Set-Up
Conditions
Calculations
Conclusions
Potpourri
100
We know what the manufacturer claims the real color distribution of skittles to be, and we want to see if they are lying. We purchase a large bag of skittles to help us test this claim. We would use a __________.
Chi-Square GOF Test
100
To conduct a Chi-Square test, we must have this type of data.
Categorical
100
We calculate the expected counts in a Chi-Square GOF Test by _________.
multiplying the sample size by the respective proportion
100

At the alpha=0.05 significance level, a Chi-Square statistic of 17.87 and a df of 6 would result in the following conclusion.

We do have sufficient evidence to reject the null hypothesis.

100
What is the shape of any Chi-Square distribution?
Skewed Right
200

We want to know if, in general, males and females get the same scores on their AP Statistics exams. We use a sample consisting of both males and females to assess this. We would use a __________.

Chi-Square Test for Homogeneity

200
Every expected count must have a value of at least ___.
5
200
__________ determines the number of Degrees of Freedom for a Chi-Square GOF Test.
(# of categories) - 1
200
At the alpha=0.01 significance level, a Chi-Square statistic of 30 and a df of 15 would result in the following conclusion.
We do not have sufficient to reject the null hypothesis. We must fail to reject.
200
The larger the Chi-Square statistic, the __________ the p-value.
Smaller
300
We want to determine whether or not musical preference among high school students has any influence on these students' preference of social media sites. We would use a __________
Chi-Square Test for Association/Independence
300
Data for Chi-Square tests must be in the form of ____.
Counts
300
__________ determines the number of Degrees of Freedom for a Chi-Square Test for Association/Independence.
(# of rows - 1)*(# of columns - 1)
300
We are testing whether or not the city of Waubun has the same age distribution as that of Ogema. Our Chi-Square value is 17.8, and our p-value is 0.003. A complete conclusion (based on a significance level of alpha=0.05) is __________.
Since our p-value (0.003) is less than our significance level of alpha=0.05, we must reject the null hypothesis. We do have sufficient evidence to suggest that there IS a difference between the distribution of age among Waubun residents and the distribution of age among Ogema residents.
300

The type of test that we run if we are exploring the relationship between 2 variables and have taken 2 samples

chi-square test for independence 

400
We want to see if hand dominance (being right-handed or left-handed) and eye color are genetically linked. The hypotheses for this test would be __________.
Ho: There is no association between hand dominance and eye color. Ha: There is an association between hand dominance and eye color.
400
The Random condition is met if _________ or _________.
the data come from a random sample or the data come from a randomized experiment
400
The formula for calculating the expected counts of a Chi-Square Test for Homogeneity is __________.
[(Row Total)*(Column Total)]/(Table Total)
400
We ran a Chi-Square test to determine whether males or females traditionally score higher on their AP Statistics exams. We got a Chi-Square statistic of 3.72 and a p-value of 0.146. A complete conclusion (based on a significance level of alpha=0.05) is __________.
Since our p-value (0.146) is greater than our significance level of alpha=0.05, we must fail to reject the null hypothesis. We do NOT have sufficient evidence to suggest that there is a difference between the distribution of AP Statistics exam scores for males and the distribution of AP Statistics exam scores for females.
400

At the alpha=0.05 significance level, a Chi-Square statistic of 17.87 and a df of 6 would result in the following conclusion.

reject the null

500
We want to know if People from Georgia and New York prefer the same brand of toothpaste, so we will take a sample of residents from each state and record the number of individuals who prefer each brand of toothpaste. The hypotheses for this test would be __________.
Ho: There is no difference between the distribution of toothpaste preference for residents of Georgia and the distribution of toothpaste preference for residents of New York. Ha: There is a difference between the distributions of toothpaste preference between the two states.
500
Before we check the conditions for a Chi-Square Test, we need to __________.
Identify the test to be used
500
The formula for calculating a Chi-Square Test Statistic is __________.
The sum of [(observed - expected)^2]/(expected)
500
We want to determine whether or not musical preference among high school students has any influence on these students' preference of social media sites. The Chi-Square statistic was 18.7 and the p-value was 0.043. A complete conclusion (based on a significance level of alpha=0.05) is __________.
Since our p-value (0.043) is less than our significance level of alpha=0.05, we must reject the null hypothesis. We have sufficient evidence to conclude that there IS an association between musical preference and social media site preference.
500

A Chi-Square Test for Homogeneity is used when we want to compare __________ with __________.

the distribution of one categorical variable for one population, the distribution of the same categorical variable for a second population 

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