I am helping a friend with an aviation research study. We have two categorical variables for our IV (Group 1 consists of low-cost airlines, Group 2 consist of legacy carriers), our dependent variable is continuous (Average Arrival Delay in minutes). For example, in a span of 5 years, airline X in group 1 might have an average arrival delay of 5 minutes for all flights, and airline Y in group 2 might have an average delay of 3 minutes for all flights. Essentially, we are trying to see if there is a difference in average delays (in minutes) between group 1 or 2. The only data we have access to for the DV is the average arrival delay in minutes for each airline. Is this enough data? Since we have a categorical IV and continuous DV should we do an ANOVA? Our DV is normally distributed.
-
$\begingroup$ An ANOVA might be appropriate if you had more than two groups of airlines. An ANOVA would work for only two groups, but software output will be more transparent if you use a two-sample t test. $\endgroup$– BruceETJun 30, 2020 at 21:30
1 Answer
I think a two-sample t test should work well because your have two groups and the variable 'avg delay' is normally distributed.
Suppose vector x1 has delays for 10 airlines in Group 1 and
vector x2 has delays for 13 airlines in Group 2.
You want to test $H_0: \mu_1 = \mu_2$ against alternative
$H_1: \mu_2 \ne \mu_2$ at the 5% level of significance.
I have used R statistical software to sample some fake data to use for a demonstration.
set.seed(2020)
x1 = rnorm(10, 3, .5); x2 = rnorm(13, 5, .6)
summary(x1); sd(x1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.602 2.560 3.105 2.948 3.317 3.880
[1] 0.6453066 # sample SD for Gp 1
summary(x2); sd(x2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
3.177 4.777 5.191 5.119 5.718 6.305
[1] 0.9372571 # sample SD for Gp 2
x = c(x1, x2); gp = rep(1:2, c(10,13))
stripchart(x~gp, ylim=c(.5,2.5), pch="|")
There is no reason to suppose that Group 1 and Group 2 airlines will have the same variability among airlines in delays, so it is best to use the Welch 2-sample t test, which does not assume equal variances.
For my (artificial) data, the P-value near 0, indicates that the null hypothesis should be rejected. That is, the difference between sample means $\bar X_1 = 2.95$ and $\bar X_2 = 5.12.$ is "statistically significant." A separate issue is whether about 2 minutes extra delay in Group2 would be of any practical importance.
t.test(x ~ gp)
Welch Two Sample t-test
data: x by gp
t = -6.5699, df = 20.81, p-value = 1.736e-06
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.858846 -1.483551
sample estimates:
mean in group 1 mean in group 2
2.948019 5.119217
Most statistical software programs will do a Welch t test, and the required formulas for computation by hand are given in most basic statistics texts. Here is output for the same test from a recent version of Minitab software:
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 10 2.948 0.645 0.20
2 13 5.119 0.937 0.26
Difference = μ (1) - μ (2)
Estimate for difference: -2.171
95% CI for difference: (-2.860, -1.482)
T-Test of difference = 0 (vs ≠):
T-Value = -6.57 P-Value = 0.000 DF = 20
In Minitab, I input the sample sizes, means, and standard deviations rather than the 10 + 13 individual data points. Also, the two programs round results differently. So answers differ very slightly between R and Minitab outputs.
Note: Following up on my Comment about the possibility (not recommendation) to use an ANOVA, here is the P-value for the one-way ANOVA test oneway.test in R, which does not assume equal variances among the
levels of the factor. (I have shown only the P-value here.)
oneway.test(x ~ gp)$p.val
[1] 1.735603e-06
