Final answer:
- a. Null hypothesis: There is no difference in preference for the three entrances; and alternative hypothesis: There is a difference in preference for the three entrances.
- b. The expected valus for entrance A,B and C are 2, 2.7 and 2 respectively
- c. The test statistic is 266.73.
- d. The degrees of freedom for the chi-square test of independence is 4.
- e. We find that there is a difference in choice across the three entrances if the null hypothesis is rejected. We find that there is no preference difference for the three entrances if the null hypothesis is not rejected.
Step-by-step explanation:
To determine if there is a difference in preference for the three entrances of the office building, we can perform a chi-square test of independence. Let's go through each step of the hypothesis testing process:
a. State null hypothesis and alternative hypothesis:
H₀: There is no difference in preference for the three entrances.
H₁: There is a difference in preference for the three entrances.
b. Find expected frequencies and place them into the table:
To find the expected frequencies, we calculate the row and column totals and use these values to calculate the expected frequencies. The expected frequency for each cell is given by (row total * column total) / total sample size.
Let's assume the observed frequencies are as follows:
| | Entrance A | Entrance B | Entrance C | Total |
|-------------|------------|------------|------------|-------|
| Observed | 60 | 80 | 60 | 200 |
We can calculate the row and column totals:
| | Entrance A | Entrance B | Entrance C | Total |
|-------------|------------|------------|------------|-------|
| Observed | 60 | 80 | 60 | 200 |
| Row Total | 60 | 80 | 60 | 200 |
| Column Total| 200 | 200 | 200 | 600 |
The expected frequencies can be calculated as follows:
| | Entrance A | Entrance B | Entrance C | Total |
|-------------|------------|------------|------------|-------|
| Observed | 60 | 80 | 60 | 200 |
| Expected | (60*200)/600 | (80*200)/600 | (60*200)/600 | |
| Row Total | 60 | 80 | 60 | 200 |
| Column Total| 200 | 200 | 200 | 600 |
c. State the calculated value of the test statistic:
The test statistic for a chi-square test of independence is calculated using the formula:
χ² = ∑ [(Observed - Expected)² / Expected]
Using the observed and expected frequencies, we calculate the test statistic:
χ² = [(60 - (60*200)/600)^2 / ((60*200)/600)] + [(80 - (80*200)/600)^2 / ((80*200)/600)] + [(60 - (60*200)/600)^2 / ((60*200)/600)] = 266.73.
d. State the critical value of the test:
The critical value of the test, denoted as χ²cr, depends on the significance level (α) and the degrees of freedom (df). Since we have 3 entrances, the degrees of freedom for the chi-square test of independence is (number of rows - 1) * (number of columns - 1) = (3-1) * (3-1) = 4.
e. Decide whether the null hypothesis is true and make a conclusion:
To decide whether to reject or fail to reject the null hypothesis, we compare the calculated test statistic to the critical value. If the calculated test statistic is greater than the critical value, we reject the null hypothesis. If it is smaller, we fail to reject the null hypothesis.
If the null hypothesis is rejected, we conclude that there is a difference in preference for the three entrances. If the null hypothesis is not rejected, we conclude that there is no difference in preference for the three entrances.