Question

An airport official wants to prove that the proportion of delayed flights for Airline A (denoted as p1) is less than the proportion of delayed flights for Airline B (denoted as p2). Random samples for both airlines after a storm showed that 51 out of 200 flights for Airline A were delayed, while 60 out of 200 of Airline B's flights were delayed. The test statistic for this problem is -1.00. The p-value for the test statistic for this problem is:

A. p = 0.3413 B. p = 0.0668 C. p = 0.1587 D. p = 0.0228

Answer 1

Answer:

The p-value for the test statistic is 0.1587.

Step-by-step explanation:

As the sample size for both the airlines are large, according to the central limit theorem the sampling distribution of sample proportion follows a normal distribution.

The test statistic for the difference between two proportions is:

$z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{P(1-P)\frac{1}{n_{1}}+\frac{1}{n_{2}} } }$

The test statistic value is, z = -1.00

The p-value of the test statistic is:

$P (Z<-1.00)=1-P(Z<1.00)=1-0.8413=0.1587$

**Use the standard normal table for the probability.

Thus, the p-value for the test statistic is 0.1587.