Assignment:
Test of Hypothesis for the Mean (Known and Unknown)
Q1) A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the packages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected periodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 packages, the mean amount dispensed is 8.159 ounces, with a sample standard deviation of 0.051 ounce.
a. Is there evidence that the population mean amount is different from 8.17 ounces? (Use a .05 level of significance.)
b. Determine the p-value and interpret its meaning.
Q2) In New York State , savings banks are permitted to sell a form of life insurance called savings bank life insurance (SBLI). The approval process consists of underwriting, which includes a review of the application, a medical information bureau check, possible requests for additional medical information and medical exams, and a policy compilation stage in which the policy pages are generated and sent to the bank for delivery. The ability to deliver approved policies to customers in a timely manner is critical to the profitability of this service. During a period of one month, a random sample of 27 approved policies is selected, and the total processing time, in days, is recorded.
73 19 16 64 28 28 31 90 60 56
22 18 45 48 17 17 17 91 92 63
50 51 69 16 17
a. In the past, the mean processing time was 45 days, at the 0.05 level of significance, is there evidence that the mean processing time has changed from 45 days?
b. What assumption about the population distribution is needed in order to conduct the t test in part (a)?
c. Construct a box plot or a normal probability plot to evaluate the assumption made in (b).
d. Do you think that the assumption needed conduct the t test in (a) is valid? Explain.
Q3) One operation of a steel mill is to cut pieces of steel into parts that are used in the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts must be cut to be within + 0.005 inch of the length specified by the automobile company. Below is a sample of 100 steel parts. The measurement reported is the difference, in inches, between the actual length of the steel part, as measured by a laser measurement device, and the specified length of the steel part. For example, a value of -.002 represents a steel part that is 0.002 inch shorter than the specified length.
a. At the .05 level of significance, is there evidence that the mean difference is not equal to 0.0 inches?
b. Construct a 95% confidence interval estimate of the population mean. Interpret this interval.
c. Compare the conclusions reached in (a) and (b).
d. Because n=100, do you have to be concerned about the normality assumption needed for the t test and t interval.
Q4) In a recent year, the Federal Communications Commission reported that the mean wait for repairs for AT&T customers was 25.3 hours. In an effort to improve this service, suppose that a new repair service process, used for a sample of 100 repairs, resulted in a sample mean of 22.3 hours and a sample standard deviation of 8.3 hours.
a. Is there evidence that the population mean amount is less than 25.3 hours? (Use a 0.05 level of significance.)
b. Determine the p-value and interpret its meaning.
c. Compare the results in (a) and (b)
Error
-0.002
0.0005
0.0025
0.001
0.002
0.001
0.005
-0.002
0
0.001
-0.0025
-0.003
0.001
-0.0005
0
-0.003
-0.001
0.0005
0.0025
-0.0025
0.002
0.001
0.001
0.001
-0.002
-0.003
-0.0015
-0.0005
0
-0.0025
-0.003
-0.001
0.002
-0.001
0
0.003
0.0015
0
0
-0.0025
0.0005
0.001
0.0005
0.001
0.0025
0.001
0
-0.0025
-0.001
0.0035
0
-0.003
-0.002
-0.001
-0.0015
-0.0015
-0.002
0.001
-0.0025
-0.0005
-0.0015
-0.0005
-0.0015
-0.0005
-0.0005
-0.001
-0.0015
0.0025
0.001
0.0005
0
-0.002
-0.0005
0.0005
0.0005
0
-0.0005
0.002
0.001
-0.0015
-0.001
-0.001
0.0005
-0.0025
0.0025
0.002
0.0025
-0.002
0
-0.0005
-0.001
0
-0.002
0
0
0.002
-0.003
-0.0005
-0.002
-0.002