You are conducting a test of independence for the claim that there is an association between therow variable and the column variable.xYZA 29 20 || 57B 31 28 16The expected observations for this table would beYThe resulting Pearson residuals are:XYWhat is the chi-square test-statistic for this data?x²Report all answers accurate to three decimal places.

Name: You are conducting a test of independence for the claim that there is
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Description: You are conducting a test of independence for the claim that there is an association between therow variable and the column variable.xYZA 29 20 || 57B 31 28 16The expected observations for this table would beYThe resulting Pearson residuals are:XYWh

The observed observations are as follows X Y Z A 29 20 57 B 31 28 16 Let's find the…

Answered: You are conducting a test of…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.1: Measures Of Center

Problem 9PPS

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You are conducting a test of independence for the claim that there is an association between the
row variable and the column variable.
xYZ
A 29 20 || 57
B 31 28 16
The expected observations for this table would be
Y
The resulting Pearson residuals are:
X
Y
What is the chi-square test-statistic for this data?
x²
Report all answers accurate to three decimal places.

Expert Solution

Step 1

The observed observations are as follows

	X	Y	Z
A	29	20	57
B	31	28	16

Let's find the row and column total.

	X	Y	Z	Total
A	29	20	57	106
B	31	28	16	75
Total	60	48	73	181

Let's find the expected observations.

The formula for expected observation is $\frac{(row total) (column total)}{grand total}$

For example:

For the A row and X column, the observed observation is 29.

The expected observation would be : $\frac{(row total) (column total)}{grand total} = \frac{106 \times 60}{181} = 35.138$

In the same way, the expected value for each cell is as follows.

	X	Y	Z	Total
A	35.138	28.110	42.751	106
B	24.862	19.890	30.249	75
Total	60	48	73	181

Hence the expected observation for this table would be:

	X	Y	Z
A	35.138	28.110	42.751
B	24.862	19.890	30.249

Step 2

The formula for Pearson residual will be as follows.

$eij = \frac{observed observation - expected observation}{\sqrt{expected observation}}$

For example:

For the A row and X column, the observed observation is 29.

The expected observation is 35.138

Then the Pearson residual for A row and X column would be : $\frac{29 - 35.138}{\sqrt{35.138}} = - 1.035$

In the same way, the Pearson residual for each cell is as follows.

	X	Y	Z
A	-1.035	-1.530	2.179
B	1.231	1.818	-2.591

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You are conducting a test of independence for the claim that there is an association between the row variable and the column variable.