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


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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

Answer:

Option b) between 0.05 and 0.01.

Step-by-step 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.

$p_1 = \dfrac{x_1}{n_1} = \dfrac{64}{100} = 0.64\\\\p_2 = \dfrac{x_2}{n_2} = \dfrac{x_2}{n_2} = \dfrac{80}{100}= 0.8$

First, we design the null and the alternate hypothesis

$H_{0}: p_1 = p_2\\H_A: p_1 \neq p_2$

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

Formula:

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

Putting all the values, we get,

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

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.