Question

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 contain 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 63 43 33 65
104 45 27 39 84
75 44 34 51 63
42 34 34 65 64
(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.)
A)H0: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
B) H0: μ1 − μ2 ≤ 0
Ha: μ1 − μ2 > 0
C)H0: μ1 − μ2 < 0
Ha: μ1 − μ2 = 0
D) H0: μ1 − μ2 ≠ 0
Ha: μ1 − μ2 = 0
E) 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______ min.
Company B______ min.
(c) Calculate the test statistic. (Round your answer to three decimal places.)
test statistic=
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?
A) Reject H0. There is statistical evidence that one airline does better than the other in terms of their population mean delay time
B) Do not Reject H0. There is statistical evidence that one airline does better than the other in
terms of their population mean delay time.
C) 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.
D) Reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Answer 1

Answer:

A)H0: μ1 − μ2 = 0

Ha: μ1 − μ2 ≠ 0

(b)  Means

Company A___50.6___ min.

Company B___52.75___ min.

c)The t-value is -0.30107.

The p-value is 0 .764815.

C) 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.

Step-by-step explanation:

A)H0: μ1 − μ2 = 0

i.e there is no difference between the means of delayed flight for two different airlines

Ha: μ1 − μ2 ≠ 0

i.e there is a difference between the means of delayed flight for two different airlines

(b)

Company A___50.6___ min.

Company B___52.75___ min.

Mean of Company A = x`1= ∑x/n =34+ 59+ 43+ 30+ 3+ 32+ 42+ 85+ 30+ 48+ 110+ 50+ 10+ 26+ 70+ 52+ 83+ 78+ 27+ 70+ 27+ 90+ 38+ 52+ 76/25

= 1265/25= 50.6

Mean of Company B = x`2= ∑x/n =

=46+ 63+ 43+ 33+ 65+ 104+ 45+ 27+ 39+ 84+ 75+ 44+ 34+ 51+ 63+ 42+ 34+ 34+ 65+ 64/20

= 1055/20= 52.75

Difference Scores Calculations

Company A

Sample size for Company A= n1= 25

Degrees of freedom for company A= df1 = n1 - 1 = 25 - 1 = 24

Mean for Company A= x`1=  50.6

Total Squared Difference (x-x`1) for Company A= SS1: 16938

s21 = SS1/(n1 - 1) = 16938/(25-1) = 705.75

Company B

Sample size for Company B= n1= 20

Degrees of freedom for company B= df2 = n2 - 1 = 20 - 1 = 19

Mean for Company B= x`2=  52.75

Total Squared Difference (x-x`2) for Company B= SS2=7427.75

s22 = SS2/(n2 - 1) = 7427.75/(20-1) = 390.93

T-value Calculation

Pooled Variance= Sp²

Sp² = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)

Sp²= ((24/43) * 705.75) + ((19/43) * 390.93) = 566.65

s2x`1 = s2p/n1 = 566.65/25 = 22.67

s2x`2 = s2p/n2 = 566.65/20 = 28.33

t = (x`1 - x`2)/√(s2x`1 + s2x`2) = -2.15/√51 = -0.3

The t-value is -0.30107.

The total degrees of freedom is = n1+n2- 2= 25+20-2=43

The critical region for two tailed test at significance level ∝ =0.05 is

t(0.025) (43) = t > ±2.017

Since the calculated value of t=  -0.30107.  does not fall in the critical region t > ±2.017, null hypothesis is not rejected that is there is no difference between the means of delayed flight for two different airlines.

The p-value is 0 .764815. The result is not significant at p < 0.05.