Question

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Performing a Chi-Square Independence Test The contingency table shows the results of a random sample of 2200 adults classified by their favorite way to eat ice cream and gender. The expected frequencies are displayed in parentheses. At a = 0.01, can you conclude that the adults' favorite ways to eat ice cream are related to gender? Favorite way to eat ice cream Gender Cup Cone Sundae Sandwich Other Male 600 288 204 24 84 Female 410 340 180 20 50 Total 1010 628 384 134 Total 1200 1000 2200 44

Answer 1

PERFORMING A CHI- SQUARE TEST OF INDEPENDENCE.

WE NEED TO TAKE TWO  VARIABLES

GENDER AND THE WAY OF EATING ICE CREAM

THEN WE NEED TO ASSUME THE NULL AND ALTERNATE HYPOTHESIS.

NULL HYPOTHESIS : BOTH VARIABLES ARE INDEPENDENT.

ALTERNATE HYPOTHESIS ; BOTH VARIABLES ARE DEPENDENT.

IF THE CALCULATED VALUE OF CHI SQUARE IS GREATER THAN THE CRITICAL VALUE OF CHI SQUARE, THEN THE NULL HYPOTHESIS WILL BE REJECTED.

AND

IF THE CALCULATED VALUE OF CHI SQUARE IS LESS THAN THE CRITICAL VALUE OF CHI SQUARE, THEN THE NULL HYPOTHESIS WILL BE ACCEPTED.

NOW WE CONDUCT THE HYPOTHESIS TEST.

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below: Column 1 Column 2 Column 3 Column 4 Column 5 Total 600 288 204 24 84 1200 410 340 180 20 50 1000 Total 1010 628 384 44 134 2200 The expected values are computed in terms of row and column totals. In fact, the formula is Eij = R: XC, where Ri corresponds to the total sum of elements in row i, C; corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values:

Expected Values Column 1 Column 2 Column 3 Column 4 Column 5 Total 1010x 1200 2200 550.909 628 x 1200 = 342.545 2200 384x 1200 2200 209.455 44 x 1200 JOT 24 134x1200 2200 = 73.091 1200 2200 1010x 1000 2200 459.091 628 x 1000 = 285.455 384 x 1000 = 174.545 44x1000 = 20 2200 134X1000 2200 = 60.909 2200 2200 1000 Total 1010 628 384 44 134 2200 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E-O)/E. The table with squared distances is shown below:

Squared Distances Column 1 Column 2 Column 3 Column 4 Column 5 (600-550.909) 550.909 4.374 (288-342.545) 342.545 8.686 (204-209.455 209.455 0.142 (24-24) = 0 (84-73.091 73.091 1.628 (410-459.091 459.091 5.249 (340-285.455) 285.455 10.423 (180-174.545) 174.545 0.17 (20-20) 20 =0 (50-60.909) 60.909 1.954 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: The two variables are independent H: The two variables are dependent This corresponds to a Chi-Square test of independence.

Rejection Region Based on the information provided, the significance level is a = 0.1, the number of degrees of freedom is df = (2-1) < (5-1) = 4, so then the rejection region for this test is R= {x? :x?> 7.779) Test Statistics The Chi-Squared statistic is computed as follows: n x? (Oi; - E;;)? 4.374 +5.249 +8.686 +10.423 +0.142 +0.17+0+0+1.628 + 1.954 = 32.627 Eij i=1 Decision about the null hypothesis Since it is observed that x = 32.627 > x = 7.779, it is then concluded that the null hypothesis is rejected.

Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.1 significance level. The corresponding p-value for the test is p = Pr(x > 32.627) = 0.