Free Printable Chi Square independence test

1.

Find the degrees of freedom.

a)

4

b)

5

c)

6

d)

7

2.

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

3.

Which test would you use to see if gender is a factor choosing a person’s favorite fruit? 900 college students were surveyed and their favorite fruit recorded.

a)

Chi-square Goodness of Fit Test

b)

Chi-Square Test for Homogeneity

c)

Chi-Square Test for Independence/Association

d)

Not a Chi-square test

4.

What are the expected counts of a female who likes Pepsi?

a)

10.5

b)

11

c)

14.5

d)

6.3

5.

A random sample of traffic tickets given to motorists in a large city is examined. The tickets are classified according to the race of the driver. The results are summarized in the table provided. Assuming H₀ is true, the expected number of Hispanic drivers who would receive a ticket is.....

(Hint: use expected number = np)

a)

10.36

b)

11

c)

11.84

d)

12

6.

What are the degrees of freedom for the following χ² test?

a)

6

b)

2

c)

5

d)

36

7.

Is nationality related to favorite drink? 300 people were observed.

a)

Chi-square Goodness of Fit Test

b)

Chi-Square Test for Homogeneity

c)

Chi-Square Test for Independence/Association

d)

Not a Chi-square test

8.

In the formula for Chi Sqare, the E stands for

a)

the sum

b)

the observed frequency

c)

The expected frequency

d)

the p value

9.

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 expected counts for spinning a 5 will be

a)

5

b)

10

c)

11.1

d)

20

10.

In the formula for Chi Sqare, the O stands for

a)

the sum

b)

the observed frequency

c)

The expected frequency

d)

the p value

11.

In the formula for Chi Sqare, the E stands for

a)

the sum

b)

the observed frequency

c)

The expected frequency

d)

the p value

12.

What is the null hypothesis?

a)

Injury type is independent of position played

b)

Injury type is dependent on the position played independent

13.

What is the

χ²_calculatedvaluefor the following situation.

a)

6.65

b)

6.66

c)

0.00991

d)

1

14.

If we are testing at a 5% significance level and our p-value is 0.00453, what is the conclusion?

a)

Accept Ho

b)

Reject Ho

c)

We need to know chi-squared

d)

You cannot do this problem without a calculator

15.

Find the chi-squared calculated value for the table.

a)

0.486170447

b)

1.442392006

c)

1.45

d)

2

16.

Given that a student is male, what is the probability that maths is their favorite?

a)

0.491

b)

0.225

c)

0.443

d)

.508

17.

For a p-value from the calculator of 1.32355E-3 what would the conclusion be if testing at 10% significance level?

a)

Accept Ho

b)

Reject H1

c)

Reject Ho

d)

Need to know the chi-squared number

18.

Have shoe brand preferences changed between last year and this year? 1000 people were observed this year and last year to see what brand of shoe they were wearing.

a)

Chi-square Goodness of Fit Test

b)

Chi-Square Test for Homogeneity

c)

Chi-Square Test for Independence/Association

d)

Not a Chi-square test

19.

All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Computer software was used to analyze the data.

The expected count of cases of lymphoma in homes with an HCC is

a)

$\frac{79\times31}{215}$

b)

$\frac{10\times21}{215}$

c)

$\frac{79\times31}{10}$

d)

$\frac{136\times31}{215}$

e)

None of the above

20.

All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Computer software was used to analyze the data.

The $\chi^2$ statistic is equal to;

(give your answer to 3 decimal places

(a)

Chi Square independence test

Total questions: 20

Worksheet time: 30mins

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