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Chapter 8: Problem 101

Yunan Corporation produces bolts that are supplied to other companies. These bolts are supposed to be 4 inches long. The machine that makes these bolts does not produce each bolt exactly 4 inches long. It is known that when the machine is working properly, the mean length of the bolts made on this machine is 4 inches. The standard deviation of the lengths of all bolts produced on this machine is always equal to \(.04\) inch. The quality control department takes a sample of 20 such bolts every week, calculates the mean length of these bolts, and makes a \(98 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(4.02\) inches or the lower limit of this confidence interval is less than \(3.98\) inches, the machine is stopped and adjusted. A recent such sample of 20 bolts produced a mean length of \(3.99\) inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the population distribution is normal.

Short Answer

Expert verified
Based on the calculated confidence interval [3.97765, 4.00235] which is within the acceptable range of 4.02 inches and 3.98 inches, it can be concluded that the machine does not need an adjustment.

Step by step solution

01

Identify the necessary values

To calculate the confidence interval, we first need to identify the values for the mean length of bolts (\( \mu \)) which is 3.99 inches, sample size (n) which is 20, standard deviation (\( \sigma \)) which is 0.04 inch, and confidence level is 98%.
02

Find the Z score for 98% confidence level

The Z score corresponding to a 98% confidence level can be obtained from the standard normal distribution table or a calculator. The z value for 98% confidence level is approximately 2.33.
03

Calculate the confidence interval

Now, we put these values into the confidence interval formula which is mean 卤 Z * (standard deviation/ sqrt(n)). Substituting the values, we get 3.99 卤 2.33*(0.04/sqrt(20)) = [3.97765, 4.00235]. This is the 98% confidence interval.
04

Comparing the confidence interval with the given values

Finally, we compare this interval with the given machine adjustment limits of 4.02 inches and 3.98 inches. If the confidence interval lies within this range the machine does not need an adjustment, if it's outside of it, the machine gets adjusted. Here, the confidence interval [3.97765, 4.00235] is within the acceptable range.

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Most popular questions from this chapter

An economist wants to find a \(90 \%\) confidence interval for the mean sale price of houses in a state. How large a sample should she select so that the estimate is within \(\$ 3500\) of the population mean? Assume that the standard deviation for the sale prices of all houses in this state is \(\$ 31,500\)

A marketing researcher wants to find a \(95 \%\) confidence interval for the mean amount that visitors to a theme park spend per person per day. She knows that the standard deviation of the amounts spent per person per day by all visitors to this park is \(\$ 11\). How large a sample should the researcher select so that the estimate will be within \(\$ 2\) of the population mean?

You are working for a supermarket. The manager has asked you to estimate the mean time taken by a cashier to serve customers at this supermarket. Briefly explain how you will conduct this study. Collect data on the time taken by any supermarket cashier to serve 40 customers. Then estimate the population mean. Choose your own confidence level.

Suppose, for a sample selected from a normally distributed population, \(\bar{x}=68.50\) and \(s=8.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=16\). b. Construct a \(90 \%\) confidence interval for \(\mu\) assuming \(n=16 .\) Is the width of the \(90 \%\) confidence interval smaller than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=25 .\) Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=25\) smaller than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=16\) calculated in part a? If so, why? Explain.

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110\) and \(\hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)
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