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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

Answer :

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Answer:

Test statistics of 1.455

P-value = 0.0728

Step-by-step explanation:

From the given information:

The test statistics can be computed as:

[tex]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} }}[/tex]

Z = 1.455

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

Null hypothesis:

[tex]H_o : p= 0.32[/tex]

Alternative hypothesis:

[tex]H_a: p>0.32[/tex]

P-value = P(Z>1.455)

= 0.0728

b) Type II error implies the error of accepting [tex]H_o \ (i.e\ \text{ the null \ hypothesis)}[/tex]when [tex]H_a \ (i.e\ \text{ the alternative \ hypothesis)}[/tex] 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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