### AP.Stat.Review.Ch.26-27

```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?
df = 20
3. Find the number of times you expect each face of a dice to come up if you roll it 90 times.
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?
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?
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
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
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
b. What kind of test is this?
c. The results of the test are shown below. State the conclusion.
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