The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer service on which airlines compete is on-time arrival. The tables below contains a sample of data from delayed flights showing the number of minutes each delayed flight was late for two different airlines, Company A and Company B.Company A34594330332428530481105010267052837827702790385276 Company B46654134671064528398674463351634334326465(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 μ1 = population mean minutes late for delayed Company A flights and μ2 = population mean minutes late for delayed Company B flights.)H0: μ1 − μ2 ≤ 0Ha: μ1 − μ2 > 0H0: μ1 − μ2 ≥ 0Ha: μ1 − μ2 < 0    H0: μ1 − μ2 = 0Ha: μ1 − μ2 ≠ 0H0: μ1 − μ2 ≠ 0Ha: μ1 − μ2 = 0H0: μ1 − μ2 < 0Ha: μ1 − μ2 = 0(b)What is the sample mean number of minutes late for delayed flights for each of these two airlines?Company A minCompany B 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?Reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.    Do not Reject H0. There is statistical evidence that one airline does better than the other in terms of their population mean delay time.Reject H0. There is statistical evidence that one airline does better than the other in terms of their population mean delay time.

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Asked Dec 7, 2019
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The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer service on which airlines compete is on-time arrival. The tables below contains a sample of data from delayed flights showing the number of minutes each delayed flight was late for two different airlines, Company A and Company B.
Company A
34 59 43 30 3
32 42 85 30 48
110 50 10 26 70
52 83 78 27 70
27 90 38 52 76
 
Company B
46 65 41 34 67
106 45 28 39 86
74 46 33 51 63
43 34 32 64 65
(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 μ1 = population mean minutes late for delayed Company A flights and μ2 = population mean minutes late for delayed Company B flights.)
H0: μ1 − μ2 ≤ 0
Ha: μ1 − μ2 > 0
H0: μ1 − μ2 ≥ 0
Ha: μ1 − μ2 < 0
    
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
H0: μ1 − μ2 ≠ 0
Ha: μ1 − μ2 = 0
H0: μ1 − μ2 < 0
Ha: μ1 − μ2 = 0
(b)
What is the sample mean number of minutes late for delayed flights for each of these two airlines?
Company A minCompany B 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?
Reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.    Do not Reject H0. There is statistical evidence that one airline does better than the other in terms of their population mean delay time.Reject H0. There is statistical evidence that one airline does better than the other in terms of their population mean delay time.
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Expert Answer

Step 1

Enter the Data in Excel as shown below and use the Formula to calculate the sample mean and standard deviation 

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1 Company A 2 34 59 43 30 3 3 32 42 85 30 48 4 110 50 10 26 70 5 52 83 78 27 70 6 27 90 38 52 76 Sample Mean |=AVERAGE(A2:E6) 8 Company B 9 46 Sample standard Deviatior =STDEV.S(A2:D6) 65 41 34 67 10 106 45 28 39 86 11 74 46 33 51 63 12 43 34 32 64 65 Sample Mean =AVERAGE(A9:E12) 13 Sample standard Deviatior =STDEV.S(A9:E12) 14 15

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Step 2

Result Obtained

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1 Company A 2 34 59 43 30 3 30 3 32 85 48 4 110 50 10 26 70 27 52 83 78 70 27 90 38 52 76 Sample Mean 50.6 8 Company B Sample standard Deviation 26.41551 34 46 65 67 10 106 45 28 39 86 11 74 46 33 51 63 12 43 34 32 64 65 Sample Mean 13 53.1 Sample standard Deviation 14 20.31048 15

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Step 3

a)

Formulation of Hypothesis

Null Hypothesis: H0: µ1-µ2 = 0

Alternative Hypothesis: Hα: µ1-µ2 ≠ 0

b)

Sample mean for Company A = 50.6

Sample Standard Deviation for company A = 26.41551

Sample size of A(nA) =20

Sample mean for Company B = 53...

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ХА-хВ t = Sp |пА пв Pooled standard deviation is calculated by |(nA – 1) x s + (ng – 1) x s; Sp ПА + Пв — 2 (20 – 1) × 26.415512 + (16 – 1) x 20.310482 V571.92 = 23.915 20 + 16 – 2 50.6-53.1 = -0.312 t = |1+1 23.915 20 16 Degrees of freedom = na+ ng - 2 = 20 +16 – 2 =34 p-value = 0.7569(From Excel =T.DIST.2T(0.312,34)) p-value > significance level (0.05)

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