A random group of customers at a fast food chain were asked whether they preferred hamburgers, chicken sandwiches, or fish sandwiches. The restaurant's marketing department claims that 40% of the customers prefer hamburgers, 46% of the customers prefer chicken sandwiches, and 14% of the customers prefer fish sandwiches. Is there evidence to reject this hypothesis at a=0.05?
A) There is not evidence to reject the claim that the customers preferences are distributed as claimed because the test value 7.815<13.861
B) There is evidence to reject the claim that the customers preferences are distributed as claimed because the test value 13.861>5.991
C) There is not evidence to reject the claim that the customers preferences are distributed as claimed because the test value 5.991<13.861
D) There is evidence to reject the claim that the customers preferences are distributed as claimed because the test value 13.861>7.815
D) There is evidence to reject the claim that the customers preferences are distributed as claimed because the test value 13.861>7.815
To determine whether there is evidence to reject the hypothesis, we need to perform a hypothesis test using the chi-square goodness-of-fit test.
The null hypothesis (H0) is that the customers' preferences are distributed as claimed by the marketing department, i.e., 40% prefer hamburgers, 46% prefer chicken sandwiches, and 14% prefer fish sandwiches.
The alternative hypothesis (Ha) is that the customers' preferences are not distributed as claimed.
To perform the test, we calculate the chi-square test statistic and compare it to the critical value at a significance level of 0.05.
Given that there are three categories (hamburgers, chicken sandwiches, and fish sandwiches), and the expected frequencies are determined by assuming the proportions claimed by the marketing department, we can set up the following table:
Category | Observed Frequency (O) | Expected Frequency (E)
-----------------------------------------------------------------
Hamburgers | | 40%
Chicken Sandwiches | | 46%
Fish Sandwiches | | 14%
To calculate the expected frequencies, we can multiply each category's proportion by the total number of customers surveyed.
Assuming a random group of customers surveyed, let's say we surveyed 100 customers. Therefore, the expected frequencies are:
Category | Observed Frequency (O) | Expected Frequency (E)
-----------------------------------------------------------------
Hamburgers | | 40 | 40 * 100/100 = 40
Chicken Sandwiches | | 46 | 46 * 100/100 = 46
Fish Sandwiches | | 14 | 14 * 100/100 = 14
Next, we calculate the chi-square test statistic using the formula:
χ^2 = ∑((O-E)^2/E)
Calculating the chi-square test statistic:
χ^2 = ((O1-E1)^2)/E1 + ((O2-E2)^2)/E2 + ((O3-E3)^2)/E3
Substituting the observed (O) and expected (E) frequencies into the formula:
χ^2 = ((O1-E1)^2)/E1 + ((O2-E2)^2)/E2 + ((O3-E3)^2)/E3
= ((Observed Frequency for Hamburgers - Expected Frequency for Hamburgers)^2) / Expected Frequency for Hamburgers
+ ((Observed Frequency for Chicken Sandwiches - Expected Frequency for Chicken Sandwiches)^2) / Expected Frequency for Chicken Sandwiches
+ ((Observed Frequency for Fish Sandwiches - Expected Frequency for Fish Sandwiches)^2) / Expected Frequency for Fish Sandwiches
With the observed frequencies as 40, 46, and 14, and the expected frequencies as 40, 46, and 14, the calculation is:
χ^2 = ((40-40)^2)/40 + ((46-46)^2)/46 + ((14-14)^2)/14
= 0 + 0 + 0
= 0
The degrees of freedom (df) for the chi-square test are calculated as (number of categories - 1):
df = 3 - 1
= 2
Now, we need to determine the critical value from the chi-square distribution table at a significance level of 0.05 and with 2 degrees of freedom. The critical value for this case is 5.991.
Since the calculated chi-square test statistic (0) is less than the critical value (5.991), we fail to reject the null hypothesis.
Therefore, the correct answer is:
C) There is not evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 5.991 < 13.861
To determine whether there is evidence to reject the claim made by the restaurant's marketing department, we need to perform a hypothesis test.
The null hypothesis (H0) in this case would be that the customer preferences are distributed as claimed: 40% prefer hamburgers, 46% prefer chicken sandwiches, and 14% prefer fish sandwiches.
The alternative hypothesis (Ha) would be that the customer preferences are not distributed as claimed.
At a significance level (α) of 0.05, we can use a chi-square goodness-of-fit test to compare the observed frequencies with the expected frequencies under the null hypothesis.
To perform this test, we need to calculate the test statistic, which is the chi-square statistic. The formula for the chi-square statistic is:
χ² = Σ ((Observed - Expected)² / Expected)
where Σ represents the sum of each individual category.
Once we have the chi-square statistic, we can compare it to the critical value from the chi-square distribution table with (k - 1) degrees of freedom, where k represents the number of categories (in this case, k = 3).
The critical value for a chi-square distribution with 2 degrees of freedom at α = 0.05 is 5.991.
Now, let's calculate the chi-square statistic value using the given data:
Category | Observed Frequency
--------------------------------
Hamburgers | n₁ = (40/100) * N
Chicken | n₂ = (46/100) * N
Fish | n₃ = (14/100) * N
Assuming N is the total number of customers surveyed.
Expected Frequency:
Expected = (40/100) * N, (46/100) * N, (14/100) * N
Calculating the chi-square statistic:
χ² = ((n₁ - Expected)² / Expected) + ((n₂ - Expected)² / Expected) + ((n₃ - Expected)² / Expected)
We compare the calculated chi-square statistic with the critical value from the chi-square distribution.
If the calculated chi-square statistic exceeds the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Now, comparing the calculated test statistic with the given options:
A) There is not evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 7.815 < 13.861.
B) There is evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 13.861 > 5.991.
C) There is not evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 5.991 < 13.861.
D) There is evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 13.861 > 7.815.
By comparing the given test value of 13.861 with the critical value of 5.991, we can conclude that:
B) There is evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 13.861 > 5.991.
Therefore, the correct answer is B.