1. (7 points) The population variance of length of a specific manufacturing part

Question

1. (7 points) The population variance of length of a specific manufacturing part is 4 mm.  

(a) Estimate the sample standard deviation for a sample size of 36 parts. (3 points)

(b) Calculate the 95 percent upper confidence limit for the mean if the average length in the sample is 30 mm. (4points)

2. (2 points) A 95% confidence interval for the mean thickness of a part in millimeters is (9.5, 11.6). Interpret this interval.

3. (6 points) State the null and alternative hypotheses in the following situations by defining the parameters used.

(a) The Postal Service wishes to prove that the mean delivery time for packages is less than 5 days.(2 points)

(b) A marketing firm believes that the average contract for a customer exceeds $50,000.(2 points)

(c) A financial institution believes that it has an average loan processing time of less than 10 days.(2 points)

4. (10 points) The population standard deviation of length of a specific manufacturing part is 5 mm.  The quality engineer selected a sample of 100 parts and measured their mean as 28 mm. The engineer would like to test whether the population average would have a mean less than 30 mm. Use a significance level of 0.05 to do the hypothesis test.

(a) What are the appropriate null and alternative hypotheses? (2 points)

(b) What is the Z-statistics for the test?(3 points)

(c) What is the rejection region?(2 points)

(d) What can you conclude about the null and alternative hypothesis (reject or fail to reject)? Why?(3 points)

Explanation / Answer

1)
a)
Sample standard deviation = sqrt(variance)
= sqrt(4) = 2

b)
z value at 95% = 1.96

Upper conidence limit = mean + z *(s/sqrt(n))
= 30 +1.96 *(2/sqrt(36))
= 30.6533

2)

We are 95% confident that the mean thickness of a part in millimeters is between 9.5 and 11.6

3)
a)
H0 : mu = 5
Ha : mu < 5

b)
H0 : mu = 50000
Ha : mu > 50000

c)
H0 : mu = 10
Ha : mu < 10

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