Below is our AP Statistics unit 8 practice test. These questions explore situations where we are interested in proportions of individuals in different categories and whether they follow certain assumptions — usually it’s that individuals are spread evenly across categories or that two categorical variables are independent. We use the Chi-Square distribution and test statistic to test assumptions.
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Question 1 |
Which is true about chi-square distributions?
Values are positive and the distribution is skewed right. | |
Values are negative and the distribution is skewed right. | |
Values are positive and the distribution is skewed left. | |
Values are positive and the distribution is symmetric. |
Question 1 Explanation:
The correct answer is (A). Chi-square distributions are positive and skewed right.
Question 2 |
A teacher has a theory that as students advance in grades, more students will have jobs. The teacher’s theory is that students having jobs will follow the ratio $1:2:2:3$ and that 12.5% of the 9th grade students have jobs. The teacher takes a sample of 100 students from each grade and finds the following number of students with jobs in each grade: 9th grade: 12, 10th grade: 22, 11th grade: 27, 12th grade: 35.
Calculate $x^2_c$ for the sample.
0.361 with df = 3 | |
0.361 with df = 4 | |
1.547 with df = 3 | |
1.547 with df = 4 |
Question 2 Explanation:
The correct answer is (C).
Based on the theoretical ratio, we would get the following numbers $(1:2:2:3$ is like $(\frac{1}{8}):(\frac{2}{8}):(\frac{2}{8}):(\frac{3}{8})$ and when we take those fractions of $100$ we have $12.5:25:25:37.5)$
$x^2_c = Σ \frac{(\text{observed} − \text{expected})^2}{\text{expected}}$
$ = \frac{(12-12.5)^2}{12.5} + \frac{(22-25)^2}{25} $ $ + \frac{(30-25)^2}{25} + \frac{(35-37.5)^2}{37.5}$
$= 1.547$
There are 4 categories, so df = 4 − 1 = 3.
Based on the theoretical ratio, we would get the following numbers $(1:2:2:3$ is like $(\frac{1}{8}):(\frac{2}{8}):(\frac{2}{8}):(\frac{3}{8})$ and when we take those fractions of $100$ we have $12.5:25:25:37.5)$
| Grade | Experiment | Theoretical |
| 9 | 12 | 12.5 |
| 10 | 22 | 25 |
| 11 | 30 | 25 |
| 12 | 35 | 37.5 |
$x^2_c = Σ \frac{(\text{observed} − \text{expected})^2}{\text{expected}}$
$ = \frac{(12-12.5)^2}{12.5} + \frac{(22-25)^2}{25} $ $ + \frac{(30-25)^2}{25} + \frac{(35-37.5)^2}{37.5}$
$= 1.547$
There are 4 categories, so df = 4 − 1 = 3.
Question 3 |
A high school principal surveyed students in grades 9–12 about their satisfaction with school lunches. The results are below:
Grade |
Satisfied |
Not Satisfied |
Total |
9 |
20 |
5 |
25 |
10 |
18 |
8 |
26 |
11 |
24 |
5 |
29 |
12 |
14 |
6 |
20 |
Total |
76 |
24 |
100 |
What is the expected count for the cell corresponding to 9th graders not satisfied with school lunch?
5 | |
6 | |
20 | |
21 |
Question 3 Explanation:
The correct answer is (B). The expected count for 9th graders not satisfied with school lunch would be:
$\dfrac{24}{100}*25 = 6$
$\dfrac{24}{100}*25 = 6$
Question 4 |
What is an appropriate null and alternative hypotheses for a chi-square test for independence?
$H_0$: There is no difference in the number of movies watched between adults and children. $H_a$: There is a difference in the number of movies watched between adults and children. | |
$H_0$: There is a difference in the number of movies watched between adults and children. $H_a$: There is no difference in the number of movies watched between adults and children. | |
$H_0$: There is no association between which movie a person chose to watch and age. $H_a$: There is an association between which movie a person chose to watch and age. | |
$H_0$: There is an association between which movie a person chose to watch and age. $H_a$: There is no association between which movie a person chose to watch and age. |
Question 4 Explanation:
The correct answer is (C). The appropriate hypotheses for a chi-square test for independence have a null hypothesis stating no association between two categorical variables and an alternative hypothesis stating that there is an association. Answer (A) would be potential null and alternative hypotheses for a test for difference in means.
Question 5 |
Which of these is NOT a condition for making statistical inferences when testing a chi-square distribution for independence or homogeneity?
Data should be collected using a simple or stratified random sample. | |
The standard deviation must be known. | |
Expected counts should be greater than 5. | |
All of these are conditions. |
Question 5 Explanation:
Answer choice (B) is NOT a condition. Standard deviation is not a concept that really applies here. This condition could help for using a Z Table instead of a T table when testing for a mean in a different scenario (not a chi square test).
Question 6 |
What is the p-value for a chi-square test for independence or homogeneity if the number of rows is 4 and the number of columns is 3 with 14.92? Use the excerpt of the table below.
df |
0.25 |
0.20 |
0.15 |
0.10 |
0.05 |
0.025 |
0.02 |
6 |
7.84 |
8.56 |
9.45 |
10.64 |
12.59 |
14.45 |
15.03 |
7 |
9.04 |
9.80 |
10.75 |
12.02 |
14.07 |
16.01 |
16.62 |
8 |
10.22 |
11.03 |
12.03 |
13.36 |
15.51 |
17.53 |
18.17 |
9 |
11.39 |
12.24 |
13.29 |
14.68 |
16.92 |
19.02 |
19.68 |
10 |
12.55 |
13.44 |
14.53 |
15.99 |
18.31 |
20.48 |
21.16 |
11 |
13.70 |
14.63 |
15.77 |
17.28 |
19.68 |
21.92 |
22.62 |
12 |
14.85 |
15.81 |
16.99 |
18.55 |
21.03 |
23.34 |
24.05 |
0.025 > $p$ > 0.02 | |
0.20 > $p$ > 0.05 | |
0.10 > $p$ > 0.05 | |
0.25 > $p$ > 0.20 |
Question 6 Explanation:
The correct answer is (A). To find the appropriate p-value range, first find $\text{df} = (\text{rows} − 1)(\text{columns} − 1)$.
In this case, $3*2=6$ so $\text{df}=6$.
In the $\text{df}=6$ row, $14.92$ falls between $0.025$ and $0.02$.
In this case, $3*2=6$ so $\text{df}=6$.
In the $\text{df}=6$ row, $14.92$ falls between $0.025$ and $0.02$.
Question 7 |
What are the degrees of freedom for the following table?
Pepperoni |
Cheese |
Sausage |
Total |
|
Grade 8 |
9 |
5 |
20 |
34 |
Grade 9 |
20 |
8 |
10 |
38 |
Grade 10 |
16 |
12 |
14 |
42 |
Grade 11 |
12 |
8 |
5 |
25 |
Grade 12 |
16 |
11 |
12 |
39 |
Total |
73 |
44 |
61 |
178 |
8 | |
10 | |
12 | |
15 |
Question 7 Explanation:
The correct answer is (A). For a chi-square test for homogeneity or independence:
$\text{df} = (r-1)(c-1) $ $ = 4*2=8$
$\text{df} = (r-1)(c-1) $ $ = 4*2=8$
Question 8 |
If the Null hypothesis is that cashiers at a particular franchise are twice as likely to be young adults as they are to be adults, and that they are just as likely to be adults as they are to be seniors, what is the chi square test statistic corresponding to that statistical test?
Young adults: 180
Adults: 85
Seniors: 135
18.25 | |
37.69 | |
80.25 | |
148.25 |
Question 8 Explanation:
The correct answer is (A). There are 400 people, and we need the number of young adults expected to be 2$x$, the number of adults expected to be $x$, and the number of seniors expected to be $x$.
So, $x$ must be 100, and we have 200, 100, and 100 expected.
$Σ \frac{(\text{observed} − \text{expected})^2}{\text{expected}} $
$=\frac{(180 − 200)^2)}{200} + \frac{(80 − 100)^2}{100} $ $ + \ \frac{(135-100)^2}{100} = 18.25$
So, $x$ must be 100, and we have 200, 100, and 100 expected.
$Σ \frac{(\text{observed} − \text{expected})^2}{\text{expected}} $
$=\frac{(180 − 200)^2)}{200} + \frac{(80 − 100)^2}{100} $ $ + \ \frac{(135-100)^2}{100} = 18.25$
Question 9 |
Which of the following statements are true about the relationship between the p-value, $α$, and the null and alternative hypotheses?
If p-value > $α$ then assume there was an error in calculation. | |
If p-value < $α$ then reject the null hypothesis. | |
If p-value > $α$ then reject the null hypothesis. | |
If p-value < $α$ then assume there was an error in calculation. |
Question 9 Explanation:
The correct answer is (B). A decision to either reject or fail to reject the null hypothesis is based on comparing the p-value to $α$. If the p-value is less than the significance level, then there is sufficient evidence to reject the null hypothesis.
Question 10 |
What would not be appropriate null and alternative hypotheses in a test for a distribution of proportions in a set of categorical data?
$H_0$: The distribution of students competing in a state track meet is consistent with the distribution of the state’s population by region.. $H_a$: The distribution of students competing in a state track meet is not consistent with the state’s population by region. | |
$H_0$: The proportion of students participating in each fall sport at a given school will match the ratio of participation across the state. $H_a$: The proportion of students participating in each fall sport at a given school will not match the ratio of participation across the state. | |
$H_0$: The distribution of flavors in a mixed candy bag will be exactly equal. $H_a$: The distribution of flavors in a mixed candy bag will match the distribution at the factory. | |
$H_0$: The proportion of pink to red flowers in a garden will match the proportion of seeds planted. $H_a$: The proportion of pink to red flowers in a garden will not match the proportion of seeds planted. |
Question 10 Explanation:
Answer choice (C) is not appropriate. The null hypothesis may or may not be exactly equal, depending on the distribution at the factor. If the factory has an equal distribution of flavors, then the alternative hypothesis should be that the distribution will not be equal. If the factory does not use an equal distribution of flavors, then the listed $H_a$ would actually be $H_0$ and the alternative hypothesis should be that the distribution in the bag will not match the distribution at the factory.
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