In a preliminary study to determine whether the installation of a camera designed to catch cars that go through an intersection on a red traffic light affects the number of violators, the number of violators that go through an intersection on a red traffic light was recorded for each day of the weak before and the week after the camera was installed. Test at a 1% level of significance whether the installation of a camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light, given the results below. Give the null and alternative hypothesis.
Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Before 12 16 31 18 20 24 16
After 8 18 24 19 16 19 16
To test whether the installation of a camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light, we can use a paired t-test.
Null hypothesis: The installation of a camera did not result in a reduction of the number of violators.
Alternative hypothesis: The installation of a camera resulted in a reduction of the number of violators.
We need to calculate the difference in the number of violators before and after the camera installation for each day:
Sunday: 12 - 8 = 4
Monday: 16 - 18 = -2
Tuesday: 31 - 24 = 7
Wednesday: 18 - 19 = -1
Thursday: 20 - 16 = 4
Friday: 24 - 19 = 5
Saturday: 16 - 16 = 0
Next, we calculate the mean difference:
Mean difference = (4 - 2 + 7 - 1 + 4 + 5 + 0) / 7 = 1.43
To perform the t-test, we also need to calculate the standard deviation of the differences.
First, we calculate the squared differences from the mean:
(4 - 1.43)^2 = 5.8449
(-2 - 1.43)^2 = 13.2849
(7 - 1.43)^2 = 19.8049
(-1 - 1.43)^2 = 5.8849
(4 - 1.43)^2 = 5.8449
(5 - 1.43)^2 = 12.9649
(0 - 1.43)^2 = 2.0449
Then, we calculate the variance by summing up the squared differences and dividing by n-1:
Variance = (5.8449 + 13.2849 + 19.8049 + 5.8849 + 5.8449 + 12.9649 + 2.0449) / 6 = 7.1134
Finally, we find the standard deviation by taking the square root of the variance:
Standard deviation = sqrt(7.1134) = 2.667
With the mean difference of 1.43 and standard deviation of 2.667, we can calculate the t-statistic:
t = (mean difference - 0) / (standard deviation / sqrt(n))
t = (1.43 - 0) / (2.667 / sqrt(7))
t = 0.535
Degrees of freedom = n - 1 = 7 - 1 = 6
We can calculate the critical t-value at a 1% level of significance for a two-tailed t-test with 6 degrees of freedom. Using a t-table or calculator, the critical t-value is approximately ±2.447.
Since the calculated t-value (0.535) is within the range of -2.447 to 2.447, we fail to reject the null hypothesis. There is not enough evidence to conclude that the installation of a camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light.
To test whether the installation of a camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light, we can use a paired t-test.
Null hypothesis: The installation of a camera did not result in a reduction of the number of violators.
Alternative hypothesis: The installation of a camera resulted in a reduction of the number of violators.
To perform the paired t-test, we first calculate the difference between the number of violators before and after the installation.
Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Before 12 16 31 18 20 24 16
After 8 18 24 19 16 19 16
Diff 4 -2 7 -1 4 5 0
Next, we calculate the mean of the differences:
Mean difference = (4 - 2 + 7 - 1 + 4 + 5 + 0) / 7 = 17 / 7 = 2.43
Then, we calculate the standard deviation of the differences:
Standard Deviation = sqrt(((4-2.43)^2 + (-2-2.43)^2 + (7-2.43)^2 + (-1-2.43)^2 + (4-2.43)^2 + (5-2.43)^2 + (0-2.43)^2) / (7-1)) = 5.34
Finally, we use the paired t-test formula to calculate the t-statistic:
t = (mean difference - expected difference) / (standard deviation / sqrt(n))
t = (2.43 - 0) / (5.34 / sqrt(7))
t = 2.43 / 2.02 = 1.20
The degree of freedom for a paired t-test is (n-1), where n is the number of observations. In this case, n = 7, so the degrees of freedom would be 6.
Using a t-table or statistical software, we can find the critical t value for a 1% level of significance for a two-tailed test with 6 degrees of freedom. Let's assume the critical t value is ±2.447 based on the table.
Since the calculated t-statistic (1.20) is less than the critical t value (±2.447), we fail to reject the null hypothesis. There is not enough evidence to suggest that the installation of the camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light.
To test whether the installation of a camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light, we can use a paired t-test.
The null hypothesis (H0) is that the installation of a camera has no effect on the number of violators. The alternative hypothesis (Ha) is that the installation of a camera results in a reduction of the number of violators.
The first step is to calculate the differences between the number of violators before and after the camera installation.
Day | Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday
Before | 12 | 16 | 31 | 18 | 20 | 24 | 16
After | 8 | 18 | 24 | 19 | 16 | 19 | 16
Difference | 4 | -2 | 7 | -1 | 4 | 5 | 0
Next, we calculate the mean of the differences:
Mean of differences (x̄) = (4 - 2 + 7 - 1 + 4 + 5 + 0) / 7 = 17 / 7 ≈ 2.43
Then, we calculate the standard deviation of the differences:
Standard deviation (s) = √[ ( (4 - 2.43)² + (-2 - 2.43)² + (7 - 2.43)² + (-1 - 2.43)² + (4 - 2.43)² + (5 - 2.43)² + (0 - 2.43)² ) / (7 - 1) ]
≈ √[ (2.7863 + 20.6463 + 21.0463 + 13.2063 + 2.7863 + 4.2463 + 6.4463 ) / 6 ]
≈ √[70.1748 / 6]
≈ √11.6958
≈ 3.42
Now, we can calculate the t-value:
t-value = (x̄ - μ) / (s / √n)
= (2.43 - 0) / (3.42 / √7)
≈ 2.43 / (3.42 / √7)
≈ 2.43 / (1.29 / 2.65)
≈ 2.43 / 0.4868
≈ 5.00
Finally, we compare the t-value obtained with the critical t-value. At a 1% level of significance and with (7 - 1 = 6) degrees of freedom, the critical t-value is approximately 3.707.
Since the t-value (5.00) is greater than the critical t-value (3.707), we can reject the null hypothesis. Therefore, we can conclude that the installation of a camera resulted in a reduction of the number of violators that go through the intersection on a red traffic light.
In summary:
Null hypothesis (H0): The installation of a camera has no effect on the number of violators.
Alternative hypothesis (Ha): The installation of a camera results in a reduction of the number of violators.