Researchers at two universities found that airlines are doing a better job of getting passengers to their destinations on time. Company A and Company B Airlines were among the leaders in 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 were these flights late

Complete question :

Formulate the hypothesis 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.)

(b) What is the sample mean number of minutes late for delayed flights for each of these two airlines?

(c) Calculate the test statistic. (Round your answer to three decimal places.)

What is the p-value? (Round your answer to four decimal places.)

Answers:

H0: μ1 − μ2 = 0

H1: μ1 − μ2 ≠ 0

Test statistic = - 0.314

Pvalue = 0.7550

Step-by-step explanation:

Mean, xbar = Σx / n

Standard deviation, s = √(Σ(x-xbar) / n-1)

Company A :

Sample size, n = 25

Mean, x1 = 50.6

Standard deviation, s1 = 26.57

Company B :

Mean, x2 = 52.82

Sample size, n2 = 20

Standard deviation, s2 = 20.02

Mean, xbar = Σx / n

Standard deviation, s = √(Σ(x-xbar) / n-1)

Company A :

Sample size, n = 25

Mean, x1 = 50.6

Standard deviation, s1 = 26.57

Company B :

Mean, x2 = 52.85

Sample size, n2 = 20

Standard deviation, s2 = 20.02

The hypothesis :

H0: μ1 − μ2 = 0

H1: μ1 − μ2 ≠ 0

The test statistic :

Test statistic:

t = (x1 - x2)/[√(s²p(1/n1 + 1/n2 )]

Pooled variance = s²p

s²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2)

s²p = (24*26.57^2) + (19*20.02^2) / 43

s²p = 571. 125

√(s²p(1/n1 + 1/n2) ;

√572.125(1/25 + 1/20) = 7.169

x1 - x2 ; 50.6 - 52.85 = - 2.25

Test statistic = - 2.25 / 7.169

Test statistic = - 0.314

Using the Pvalue from t score calculator :

Pvalue(-0.314, 43) = 0.755

Pvalue > α ; We fail to reject the Null

the answer to your question is b.