To test the null hypothesis that the cereals on the market would be equally popular, we can use the chi-square test for independence.
First, let's set up the contingency table with the observed frequencies:
Cereal A Cereal B Total
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Test Panel 1 52 123 175
Test Panel 2 35 115 150
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Total 87 238 325
We need to calculate the expected frequencies under the assumption of equal popularity of the cereals.
The expected frequency for each cell can be calculated as:
Expected Frequency = (row total * column total) / grand total
Let's calculate the expected frequencies:
Expected Frequency (Cereal A - Test Panel 1) = (175 * 87) / 325 = 46.2
Expected Frequency (Cereal A - Test Panel 2) = (150 * 87) / 325 = 40.2
Expected Frequency (Cereal B - Test Panel 1) = (175 * 238) / 325 = 127.6
Expected Frequency (Cereal B - Test Panel 2) = (150 * 238) / 325 = 110.4
Now, we can calculate the chi-square test statistic:
Chi-square = Σ [(Observed Frequency - Expected Frequency)^2 / Expected Frequency]
For our contingency table, the calculation would be:
Chi-square = [(52 - 46.2)^2 / 46.2] + [(123 - 127.6)^2 / 127.6] + [(35 - 40.2)^2 / 40.2] + [(115 - 110.4)^2 / 110.4]
Chi-square = (5.8^2 / 46.2) + (-4.6^2 / 127.6) + (-5.2^2 / 40.2) + (4.6^2 / 110.4)
Chi-square = 1.07 + 0.153 + 1.366 + 0.191
Chi-square = 2.779
Finally, we can compare the calculated chi-square value to the critical chi-square value at a significance level of 5% and degrees of freedom equal to (number of rows - 1) * (number of columns - 1). In this case, degrees of freedom = (2 - 1) * (2 - 1) = 1.
Using a chi-square distribution table or a calculator, the critical chi-square value at a 5% significance level and 1 degree of freedom is approximately 3.841.
Since the calculated chi-square value (2.779) is smaller than the critical chi-square value (3.841), we fail to reject the null hypothesis. There is not enough evidence to conclude that the cereals on the market are significantly different in popularity.
Fifty two of a test panel of 175 consumers say that they would buy cereal A if it is presented on the market and 35 of another test panel of 150 consumers say that they would buy cereal B. Test the null hypothesis that the cereals on the market would be equally popular. Use a 5% significance level. Calculate the test statistic
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