Chpt 11 - Chi-Square Tests

Question 1

What test can be used to find out if a die is claimed to be unfair?

Answer 1

Chi squared test

Question 2

What is the chi square distribution

Answer 2

Special type of right-skewed curve which depends on its degrees of freedom

Starts at 0 on the horizontal axis and extends indefinitely to the right, approaching, but never touching, the horizontal axis

Question 3

What is the total area under the chi squared curve?

Answer 3

Equal to 1

Question 4

What are the effects of degrees of freedom on a chi-squared curve?

Answer 4

The larger the degrees of freedom, the more the X2 curve looks like normal curves

Question 5

What is the value of X2α?

Answer 5

The area of α to its right under the chi-square curve

Question 6

Which table do we use to find the X2α value?

Answer 6

Table VII

Question 7

What is the formula to determine the expected frequency for a chi-squared test?

Answer 7

E = np

E is the expected frequency
n is the sample size
p is the probability specified by Ho

Question 8

What is the test statistic for a chi-squared test?

Answer 8

X2 = Σ(O-E)squared/E

O is observed frequency

E is expected frequency

Question 9

What are the steps to a goodness-of-fit test?

Answer 9

1. Set up the hypotheses
2. Check the assumptions
3. Decide significance level and find critical value
4. Calculate the test statistic
5. Compare the test statistic with critical value
6. Interpret the result in the context of the question

Question 10

What are the assumptions that must be checked for a chi-square goodness-of-fit test?

Answer 10

All expected frequencies are at least 1

At most 20% of the expected frequencies are less than 5

Simple random sample

Question 11

What are the basics for the hypotheses for a chi-square goodness-of-fit test?

Answer 11

Ho: The variable has the specified distribution

Ha: The variable does not have the specified distribution

Question 12

How do we determine degrees of freedom for a chi-squared goodness-of-fit test?

Answer 12

c-1

the number of categories - 1

Question 13

If we are using the P-value to solve a chi-squared goodness-of-fit test, how do we determine if we reject the Ho?

Answer 13

If:

α > p value -> we reject Ho

α < p value -> we DO NOT reject ho

Question 14

What is the rejection region of a chi-squared test and how do we determine if we reject Ho?

Answer 14

The critical value is X2α with df=C-1

The rejection region is the area to the right of the critical value

If the test statistic is larger than the X2a value, we reject Ho

If the test statistic is smaller than the X2a value, we DO NOT reject Ho

Question 15

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

Set up the hypotheses

Answer 15

Ho: The distribution of the outcome of rolling this die is P(X=x) = 1/6, x = 1, 2, 3, 4, 5, 6

Ha: the distribution is not the one as shown above

Question 16

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

Determine the significance level and critical value

Answer 16

α = 5% = 0.05

df = 6 category (one for each die) -1 = 5

Z2α = 11.070

Question 17

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

What are the expected outcomes?

Answer 17

E = np = 1200 x 1/6 = 200

Each side (1, 2, 3, 4, 5, 6) all have the same expected outcome because we expect Ho to be true, so all sides should be equal

Question 18

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

Check the assumptions

Answer 18

simple random sample ✓

all expected frequencies are at least 1 ✓

at most 20% of the expected frequencies are less than 5 ✓

(all expected outcomes should be 200)

Question 19

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

If the observed frequency of 3 was 183, what is the statistic for this line?

How do we determine the test statistic?

Answer 19

(O-E )squared/E

(183-200) squared/200 = 1.445

The test statistic is the sum of this value for each category (so the dice sides 1-6)

Question 20

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

Compare:

Test statistic 11.38
Critical value 11.070

Answer 20

The test statistic value is greater than the critical value, so we reject Ho

Question 21

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

Interpret

The test statistic value is greater than the critical value, so we reject Ho

Answer 21

At the 5% significance level, the data provides sufficient evidence that the die is unfair

Question 22

A six sided die is claimed to be unfair so we rolled the die 1200 times and observed the results. We are using a significance level of 5%.

Compare:

P(X2 < 11.38) = 0.9557

Answer 22

The p-value given by the software is the area to the left, for chi-squared goodness-of-fit test, we need the area to the right so:

1-0.9557 = 0.0444

α = 0.05

α > p value so we reject Ho

Question 23

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level.

Set up the hypotheses

Answer 23

Ho: distributions of blood type is

P(O) = 0.46
P(A) = 0.42
P(B) = 0.09
P(AB) = 0.03

Ha: the distribution is not the one as shown above

Question 24

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level.

Check assumptions

Answer 24

simple random sample ✓

all expected frequencies are at least 1 ✓

at most 20% of the expected frequencies are less than 5 ✓

(the smallest expected outcome is 3% of 200=6)

Question 25

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level. Find critical value

Answer 25

α = 1% = 0.01 degrees of freedom = number of blood types - 1 = 4-1 = 3 Critical value is 11.345

Question 26

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level. If the number of people with A blood was actually 76 in the community, what is the calculation for this line? How do we use this information to determine the test statistic?

Answer 26

Expected: np = 200*0.42 = 84 (O-E)squared/E = (76-84) squared / 84 = 0.7619 Test statistic is the sum of this equation for each category

Question 27

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level. Compare: Test statistic 3.7559 critical value 11.345

Answer 27

The test statistic is less than the critical value, we DO NOT reject Ho

Question 28

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level. Interpret: The test statistic is less than the critical value, we DO NOT reject Ho

Answer 28

At the 1% significance level, the data does not provide sufficient evidence that the proportions in this community differ from those in the general population

Question 29

The proportions of blood types O, A, B, and AB in the general population are known to be 46%, 42%, 9%, 3% correspondingly. A research team, investigating a small isolated community in Canada of 200, obtained the following frequencies of blood type. Test that the proportions in this community differ significantly from those in the general population at 1% significance level. Compare: P value is P(X2 < 3.7558) = 0.7109

Answer 29

The value given is the area to the left and we need the area to the left so: 1-0.7109 = 0.2891 α = 0.01 α < p value so we DO NOT reject Ho

Question 30

What are the 6 steps of a chi-square independence test?

Answer 30

1. Set up the hypotheses 2. Check the assumptions 3. Decide significance level and find critical value 4. Calculate the test statistic 5. Compare the test statistic with critical value 6. Interpret the result in the context of the question

Question 31

What are the assumptions for a chi-square independence test?

Answer 31

All expected frequencies are at least 1 At most 20% of the expected frequencies are less than 5 Simple random sample

Question 32

What are the basics for the hypotheses for a chi-square goodness-of-fit test?

Answer 32

Ho: the two variables are not associated (independent) Ha: the two variables are associated (not independent)

Question 33

How are expected frequencies determined for chi-independence tests?

Answer 33

E = RC/n E - expected frequencies R - Row frequency C - column frequency n - sample size

Question 34

How is a test statistic calculated for a chi-independence test?

Answer 34

X2 = Σ(O-E)squared/E O is observed frequency E is expected frequency Degrees of freedom = (r-1)(c-1) r - number of row variables c - number of column variables As a reminder: E = RC/n E - expected frequencies R - Row frequency C - column frequency n - sample size

Question 35

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. Set up the hypotheses

Answer 35

Ho: gender and age are not associated (independent) Ha: gender and age are associated (not independent)

Question 36

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. Check the assumptions

Answer 36

To do this you would have to set up the table to determine expected frequencies first, and when we do, all are over 5, so: simple random sample ✓ all expected frequencies are at least 1 ✓ at most 20% of the expected frequencies are less than 5 ✓

Question 37

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. We are going to group the ages into 20-34, 35-44, and 45-64 Determine the critical value

Answer 37

α = 5% = 0.05 df = (r-1)(c-1) = (2-1)(3-1) = 1*2 = 2 Critical value is 5.991

Question 38

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. We are going to group the ages into 20-34, 35-44, and 45-64. Determine the statistic for this line: --------------20-34----Total Women---2768----9445 Total-------5166----17890 How do we use this to determine the test statistic for the chi independence test?

Answer 38

Expected = RC/n = (9445*5166)/17890 = 2727.382 (O-E)squared/E = (2768-2727.382)squared/2727.382 = 0.6049 For the test statistic, we take the sum of all of the above values

Question 39

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. We are going to group the ages into 20-34, 35-44, and 45-64. Compare Test statistic is 8.4567 critical value is 5.991

Answer 39

Test statistic is larger than the critical value, so we reject Ho

Question 40

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. We are going to group the ages into 20-34, 35-44, and 45-64. Interpret: Test statistic is larger than the critical value, so we reject Ho

Answer 40

At the 5% significance level, the data provides sufficient evidence that the two variables of gender and age are associated

Question 41

We are interested in a population with a regular doctor by age group and gender in Canada. We randomly selected 17 890 Canadians. At the 5% significance level, we want to test the claim that gender and age are associated. We are going to group the ages into 20-34, 35-44, and 45-64. Compare P-value is P(X2, df=2 < 0.9854)

Answer 41

The value given is to the left and we need the area to the right so 1-0.9854 = 0.0146 α = 0.05 α > p value so we reject Ho