Question

DATAfile: AirDelay You may need to use the appropriate technology to answer this question Researchers at Purdue University and Wichita State University found that airlines are doing a better job Southwest Airlines were among the leaders were these flights late? Sample data showing the number of minutes that delayed flights were late are provided in the file named AirDelay. Data are shown for both airlines. Assume getting passengers to their destinations on time.t AirTran Airways and on-time arrivals with both having 88 % of their flights arriving on time. But for the 12 % of flights that were delayed, how many minutes population variances are unknown and unequal. Southwest AirTran 34 59 43 зо 3 45 64 42 33 42 85 45 28 38 32 3о 48 105 85 110 50 10 25 70 75 45 33 50 63 42 65 52 83 78 27 70 35 33 64 90 38 27 52 76 (a) Formulate the hypotheses that can be used to test for a difference between the population mean minutes late for delayed flights by these two airlines. (Let u, = population mean minutes late for delayed Airtran flights and u, = population mean minutes late for delayed Southwest flights.) о но м, - м, - о - и :0н Hot M, - м, < 0 H 0 (b) What is the sample mean number of minutes late for delayed flights for each e F these two airlines? AirTran min Southwest min (c) Calculate the test statistic. (Round your answer to three decimal places.) What is the p-value? (Round your answer to four decimal places.) p-value Using a 0.05 level of significance, what is your conclusion? Do not Reject Ho. There is statistical evidence that one airline does better than the other in terms of their population mean delay time. O Do not reject H. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time. O Reject Hn. There is statistical evidence that one airline does better than the other in terms of their population mean delay time. OReject H There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Answer 1

Following is the output of descriptive statistics generated by excel:

Descriptive statistics
	AirTran, X1	SouthWest, X2
count	25	20
mean	50.60	52.80
sample standard deviation	26.57	20.11
sample variance	705.75	404.38
minimum	3	28
maximum	110	105
range	107	77

Here we have х, -50.6, 5, -26.57, п -25 -52.8, s,-20.11, n-20 Hypotheses are - m: °н 0 0 Test is two tailed. Level of significance: a-0.05 Since we can assume that variances are equal so degree of freedom of t test will be: df n n-2 =43 The pooled standard deviation is: (4-1)s, +(n,-1)s, -23.9316 п +п, -2 Test Statistics: )-) ((50.6-52.8)-0)/(23.93 16* sqrt((1/25)+(1/20)) t= _ 1 1 -0.306 п п, Critical value: 2.017 Rejection Region: If It >2.017, Reject H0 Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value: P-value-0.7611 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis. =TINV (0.05,43) =TDIST (0.306,43,2) Excel function for critical value: Excel function for p-value:

Answers:

(a) First option is correct.

(b)

Means:

For Airtran = 50.60

For Southwest = 52.80

(c)

The test statistics is

t = -0.306

p-value = 0.7611

(d)

Second option is correct.