Delineate the basic ideas behind the chi-square test of independence. Define a contingency table . Provide the null and alternative hypotheses. Explain how the test statistic is constructed. Give the basic assumptions that underlie the application of the chi-square distribution.

Question

Want to see more full solutions like this?

Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Harvard University California Institute of Technology Massachusetts Institute of Technology Stanford University Princeton University University of Cambridge University of Oxford University of California, Berkeley Imperial College London Yale University University of California, Los Angeles University of Chicago Johns Hopkins University Cornell University ETH Zurich University of Michigan University of Toronto Columbia University University of Pennsylvania Carnegie Mellon University University of Hong Kong University College London University of Washington Duke University Northwestern University University of Tokyo Georgia Institute of Technology Pohang University of Science and Technology University of California, Santa Barbara University of British Columbia University of North Carolina at Chapel Hill University of California, San Diego University of Illinois at Urbana-Champaign National University of Singapore McGill…

Name Harvard University California Institute of Technology Massachusetts Institute of Technology Stanford University Princeton University University of Cambridge University of Oxford University of California, Berkeley Imperial College London Yale University University of California, Los Angeles University of Chicago Johns Hopkins University Cornell University ETH Zurich University of Michigan University of Toronto Columbia University University of Pennsylvania Carnegie Mellon University University of Hong Kong University College London University of Washington Duke University Northwestern University University of Tokyo Georgia Institute of Technology Pohang University of Science and Technology University of California, Santa Barbara University of British Columbia University of North Carolina at Chapel Hill University of California, San Diego University of Illinois at Urbana-Champaign National University of Singapore…

A company found that the daily sales revenue of its flagship product follows a normal distribution with a mean of $4500 and a standard deviation of $450. The company defines a "high-sales day" that is, any day with sales exceeding $4800. please provide a step by step on how to get the answers in excel Q: What percentage of days can the company expect to have "high-sales days" or sales greater than $4800? Q: What is the sales revenue threshold for the bottom 10% of days? (please note that 10% refers to the probability/area under bell curve towards the lower tail of bell curve) Provide answers in the yellow cells

Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Answer 6

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Answer 7

Chapter 10, Problem 1LC

(a)

To determine

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

Answer 8

(a)

Expert Solution

Explanation of Solution

Chi-square test of independence:

A single simple random sampling is done with each individual being classified based on the two categorical variables. Then, the hypothesis testing is carried out to check whether there is any relationship between the two categorical variables. This test is called chi-square test of independence.

Contingency tables:

The data from the two variables of a population are known as bivariate data and the frequency distribution for the bivariate data is known as the contingency table.

The test hypotheses are given below:

H₀: The variables are independent.

H₁: The variables are not independent.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in the contingency table.

R= number of rows in a contingency table.

C= number of columns in a contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumptions underlie the application of the chi-square as given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

Answer 13

(b)

To determine

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

Answer 14

(b)

Expert Solution

Explanation of Solution

Chi-square test of goodness of fit:

The test of goodness of fit enables people to see how well does a specific (theoretical) distribution fits the observed data. Then, the hypothesis testing is carried out to check whether a frequency distribution follows a claimed distribution or not.

Chi-square test of goodness of fit is a test that provides conformation between the observed frequencies (O_i) the expected frequencies (E_i) for several categories.

In a goodness of fit, the following is obtained:

O= Observed frequency count of a category.

E= Expected frequency of a category.

The value of the expected frequency will be as given below:

E=(Sample size,n)×(Probability assigned to category)=n×p

The test hypotheses are given below:

H₀: The population fits the specified distribution of categories.

H₁: The population has a different distribution.

Test statistic for the chi-square test of goodness of fit:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(k−1)

Where, the sum is over all the cells in the contingency table and k= number of categories.

The basic assumption that underlies the application of the chi-square is as given below:

In order to apply a chi-square distribution, the sample size should be large so that for each cell, E≥5.

Similarities between the tests of independence and the tests of goodness of fit:

Normally, the chi-square test is used to compare the frequencies of one category variable to the theoretical distribution (test of goodness of fit) and to compare the frequencies of one category variable to the different values of a second categorical variable (test of independence).

Therefore, both tests take a categorical variable to perform the hypothesis.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

Answer 19

(c)

To determine

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

Answer 20

(c)

Expert Solution

Explanation of Solution

If the standard deviation of a population standard deviation is equal to a specified value, then the chi-square method of testing and estimating a standard deviation is used. It is applicable only when random variable x has a normal distribution. In testing σ2, the normality assumption must be strictly followed.

The basic assumptions that underlie the application of the chi-squareis are given below:

Sample selected is a random sample.
The distribution of observation should be approximately normal.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Answer 25

(d)

To determine

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

Answer 26

(d)

Expert Solution

Explanation of Solution

Chi-square test of homogeneity:

Independent simple random samples are taken from two or more populations, where every individual is being classified based on one categorical variable. Here, the hypothesis testing is done to check whether any homogeneity exists. In other words, the homogeneity tests claim that different populations are from the same proportions of specified characteristics. This test is called the chi-square test of homogeneity.

The test hypotheses are given below:

H₀: The proportion of each population sharing specified characteristics is the same for all populations.

H₁: The proportion of each population sharing specified characteristics is not the same for all populations.

Test statistic for the chi-square test of independence:

χ2=∑((O−E)2E) with degrees of freedom d.f.=(R−1)(C−1)

Where, the sum is over all the cells in a contingency table.

R= number of rows in the contingency table.

C= number of columns in the contingency table.

E= expected frequency.

For each cell, the value of the expected frequency is obtained as given below:

E=(Row total)(Column total)Sample size

The basic assumption that underlies the application of the chi-square is given below:

In order to apply the chi-square distribution, the sample size should be large so that for each cell, E≥5.

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Ch. 10.1 - Statistical Literacy In general, are chi-square...Ch. 10.1 - Statistical Literacy For chi-square distributions,...Ch. 10.1 - Prob. 3P Ch. 10.1 - Prob. 4P Ch. 10.1 - Prob. 5P Ch. 10.1 - Prob. 6P Ch. 10.1 - Prob. 7P Ch. 10.1 - Interpretation: Test of Independence Consider...Ch. 10.1 - Prob. 9P Ch. 10.1 - For Problems 919, please provide the following...

Delineate the basic ideas behind the chi-square test of independence. Define a contingency table . Provide the null and alternative hypotheses. Explain how the test statistic is constructed. Give the basic assumptions that underlie the application of the chi-square distribution.

Concept explainers

Videos

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

Explanation of Solution

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

Explanation of Solution

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

Explanation of Solution

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions

Delineate the basic ideas behind the chi-square test of independence. Define a contingency table . Provide the null and alternative hypotheses. Explain how the test statistic is constructed. Give the basic assumptions that underlie the application of the chi-square distribution.

Concept explainers

Videos

Delineate the basic ideas behind the chi-square test of independence. Define a contingency table. Provide the null and alternative hypotheses. Explain how the test statistic is constructed. Give the basic assumptions that underlie the application of the chi-square distribution.

Explanation of Solution

Delineate the basic ideas behind the chi-square test of goodness of fit. Provide the null and alternative hypotheses. Explain how the test statistic is constructed. Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

Explanation of Solution

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation. Give the basic assumptions that underlie the application of the chi-square distribution.

Explanation of Solution

Delineate the basic ideas behind the chi-square test of homogeneity. Provide the null and alternative hypotheses. Explain how the test statistic is constructed. Give the basic assumptions that underlie the application of the chi-square distribution.

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions

Delineate the basic ideas behind the chi-square test of independence.

Define a contingency table.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.

Delineate the basic ideas behind the chi-square test of goodness of fit.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Discuss and summarize the similarities between the tests of independence and tests of goodness of fit.

Delineate the basic ideas behind the chi-square method of testing and estimating a standard deviation.

Give the basic assumptions that underlie the application of the chi-square distribution.

Delineate the basic ideas behind the chi-square test of homogeneity.

Provide the null and alternative hypotheses.

Explain how the test statistic is constructed.

Give the basic assumptions that underlie the application of the chi-square distribution.