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
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.
To test the null hypothesis that the cereals on the market would be equally popular, we can use a chi-square test for independent proportions. Here's how you can calculate the test statistic:
Step 1: Set up the null and alternative hypotheses:
Null hypothesis: The cereals on the market are equally popular.
Alternative hypothesis: The cereals on the market are not equally popular.
Step 2: Calculate the expected frequencies under the null hypothesis:
To calculate the expected frequencies, we need to assume that the cereals are equally popular. We can find the expected frequency for each cereal by multiplying the proportion of the total consumers who would buy that cereal by the total sample size for each cereal.
For cereal A:
Expected frequency for cereal A = (Number of consumers who would buy cereal A / Total consumers) * Total sample size for cereal A
= (52 / 175) * 175
= 52
For cereal B:
Expected frequency for cereal B = (Number of consumers who would buy cereal B / Total consumers) * Total sample size for cereal B
= (35 / 150) * 150
= 35
Step 3: Calculate the chi-square test statistic:
The chi-square test statistic can be calculated using the following formula:
χ² = ∑ (O - E)² / E
where,
O is the observed frequency
E is the expected frequency
Using the given information, we can calculate the test statistic as follows:
χ² = ((52 - 52)² / 52) + ((35 - 35)² / 35)
= (0² / 52) + (0² / 35)
= 0 + 0
= 0
Step 4: Determine the critical value:
Since the significance level is 5%, we need to find the critical value for chi-square at a 5% significance level with (2-1) = 1 degree of freedom. Looking up in the chi-square distribution table, the critical value is approximately 3.841.
Step 5: Compare the test statistic with the critical value:
Since the test statistic (0) is less than the critical value (3.841), we fail to reject the null hypothesis.
Conclusion: Based on the test statistic, we do not have enough evidence to conclude that the cereals on the market are not equally popular.
To test the null hypothesis that the cereals on the market would be equally popular, we can use the chi-square test for independence. The test statistic for the chi-square test is calculated using the formula:
χ² = Σ((O - E)² / E)
where χ² is the test statistic, O is the observed frequency, and E is the expected frequency.
In this case, we have two test panels: one for cereal A and another for cereal B. We need to calculate the expected frequencies to compare with the observed frequencies.
To calculate the expected frequencies, we first need to find the overall proportions for each cereal.
Proportion for cereal A:
p(A) = 52/175 = 0.2971
Proportion for cereal B:
p(B) = 35/150 = 0.2333
Next, we can calculate the expected frequencies:
Expected frequency for cereal A:
E(A) = p(A) * total panel size = 0.2971 * 175 = 52.03
Expected frequency for cereal B:
E(B) = p(B) * total panel size = 0.2333 * 150 = 35.00
Now we can calculate the test statistic:
χ² = ((O(A) - E(A))² / E(A)) + ((O(B) - E(B))² / E(B))
Substituting the observed frequencies:
χ² = ((52 - 52.03)² / 52.03) + ((35 - 35.00)² / 35.00)
Calculating the numerator:
((52 - 52.03)² + (35 - 35.00)²) = (0.0009 + 0) = 0.0009
Calculating the test statistic:
χ² = 0.0009 / 52.03 + 0 / 35.00 = 0.000017
So, the test statistic (χ²) is approximately 0.000017.
To interpret this result, we would compare it to the critical value from the chi-square distribution with (number of rows - 1) * (number of columns - 1) degrees of freedom at a 5% significance level.
In this case, since we are comparing two proportions (rows: A, B and columns: buy, not buy), we have (2-1)*(2-1) = 1 degree of freedom.
Looking up the critical value for chi-square distribution with 1 degree of freedom and a 5% significance level, it is approximately 3.841.
Since the test statistic (0.000017) is less than the critical value (3.841), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the cereals on the market are not equally popular.