We fail to reject the null hypothesis that the mean departure delay time for the three flights is the same. In other words, there is not enough evidence to conclude that the mean departure delay time for the three flights is different.
a. Identify the null and alternative hypotheses.
The null hypothesis is that the mean departure delay time for the three flights is the same. The alternative hypothesis is that the mean departure delay time for the three flights is not the same.
b. Find ns and s.
ns is the number of flights in each sample. n1 = n2 = n3 = 3.
s is the sample standard deviation. To calculate s, we first need to calculate the squared deviations from the mean for each flight.
Flight 1: (72 - 0)^2 = 5184
Flight 19: (0 - 0)^2 = 0
Flight 21: (72 - 0)^2 = 5184
The total squared deviations from the mean is 10368. The sample standard deviation is calculated by taking the square root of the total squared deviations from the mean divided by the number of samples minus one.
s = sqrt(10368 / (3 - 1)) = 60
c. Find the value of the test statistic.
The test statistic for this problem is the F-statistic. The F-statistic is calculated by taking the ratio of the variance between the groups to the variance within the groups.
F-statistic = (variance between groups) / (variance within groups)
The variance between the groups is calculated by taking the sum of the squared deviations between the group means and the overall mean, divided by the number of groups minus one.
Variance between groups = [(72 - 0)^2 + (0 - 0)^2 + (72 - 0)^2] / (3 - 1)
Variance between groups = 10368 / 2
Variance between groups = 5184
The variance within the groups is calculated by taking the sum of the squared deviations from the mean for each flight, divided by the total number of samples minus one.
Variance within groups = (5184 + 0 + 5184) / (3 * 3 - 1)
Variance within groups = 10368 / 8
Variance within groups = 1296
Now we can calculate the F-statistic.
F-statistic = 5184 / 1296 = 4
d. Find the p-value or critical value. What should we conclude about the null hypothesis?
To find the p-value, we need to look up the F-statistic in a table of F-statistics. The p-value is the probability of obtaining an F-statistic that is greater than or equal to the observed F-statistic, assuming that the null hypothesis is true.
The degrees of freedom for the numerator and denominator of the F-statistic are 2 and 6, respectively. Looking up the F-statistic of 4 in a table of F-statistics, we find that the p-value is 0.049.
The critical value for the F-statistic at a significance level of 0.05 is 5.14. Since the observed F-statistic of 4 is less than the critical value of 5.14, we fail to reject the null hypothesis.
Question
Listed below are departure delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that departed early. Use a 0.05 significance level to test the claim that the different flights have the same mean departure delay time. Flight 1 -2 -1 -2 2 -2 0 -2 -3 Flight 19 19 -4 -5 -1 73 0 1 Flight 21 18 60 142 -1 -11 -1 47 13 Flight 1 Flight 19 Flight 21 n Y s2
a. Identify the null and alternative hypotheses.
b. Find ns and s. ns =
c. Find the value of the test statistic. --
d. Find the p-value or critical value. What should we conclude about the null hypothesis?
e. Conclusion.