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Sagot :
### Question 2: Hypothesis Test for Number of Riders
Objective:
We want to test whether the number of riders on a bus route is the same for each day from Monday through Friday. The null hypothesis is that the number of riders per day is the same.
Given Data:
- Number of riders on:
- Monday: 13
- Tuesday: 16
- Wednesday: 28
- Thursday: 17
- Friday: 16
Total Riders: 90
Proportions Under Null Hypothesis:
- Since there are five days, the expected proportion for each day is 0.20 (i.e., 20% of the total riders should be on each day if the null hypothesis is true).
Expected Frequencies:
- For each day, given the total number of riders (90), the expected frequency is:
[tex]\[ \text{Expected Riders per Day} = 90 \times 0.20 = 18 \][/tex]
So, the expected number of riders for each day is:
- Monday: 18
- Tuesday: 18
- Wednesday: 18
- Thursday: 18
- Friday: 18
Observed Frequencies:
- Monday: 13
- Tuesday: 16
- Wednesday: 28
- Thursday: 17
- Friday: 16
Chi-Square Test Statistic:
The chi-square test statistic is given by:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
where [tex]\(O_i\)[/tex] is the observed frequency and [tex]\(E_i\)[/tex] is the expected frequency.
Computing this for each day:
[tex]\[ \begin{aligned} \chi^2 &= \frac{(13 - 18)^2}{18} + \frac{(16 - 18)^2}{18} + \frac{(28 - 18)^2}{18} + \frac{(17 - 18)^2}{18} + \frac{(16 - 18)^2}{18} \\ &= \frac{(-5)^2}{18} + \frac{(-2)^2}{18} + \frac{(10)^2}{18} + \frac{(-1)^2}{18} + \frac{(-2)^2}{18} \\ &= \frac{25}{18} + \frac{4}{18} + \frac{100}{18} + \frac{1}{18} + \frac{4}{18} \\ &= 1.389 + 0.222 + 5.556 + 0.056 + 0.222 \\ &= 7.444 \end{aligned} \][/tex]
So, [tex]\(\chi^2 \approx 7.44\)[/tex].
Degrees of Freedom:
The degrees of freedom for this test is given by the number of categories minus 1:
[tex]\[ \text{Degrees of Freedom} = 5 - 1 = 4 \][/tex]
p-value:
Using the chi-square distribution table or a chi-square calculator, with 4 degrees of freedom and [tex]\(\chi^2 \approx 7.44\)[/tex]:
[tex]\[ p \approx 0.114 \][/tex]
Conclusion:
- Compare the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex]. If [tex]\(p > \alpha\)[/tex], we fail to reject the null hypothesis.
- Here, [tex]\(p \approx 0.114\)[/tex], which is greater than 0.05.
Therefore, we fail to reject the null hypothesis. There is not enough evidence at the 5% significance level to conclude that the number of riders is different on different days. The transit authority's assumption that the number of riders is the same on every day from Monday through Friday is not contradicted by the sample data.
Objective:
We want to test whether the number of riders on a bus route is the same for each day from Monday through Friday. The null hypothesis is that the number of riders per day is the same.
Given Data:
- Number of riders on:
- Monday: 13
- Tuesday: 16
- Wednesday: 28
- Thursday: 17
- Friday: 16
Total Riders: 90
Proportions Under Null Hypothesis:
- Since there are five days, the expected proportion for each day is 0.20 (i.e., 20% of the total riders should be on each day if the null hypothesis is true).
Expected Frequencies:
- For each day, given the total number of riders (90), the expected frequency is:
[tex]\[ \text{Expected Riders per Day} = 90 \times 0.20 = 18 \][/tex]
So, the expected number of riders for each day is:
- Monday: 18
- Tuesday: 18
- Wednesday: 18
- Thursday: 18
- Friday: 18
Observed Frequencies:
- Monday: 13
- Tuesday: 16
- Wednesday: 28
- Thursday: 17
- Friday: 16
Chi-Square Test Statistic:
The chi-square test statistic is given by:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
where [tex]\(O_i\)[/tex] is the observed frequency and [tex]\(E_i\)[/tex] is the expected frequency.
Computing this for each day:
[tex]\[ \begin{aligned} \chi^2 &= \frac{(13 - 18)^2}{18} + \frac{(16 - 18)^2}{18} + \frac{(28 - 18)^2}{18} + \frac{(17 - 18)^2}{18} + \frac{(16 - 18)^2}{18} \\ &= \frac{(-5)^2}{18} + \frac{(-2)^2}{18} + \frac{(10)^2}{18} + \frac{(-1)^2}{18} + \frac{(-2)^2}{18} \\ &= \frac{25}{18} + \frac{4}{18} + \frac{100}{18} + \frac{1}{18} + \frac{4}{18} \\ &= 1.389 + 0.222 + 5.556 + 0.056 + 0.222 \\ &= 7.444 \end{aligned} \][/tex]
So, [tex]\(\chi^2 \approx 7.44\)[/tex].
Degrees of Freedom:
The degrees of freedom for this test is given by the number of categories minus 1:
[tex]\[ \text{Degrees of Freedom} = 5 - 1 = 4 \][/tex]
p-value:
Using the chi-square distribution table or a chi-square calculator, with 4 degrees of freedom and [tex]\(\chi^2 \approx 7.44\)[/tex]:
[tex]\[ p \approx 0.114 \][/tex]
Conclusion:
- Compare the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex]. If [tex]\(p > \alpha\)[/tex], we fail to reject the null hypothesis.
- Here, [tex]\(p \approx 0.114\)[/tex], which is greater than 0.05.
Therefore, we fail to reject the null hypothesis. There is not enough evidence at the 5% significance level to conclude that the number of riders is different on different days. The transit authority's assumption that the number of riders is the same on every day from Monday through Friday is not contradicted by the sample data.
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