Question

A simple random sample of 100 flights Airline 1 showed that 64 were on time. A simple random sample of 100 flights of Airline 2 showed that 80 were on time. Let p1 and p2 be the proportions of all flights that are were time for these two airlines. Is there evidence of a difference in the on-time rate for the two airlines? To determine this, test the hypotheses H0: p1 = p2 versus Ha: p1 ≠ p2 at a 10% significance level. What would our decision be, and what type of mistake might we have made?

a. reject H0; Type I error

b. fail to reject H0; Type I error

c. fail to reject H0; Type II error

d. reject H0; Type II error

Answer 1

For testing H0: p1=p2 against H1: p1 is not equal to p2, our test statistic is -. (p1° - p2°) - 0 / √ ((p°×(1-p°)) × ((1/n1)+ (1/n2)))

Where p° = (x1+x2)/(n1+n2)

Under H0, this follows standard normal distribution.

Here, p1° = 64/100 = 0.64, p2° = 80/100 = 0.8,. p° = (64+80) /(100+100) = 144 / 200 = 0.72

The value of the test statistic is = - 2.52, we reject the test statistic if |z|<= z-(alpha/2), here alpha = 0.10. Now alpha/2=0.05, and z (0.05) = 1.645. Obviously, 2.52> 1.645. So, we reject the null hypothesis of two proportions being equal.

Here, we are rejecting the null hypothesis but in reality it might be true. In that case, we might have committed a type 1 error.

So, out of the four options, option a is correct.