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In: Statistics and Probability

There was an SRS of 100 flights on a large airline (airline 1) that showed that...

There was an SRS of 100 flights on a large airline (airline 1) that showed that 64 of the flights were on time. An SRS of 100 flights of another large airline (airline 2) showed that 80 of the flights were on time. Let p1 and p2 be the proportion of all flights that are on time for these two airlines.

What is a 95% confidence interval for the difference p1-p2?

(-.222, -.098)

(-.263, -.057)

(-.218, -.102)

(-.283, -.038)

(.098, .222)

Solutions

Expert Solution

Solution :

Given that,

= x1 / n1 = 64 / 100 = 0.64

1- = 0.36

= x2 / n2 = 80 / 100 = 0.80

1 - = 0.20

At 95% confidence level the z is ,

Z/2 = 1.96

95% confidence interval for p1 - p2 is ,

( - )   Z/2  * [(1- ) / n1 + (1 - ) / n2]

(0.64 - 0.80)   1.96 * [(0.64 * 0.36 ) / 100 + (0.80 * 0.20) / 100]

-0.283 < p1 - p2 < -0.038

(-0.283 , -0.038)

orchestra answered 1 year ago
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