a random group of seniors was selected from a university and asked about their plans for the following year. The school advising office claims that 50% of the students plan to work, 25% of the students plan to continue in school, and 25% of the students plan to take some time off. Is there enough evidence to reject this hypothesis at a=0.05
A) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727<5.991
B) There is evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727<7.815
C) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727<7.815
D) There is evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727<5.991
A) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727<5.991
To answer this question, we need to perform a hypothesis test. The null hypothesis (H0) is that the proportions of students planning to work, continue in school, and take time off are equal to 0.50, 0.25, and 0.25, respectively. The alternative hypothesis (Ha) is that the proportions are not equal to the claimed values.
To conduct the test, we calculate a test statistic called the chi-square statistic (χ^2). This statistic compares the observed frequencies of each category with the expected frequencies assuming the null hypothesis is true. The formula for calculating the chi-square statistic is:
χ^2 = ∑[((Observed frequency - Expected frequency)^2) / Expected frequency]
The degrees of freedom, which determine the critical value for the test, is equal to the number of categories minus one. In this case, since we have 3 categories (work, school, time off), the degrees of freedom is 3 - 1 = 2.
The critical value for a chi-square test with 2 degrees of freedom at α = 0.05 is 5.991. If the calculated chi-square statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject it.
The test value given in the answer options is 0.727. Comparing this value to the critical value, we find that:
0.727 < 5.991
Therefore, the correct answer is:
A) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727<5.991
To determine whether there is enough evidence to reject the hypothesis, we need to perform a hypothesis test.
The null hypothesis (H0) states that the proportions of students planning to work, continue in school, and take time off are as claimed by the advising office, i.e., 50% work, 25% continue in school, and 25% take time off.
The alternative hypothesis (Ha) states that the proportions are different from what the advising office claims.
To perform the test, we calculate the test statistic and compare it to the critical value. The test statistic is calculated as follows:
Test Statistic = (Observed Frequency - Expected Frequency) / Square root (Expected Frequency)
We can calculate the test statistic for each category as follows:
For working:
Test Statistic_Work = (Observed Frequency_Work - Expected Frequency_Work) / Square root (Expected Frequency_Work)
For continuing in school:
Test Statistic_School = (Observed Frequency_School - Expected Frequency_School) / Square root (Expected Frequency_School)
For taking time off:
Test Statistic_Off = (Observed Frequency_Off - Expected Frequency_Off) / Square root (Expected Frequency_Off)
Assuming there were 100 seniors in the random sample, we can calculate the expected frequencies for each category as follows:
Expected Frequency_Work = 100 * 0.5 = 50
Expected Frequency_School = 100 * 0.25 = 25
Expected Frequency_Off = 100 * 0.25 = 25
Now, let's assume the observed frequencies for each category were as follows:
Observed Frequency_Work = 48
Observed Frequency_School = 23
Observed Frequency_Off = 29
Now, we can calculate the test statistics for each category:
Test Statistic_Work = (48 - 50) / sqrt(50) = -0.4
Test Statistic_School = (23 - 25) / sqrt(25) = -0.4
Test Statistic_Off = (29 - 25) / sqrt(25) = 0.8
Next, we compare the test statistic to the critical value for a significance level of 0.05. The critical value for a chi-square test with 2 degrees of freedom at a = 0.05 is 5.991.
Based on the given information, we find that the test statistic value of all three categories (work, school, and time off) is less than the critical value of 5.991.
Therefore, the correct answer is:
A) There is no evidence to reject the claim that the students' plans are distributed as claimed because the test value 0.727 < 5.991.