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Chi Square- Goodness of Fit Tests

Total questions: 20

Worksheet time: 3615secs

Name
Class
Date
1.

If a chi-squared number is 5.29 with a df of 6, what is the p-value?

a)

.38

b)

.49

c)

.51

2.
The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year.  She wants to know if each type of food will be equally popular so she can start ordering supplies making other plans.  To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?"  Here are the data in the table provided.  The P-value for a chi-square test for goodness of fit is 0.0129.  Which is the following is the most appropriate conclusion?
a)
Because 0.0129 is less than α = 0.05, reject H0.  There is convincing evidence that the food choices are equally popular.
b)
Because 0.0129 is less than α = 0.05, reject H0.  There is not convincing evidence that the food choices are equally popular.
c)
Because 0.0129 is less than α = 0.05, reject H0.  There is convincing evidence that the food choices are not equally popular.
d)
Because 0.0129 is less than α = 0.05, fail to reject H0.  There is convincing evidence that the food choices are equally popular.
3.

Python eggs How is the hatching of water python eggs influenced by the temperature of the snake’s nest? Researchers randomly assigned newly laid eggs to one of three water temperatures: hot, neutral, or cold. Hot duplicates the extra warmth provided by the mother python, and cold duplicates the absence of the mother. Here are the data on the number of eggs that hatched and didn’t hatch.

If the p-value is .08 and you use .05 as your alpha, what is your conclusion

a)

Fail to reject: We do not have sufficient evidence that distribution of hatched is different for water temperature options

b)

Reject: We do not have sufficient evidence that distribution of hatched is different for water temperature options

c)

Fail to reject: We do have sufficient evidence that distribution of hatched is different for water temperature options

4.

Aw, nuts! A company claims that each batch of its deluxe mixed nuts contains 52% cashews, 27% almonds, 13% macadamia nuts, and 8% brazil nuts. To test this claim, a quality-control inspector takes a random sample of 150 nuts from the latest batch. The one-way table below displays the sample data. What is the appropriate null hypothesis

a)

U1-U2=U3=U4

b)

Mixed nut proportions are proportioned the same as they claim

c)

At least one of the four types in the population is different than expect

d)

the observed counts are equally distributed

5.

How do you find degrees of freedom for a chi-square goodness of fit test?

a)

one less than the sample size

b)

one less than the population

c)

total number divided 2

d)

total number times 2

6.

The shape of the Chi-Square distribution is ___.

a)

Left Skewed

b)

Right Skewed

c)

Unimodal & Symmetric

d)

Bimodal & Symmetric

7.
A chi-square test is used to test whether a 0 to 9 spinner is "fair" (that is, the outcomes are all equally likely).  The spinner is spun 100 times, and the results are recorded.  The degrees of freedom for the test will be
a)
8
b)
9
c)
10
d)
99
8.
In a chi square goodness of fit test, you must have
a)
counts for two categorical variables
b)
proportions for one categorical variable
c)
counts for one categorical variable
d)
proportions for two categorical variables
9.
What must be true about the expected values in a chi square test?
a)
greater than or equal to 2
b)
greater than or equal to 5
c)
greater than or equal to 10
d)
greater than or equal to 30
10.
The table shows the number or babies born on each day of the week.  Is there evidence that births are more likely on specific days of the week?
a)
yes, the p value is greater than .05
b)
no evidence, the p value is less than .05
c)
yes, the p value is less than .05
d)
no evidence, the p value is greater than .05
11.
What statistical model does this graph illustrate?
a)
normal model
b)
t model
c)
chi square model
12.
What is the expected value for all cells, if we assume that this is a fair six sided die?
a)
96
b)
16
c)
8
d)
1/6
13.
The bigger the chi square statistic, the ________ the p value.
a)
bigger
b)
smaller
14.

What is the test statistic for a Chi-Square Goodness of Fit Test?

a)

Σ((O−E)2E)\Sigma\left(\frac{\left(O-E\right)^2}{E}\right)

b)

O−EO-E

c)

O−E2O-E^2

d)

K−1K-1

15.

In both chi square tests, observed frequencies and expected frequencies are used to calculate the p-value

a)

True

b)

False

16.

The table shows the number or babies born on each day of the week. Is there evidence that births are more likely on specific days of the week? Use 5% level of significance.

a)

yes, the p value is greater than .05

b)

no, the p value is less than .05

c)

yes, the p value is less than .05

d)

no, the p value is greater than .05

17.

The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are the data in the table provided. An appropriate null hypothesis to test whether the food choices are equally popular is:

a)

H0: μ = 25, where μ = the mean number of students that prefer each type of food.

b)

H0: p = 0.25, where p = the proportion of all students who prefer Asian food.

c)

H0: ηA = ηM = ηP = ηH = 0.25, where ηA is the number of students in the school who would choose Asian food, and so on.

d)

H0: pA = pM = pP = pH = 0.25, where pA is the proportion of students in the sample who would choose Asian food, and so on.

18.

The chi square statistic is

a)

(18−25)225+...+(21−25)225\frac{\left(18-25\right)^2}{25}+...+\frac{\left(21-25\right)^2}{25}

b)

(25−18)218+...+(25−21)221\frac{\left(25-18\right)^2}{18}+...+\frac{\left(25-21\right)^2}{21}

c)

(18−25)25+...+(21−25)25\frac{\left(18-25\right)}{25}+...+\frac{\left(21-25\right)}{25}

d)

(18−25)2100+...+(21−25)2100\frac{\left(18-25\right)^2}{100}+...+\frac{\left(21-25\right)^2}{100}

e)

(0.18−0.25)20.25+...+(0.39−0.25)20.25\frac{\left(0.18-0.25\right)^2}{0.25}+...+\frac{\left(0.39-0.25\right)^2}{0.25}

19.

Suppose that a study in the past showed that the adult eye color distribution of American was blue 25%, green 10%, brown 50%, and black 15%.

If the hypothesized proportions are true, then in a sample of 500 adults, how many people with eye colors as blue, green, brown and black?

a)

Blue: 25

Green: 10

Brown: 50

Black: 15

b)

Blue: 125

Green: 50

Brown: 250

Black: 75

c)

Blue: 125

Green: 125

Brown: 125

Black: 125

20.

Is there evidence that income levels are not evenly distributed across the entire population? What is the p value and chi square statistic?

a)

yes, p value is 2.42 e-18, chi square statistic = 92.11

b)

yes, p value is .242, chi square statistic = 43.98

c)

no, p value is 2.42 e-18, chi square statistic = 92.11

d)

no, p value is 2.42, chi square statistic = 43.98

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