Q. R 11.3 Fewer TVs? The United States Ene... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 11: Q. R 11.3 (page 763)

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, $1, 2, 3, 4, o r 5$ or more). The computer output compares the distribution of number of TVs for households in $2009 a n d 2015$ .
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 $2009 / N o n e$ 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

Expert verified

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

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

(c) The most contributing chi-square test statistics cells are role="math" localid="1654195739873" $N o n e / 2015, 5 o r m o r e / 2015, 1 / 2015$ 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 $2009 / N o n e$ cell.

Part (b) Step 2: Explanation

The first value in the $2009 / N o n e$ 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 $" 2009 " i s 14557$ , 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.

$蠂^{2} = 14.113$

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 $N o n e / 2015, 5 o r m o r e / 2015, 1 / 2015$ , respectively, had the largest contributions of $31.034, 27.628, 23.258$ .

Furthermore, in the cells $N o n e / 2015, 1 / 2015$ , the expected count surpasses the observed count, whereas, in the cell $5 o r m o r e / 2015$ , 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).