Question

In the airline business, "on-time" flight arrival is important for connecting flights and general customer satisfaction. Is there a difference between summer and winter average on-time flight arrivals? Let x1 be a random variable that represents percentage of on-time arrivals at major airports in the summer. Let x2 be a random variable that represents percentage of on-time arrivals at major airports in the winter. A random sample of n1 = 16 major airports showed that x1 = 74.5%, with s1 = 5.2%. A random sample of n2 = 18 major airports showed that x2 = 69.9%, with s2 = 8.5%. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative" answer. (a) Does this information indicate a difference (either way) in the population mean percentage of on-time arrivals for summer compared to winter? Use α = 0.05. (i) What is the level of significance? State the null and alternate hypotheses. H0: μ1 = μ2; H1: μ1 > μ2 H0: μ1 > μ2; H1: μ1 = μ2 H0: μ1 = μ2; H1: μ1 ≠ μ2 H0: μ1 = μ2; H1: μ1 < μ2 (ii) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? (Round your answer to three decimal places.) (iii) Find (or estimate) the P-value. P-value > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010 Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot (iv) Based on your answers in parts (i) to (iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.05 level that there was a difference in the population mean proportion of on-time arrivals in summer versus winter. There is insufficient evidence at the 0.05 level that there was a difference in the population mean proportion of on-time arrivals in summer versus winter. (b) Find a 95% confidence interval for μ1 − μ2. (Round your answers to two decimal places.) lower limit % upper limit % (c) What assumptions about the original populations have you made for the methods used? We assume that x1 and x2 are skewed right. We assume that x1 and x2 are skewed left. We assume that x1 and x2 are approximately normal or at least mound-shaped and symmetric.

Answer 1

a) H0: μ1 = μ2; H1: μ1 ≠ μ2

ii)

The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.

Point estimate =x1-x2=		4.600
std error =√(S21/n1+S22/n2)=		2.3883
test stat t =(x1-x2-Δo)/Se =		1.926

iii)

0.050 < P-value < 0.100

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

. There is insufficient evidence at the 0.05 level that there was a difference in the population mean proportion of on-time arrivals in summer versus winter

b)

Point estimate of differnce =x1-x2     =		4.600
for 95 % CI & 15 df value of t=		2.131	from excel: t.inv(0.975,15)
margin of error E=t*std error                   =		5.089
lower bound=mean difference-E     =		-0.49
Upper bound=mean differnce +E      =		9.69
c) We assume that x1 and x2 are approximately normal or at least mound-shaped and symmetric.