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san4es73 [151]
2 years ago
11

In an effort to reduce boarding time, an airline tries a new method of boarding its planes. Historically, only 32% of passengers

were satisfied with the boarding process random sample of 32 passengers using the new boarding process found 44% who said they were satisfied with the boarding process, yielding a test statistic of _______.
(a) What is the p-value of the hypothesis test to determine if this new boarding process resulted in improved passenger satisfaction?______ (4 decimals)
(b) What is the meaning of a Type 11 error in this context? The airline concludes a greater proportion of passengers were satisfied, when in reality, the same proportion of passengers were satisfied.
(c) Which significance level would minimize the likelihood of making a Type II error?
a = 0.01
a = 0.10
Mathematics
1 answer:
Nesterboy [21]2 years ago
4 0

Answer:

Test statistics of <u>1.455 </u>

P-value = 0.0728

Step-by-step explanation:

From the given information:

The test statistics can be computed as:

Z = \dfrac{\hat p - p }{\sqrt{\dfrac{p(1-p)}{n} }} \\ \\  \\ Z = \dfrac{0.44 -0.32}{\sqrt{\dfrac{0.32(1-0.32)}{32} }}

Z = 1.455

We want to test if the customer satisfaction increased significantly(one-tailed test)

Null hypothesis:

H_o : p= 0.32

Alternative hypothesis:

H_a: p>0.32

P-value = P(Z>1.455)

= 0.0728

b) Type II error implies the error of accepting H_o \ (i.e\ \text{ the null \ hypothesis)}when H_a \ (i.e\ \text{ the alternative \ hypothesis)} is true.

This implies inferring that there is no huge improvement in passenger's satisfaction when there is.

c) Type 1 and Type II errors are inversely proportional. In this situation, as one increases, the other definitely decreases.

∴ A Smaller value of Type II error will be achieved by a higher type I error.

⇒ 0.10

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