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Sagot :
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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