through high school.
20-29
30-39
40-49
50-59
60 & over
10
12
17
20
22
Did not succeed in
20
lowering blood pressure
18
13
10
8
Succeeded in lowering
blood pressure
Back to our blood pressure medication example. Test whether the blood pressure medicine
Is equally effective in treating all age groups:
P:
The distribution of effectiveness of blood pressure medication across different
age groups
H:
Ho:
Ha:
The medicine is equally effective for all age groups
The medicine is not equally effective for all age groups
20-29
30-39
40-49
50-59
60 & over
16
16
16
16
16
Did not succeed in
14
lowering blood pressure
14
14
14
14
Succeeded in lowering
blood pressure
A:
All ECF’s are greater than 5, data is counted, sample is representative
N:
Chi-square test for homogeneity
T:
ECF’s shown above, df = 4
O:
(10  16) 2
(22  16) 2
 
 ... 
 14.0633
16
16
2
M:
Reject Ho at the .05 level
S:
There is strong evidence that the medicine is more effective for different age
groups. It appears that the medicine is more effective in treating older patients.
A group of students volunteered for a students where they drank a randomly assigned number
of cans of beer. 30 minutes later, a police officer measured their blood alcohol content (BAC).
A linear regression is run & the output is shown below:
1. Write the equation of the LSRL
2. Test the hypothesis for a linear association between beers consumed & BAC
3. Perform a 95% CI to predict the rate of change between beers consumed & BAC.
1. Predicted BAC = -.0127006 + .0179638(beers consumed)
2. P: β = linear association between beers consumed & BAC
H:
Ho: β = 0
A:
The scatterplot is straight. The residual plot is random. The residuals are normally
distributed. Each student’s BAC is independent.
N:
Lin Reg t-test
T:
n = 16, df = 14, b1 = .0179628, SE = .002402
O:
t
Ha: β ≠ 0
.0179638  0 Draw distribution with t = -7.48 & 7.48, P < .0001
.002402
M:
Reject Ho at the .05 level
S:
There is strong evidence of a linear association between beers consumed & BAC
3. N: Lin Reg CI
I: CI = .0179638 +- 2.145(.002402)
CI = .0179638 +- .0052
(.0128, .0232)
We are 95% confident that as BAC will increase between .01238 & .0232 for each
additional beer consumed.