Answered

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The management of KSmall Industries is considering a new method of assembling a computer. The current assembling method requires a mean time of 66 minutes with a standard deviation of 3.0 minutes. Using the new method, the mean assembly time for a random sample of 24 computers was 61 minutes.
a. Using the 0.10 level of significance, can we conclude that the assembly time using the new method is faster?
b. What is the probability of a Type II error? (Round "Critical values" to 2 decimal places and round your final answer to 4 decimal places.)

Sagot :

akiran007

Answer:

Part a:

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)=  0.6628

Step-by-step explanation:

The null and alternate hypothesis are

H0: μ1 ≥ μ2   against the claim Ha: μ1 < μ2

This is left tailed test

The critical region for this test at 0.10 ∝ is z < ± 1.28

The test statistic

z= x1-x2/ s/√n

z= 66-61/ 3/√24

z= 5/0.61238

z= 8.16486

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)  = P ( accept H0/ H0 is false)

z ( ∝)= X-u/s/√n

1.28= X-61/0.61238

X= 61.7838

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 61.7838

 The probability of a Type II error is given by

P (z> 61.7838-66/24)

P (z>- 0.42161)

From the tables

P (z>- 0.42161)=  0.6628

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