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 60% of the students plan to work, 30% of the students plan to continue in school, and 10% of the students plan to take some time off. Is there evidence to reject this hypothesis at = 0.05?
What hypothesis?
To determine whether there is evidence to reject the hypothesis at a significance level of α = 0.05, we need to perform a hypothesis test.
Let's state the null and alternative hypotheses:
Null hypothesis (H0): The proportions of seniors planning to work, continue in school, and take time off are equal to the claims made by the school advising office.
Alternative hypothesis (Ha): The proportions of seniors planning to work, continue in school, or take time off are not equal to the claims made by the school advising office.
To conduct the hypothesis test, we can use a chi-square goodness-of-fit test.
Step 1: Identify the observed frequencies.
We need to collect data on the actual proportions of seniors' plans from the random group surveyed.
Step 2: Calculate the expected frequencies.
Under the null hypothesis, the expected frequencies for each category can be calculated by multiplying the total number of seniors surveyed by the claimed proportions (60%, 30%, and 10%).
Step 3: Calculate the chi-square test statistic.
The chi-square test statistic (χ^2) can be calculated using the formula:
χ^2 = Σ((O - E)^2 / E), where O is the observed frequency and E is the expected frequency.
Step 4: Determine the degrees of freedom.
The degrees of freedom (df) is equal to the number of categories minus one. In this case, since there are three categories (work, continue in school, take time off), df = 2.
Step 5: Determine the critical value.
At α = 0.05 and df = 2, the critical value from the chi-square distribution table is approximately 5.99.
Step 6: Compare the test statistic with the critical value.
If the chi-square test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Keep in mind that without actual data, it is not possible to perform this specific test. You would need to collect the data on the seniors' plans and follow the steps outlined above to determine if there is evidence to reject the claim made by the school advising office.
To determine whether there is evidence to reject the hypothesis at a significance level of α = 0.05, we would need to conduct a hypothesis test using the data from the randomly selected group of seniors.
First, we need to define the null and alternative hypotheses. In this case, the null hypothesis (H₀) would be that the proportions stated by the school advising office are true (i.e., 60% working, 30% continuing in school, and 10% taking time off). The alternative hypothesis (H₁) would be that the proportions are different from those claimed.
To conduct the hypothesis test, we can perform a chi-square goodness-of-fit test, which tests whether observed frequencies differ significantly from the expected frequencies.
Here is the step-by-step process to perform the hypothesis test:
1. Gather the data: Record the number of seniors who plan to work, continue in school, and take time off from the randomly selected group.
2. Determine the expected frequencies: Calculate the expected frequencies for each category under the assumption of the null hypothesis. In this case, we would expect 60% of the total number of students to work, 30% to continue in school, and 10% to take time off.
3. Calculate the chi-square test statistic: Compute the chi-square test statistic using the formula: χ² = Σ [(Observed frequency - Expected frequency)² / Expected frequency]. Sum the values for each category.
4. Determine the degrees of freedom: Calculate the degrees of freedom (df) for the chi-square test, which is equal to the number of categories minus 1. In this case, since there are three categories (work, continue in school, take time off), the df would be 2.
5. Find the critical value: Find the critical value from the chi-square distribution table using the chosen significance level (α) and the degrees of freedom (df). For α = 0.05 and df = 2, the critical value is 5.991.
6. Compare the test statistic and critical value: If the chi-square test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Performing these steps will help us determine whether there is evidence to reject the hypothesis at α = 0.05 based on the randomly selected group of seniors' responses.