Question

Fewer TVs? The United States Energy Information Administration periodically surveys a random sample of U.S. households to determine how they use energy. One of the variables they track is how many TVs are in a household (None, $1,2,3,4,or5$or more). The computer output compares the distribution of number of TVs for households in $2009and2015$.

Cell Contents: Count

Expected count

Contribution to Chi-square

Chi-Squarerole="math" localid="1654195309908" $=137.137$, DF $=5$, P-Value $=0.000$

a. Which chi-square test is appropriate to analyze these data? Explain your answer.

b. Show how the numbers $252$and $14.113$were obtained for the $2009/None$cell.

c. Which $3$cells contribute most to the chi-square test statistic? How do the observed and expected counts compare for these cells?

Short answer

(a) The chi-square test for homogeneity is the most appropriate test for analyzing homogeneity.

(b) The numbers obtained from the cell are $E=252,{\chi }^2=14.113$

(c) The most contributing chi-square test statistics cells are role="math" localid="1654195739873" $None/2015,5ormore/2015,1/2015$And the expected count is greater than observed count.

Step-by-step solution

Part (a) Step 1: Given information

We need to find out the most appropriate chi-square test for analyzing data.

Part (a) Step 2: Explanation

We know that

• A chi-square goodness-of-fit test will be used if we are only interested in the distribution of one variable.
• A chi-square test for homogeneity is used when we are interested in the distribution of two variables and there are several independent samples.
• A chi-square test for independence is used when we are interested in the distribution of two variables and there is only one sample.

Two variables are of interest: the number of televisions in the home and the year. We should also notice that we have two independent samples (one for $2009$and one for $2015$), hence the chi-square test for homogeneity should be used.

Part (b) Step 1: Given information

We need to find out the reason for the numbers obtained for the $2009/None$cell.

Part (b) Step 2: Explanation

The first value in the $2009/None$cell is $192$, indicating that $192$is the observed count.

The product of the column and row totals, divided by the table total, yields the expected frequencies role="math" $E$. The row total for the row "None" is $366$, the column total for the column $"2009"is14557$, and the total for the table is $21177$.

$E=252$

The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

${\chi }^2=14.113$

Part (c) Step 1: Given information

We need to find the most contributing cells to chi-square test statistics.

Part (c) Step 2: Explanation

The third number in a cell is the cell's contribution to the chi-square test statistic. The cells $None/2015,5ormore/2015,1/2015$, respectively, had the largest contributions of $31.034,27.628,23.258$.

Furthermore, in the cells $None/2015,1/2015$, the expected count surpasses the observed count, whereas, in the cell $5ormore/2015$, the observed count exceeds the observed count (expected count is the second number in the cell and the observed count is the first number in the cell).