1 A production process is checked periodically by a quality

1. A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weight . If test results over a long period of time show that 5% of the  values are over 2.1 pounds and 5% are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process?

a. What is the population mean? (to 1 decimal)

b. What is the population standard deviation (to 2 decimals)?

2. The following sample data are from a normal population: 10, 9, 12, 14, 13, 11, 6, 5

a. What is the point estimate of the population mean?

b. What is the point estimate of the population standard deviation (to 2 decimals)?

c. With 95% confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?

d.  What is the 95% confidence interval for the population mean (to 1 decimal)?

3. Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 75 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

a. 90% Confidence, to 2 decimals

b. 95% Confidence, to 2 decimals

4. Refer to the Scheer Industries example in Section 8.2. Use 6.84 days as a planning value for the population standard deviation.

a. a. Assuming 95% confidence, what sample size would be required to obtain a margin of error of 1.5 days (round up to the next whole number)?

b. Assuming 90% confidence, what sample size would be required to obtain a margin of error of 2 days (round up to the next whole number)?

Solution

1.

Note that

x = u + z*sigma/sqrt(n)

For 5% right tailed area, the z score corresponding to it is, by table/technology,

z = 1.644853627

Also, for a left tailed area of 5%, the corresponding z score is

z = -1.644853627

Thus,

1.9 = u + (-1.644853627)*sigma/sqrt(30)
2.1 = u + (1.644853627)*sigma/sqrt(30)

Hence,

u = 2.0 [ANSWER, PART A]

sigma = 0.332991671 = 0.33 [ANSWER, PART B]

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