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Answer:
There is a 0.57% probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, we have that:
Desired outcomes:
The number of male nannies selected. 24 of the nannies placed were men. So the number of desired outcomes is 24.
Total outcomes:
The number of nannies selected. 4,176 nannies were placed in a job in a given year. So the number of total outcomes 4176.
Find the probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
There is a 0.57% probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
Solution:-
- The population proportion, P = 0.03 ( 3 % )
- The sample proportion, p = 0.058 ( 5.8 % )]
- The sample size , n = 85.
- State the hypothesis:
Null hypothesis : P = 0.03 ( 3 % )
Alternate hypothesis : P ≥ 0.03 ( 3 % )
- The rejection region is defined by the significance level α:
significance α = 0.05
Z-critical = Z_α = Z_0.05
p-value = Z_0.05 = 1.645
- The Z-statistics:
- Compare against the rejection value ( p - value ):
Z-critical = 1.645
Z-test = 1.5132
Z-test < Z-critical ... ( Null not rejected )
- We fail to reject the Null hypothesis, there is not enough evidence to support the manager's claim that the production process is out of control. Hence, we can support that the production process is not out of control.