Question 857:

Asked on March 17, 2010

Tags: Chi-Square Test of Independence

1

Answer:

No answer provided yet.We will use a Chi-Square test of independence to determine if there is a relationship between packaging size and economic status. The Null Hypothesis for this test is that there is no relationship between the two variables (they are independent). We will reject the Null Hypothesis if we obtain a p-value of less than .05.

The test statistic, Chi-Square, is found by subtracting the expected value from the observed value, squaring this difference, then dividing by the expected value. You add this value up for all the cells in the table.

First we compute the column totals and row totals.

	Lower	Middle	Upper	Row Total
Small	24	22	18	64
Medium	23	28	19	70
Large	18	27	29	74
Jumbo	16	21	33	70
	81	98	99	278

Expected Values
Expected values are found by multiplying the row total by the column total and dividing by the total for the table.

	Lower	Middle	Upper
Small	18.647	22.56	22.79
Medium	20.39	24.67	24.93
Large	21.56	26.08	26.35
Jumbo	20.39	24.67	24.92

Now we subtract the expected counts from the actual counts from the first table, square it and divide by the expected.

	Lower	Middle	Upper
Small	1.53	.013	1.007
Medium	.3325	.447	1.41
Large	.588	.032	.2659
Jumbo	.947	.547	2.61

Finally we add up all the values in the 3rd table to get the Chi-Square test statistic of 9.74. We evaluate its significance using the degrees of freedom found as the (# of rows-1)*(# of cols -1) = 2*3 = 6. Looking up the significance of the chi-square value using a table or the excel function = CHIDIST(9.74,6) we get the p-value of .1359.

Since the p-value is above the alpha cutoff of .05 we FAIL to reject the Null and conclude there is not enough evidence to conclude the variables are related. In other words, there is not a significant relationship between packaging preference and economic status.