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.
I think a two-sample t test should work well because your have two groups and the variable 'avg delay' is normally distributed.
x1 has delays for 10 airlines in Group 1 and
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  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  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  1.735603e-06