Question

Suppose the Federal Aviation Administration (FAA) would like to compare the on-time performances of different airlines on domestic, nonstop flights. The following table shows two different airlines and the frequency of flights that arrived early, on-time, and late for each. To test at the 0.1 level to determine if Airline and Status are dependent, calculate the chi-square test statistic and p-value.

Chart:

26 14 16

16 22 6

Question 4 options:

	1) 7.37, the degrees of freedom is 6, and the p-value is 0.2880.
	2) 5.236, the degrees of freedom is 2, and the p-value is 0.9749.
	3) 7.37, the degrees of freedom is 2, and the p-value is 0.1000.
	4) 7.37, the degrees of freedom is 2, and the p-value is 0.0251.
	5) None of the other options are fully correct.

Answer 1

Solution:- (4) 7.37, the degrees of freedom is 2, and the p-value is 0.0251.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H₀: Airline and Status are dependent are independent.
H_a: Airline and Status are dependent are not independent.

Formulate an analysis plan. For this analysis, the significance level is 0.10.

Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1)
D.F = 2
E_r,c = (n_r * n_c) / n


Χ² = 7.37

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 7.37.

We use the Chi-Square Distribution Calculator to find P(Χ² > 7.37) = 0.0251

Interpret results. Since the P-value (0.025) is less than the significance level (0.10), we cannot accept the null hypothesis. Thus, we conclude that there is a relationship between Airline and Status.