Question

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

Reference: Ref 21-3


-
Is there evidence of a difference in the on-time rate for the two airlines? To determine this, you test the hypotheses
H 0: p 1 = p 2, Ha: p 1 not equal to p 2

The P-value of your test of the hypotheses given is
Select one:
a. between 0.10 and 0.05.
b. between 0.05 and 0.01.
c. between 0.01 and 0.001.
d. below 0.001.

Answer 1

Option b) between 0.05 and 0.01.

Explanation:

We are given the following in the question:

`x_1 = 64\\n_1 = 100\\x_2 = 80\\n_2 = 100`

Let
`p_1` and
`p_2` be the proportion of all parts from suppliers 1 and 2, respectively, that are defective.

`p_1 = (x_1)/(n_1) = (64)/(100) = 0.64\\\\p_2 = (x_2)/(n_2) = (x_2)/(n_2) = (80)/(100)= 0.8`

First, we design the null and the alternate hypothesis

`H_(0): p_1 = p_2\\H_A: p_1 \\eq p_2`

We use Two-tailed z test to perform this hypothesis.

Formula:

`\text{Pooled P} = (x_1+x_2)/(n_1+n_2)\\\\Q = 1 - P\\\\Z_(stat) = \frac{p_1-p_2}{\sqrt{PQ((1)/(n_1) + (1)/(n_2))}}`

Putting all the values, we get,

`\text{Pooled P} = (80+64)/(100+100) = 0.72\\\\Q = 1 - 0.72 = 0.28\\\\Z_(stat) = \frac{0.64-0.8}{\sqrt{0.72* 0.28((1)/(100) + (1)/(100))}} = -2.5198`

Now, we calculate the p-value from the table at 0.05 significance level.

P-value = 0.01174

Since,

`0.05 > 0.01174 > 0.01`

Thus, the correct answer is

Option b) between 0.05 and 0.01.