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Chapter 8: Problem 12

The diameter of holes for a cable harness is known to have a normal distribution with \(\sigma=0.01\) inch. A random sample of size 10 yields an average diameter of 1.5045 inch. Find a \(99 \%\) two-sided confidence interval on the mean hole diameter.

Short Answer

Expert verified
The \(99\%\) confidence interval for the mean hole diameter is \((1.4964, 1.5126)\) inches.

Step by step solution

01

Identify Given Variables

We are given the population standard deviation \( \sigma = 0.01 \) inch, the sample size \( n = 10 \), and the sample mean \( \bar{x} = 1.5045 \) inch. We need to find the \(99\%\) confidence interval for the mean hole diameter.
02

Understand the Confidence Interval Formula

Since the population standard deviation is known and the sample size is small, we use the formula for the confidence interval of a normal distribution: \[ \bar{x} \pm z^* \frac{\sigma}{\sqrt{n}} \]where \( z^* \) is the critical value that corresponds to the desired confidence level.
03

Determine the Critical Value

For a \(99\%\) confidence level, the critical value \( z^* \) corresponds to the upper \(0.5\%\) of the standard normal distribution in each tail. Using a standard normal distribution table or calculator, we determine \( z^* = 2.576 \).
04

Calculate the Margin of Error

The margin of error \( ME \) is calculated using the formula:\[ ME = z^* \frac{\sigma}{\sqrt{n}} = 2.576 \frac{0.01}{\sqrt{10}} \approx 0.0081 \]
05

Construct the Confidence Interval

Using the sample mean and the margin of error, the \(99\%\) confidence interval for the mean is:\[ (\bar{x} - ME, \bar{x} + ME) = (1.5045 - 0.0081, 1.5045 + 0.0081) = (1.4964, 1.5126) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
Normal distribution is a probability distribution that is symmetric about the mean. It is often referred to as the bell curve due to its bell-shaped appearance.
In a normal distribution:
  • Most of the data points fall close to the mean.
  • The mean, median, and mode of the distribution are equal.
  • Parameters include the mean (which indicates the center of the distribution) and the standard deviation (which measures the spread or width of the curve).
This type of distribution is important when calculating confidence intervals, as it allows us to use specific statistical methods to estimate the range where the true population parameter is expected to fall.
Critical Value
The critical value is a factor used to compute the margin of error in a confidence interval. It is chosen based on the desired confidence level of the interval.
For example:
  • At a 95% confidence level, the critical value is typically 1.96 when using a normal distribution.
  • At a 99% confidence level, as in this exercise, the critical value is 2.576.
To find the critical value, you consult a z-table, which gives you the critical value corresponding to the tail area of the normal distribution's curve.
The smaller the p-value in each tail, the larger the critical value, which means a wider confidence interval.
Margin of Error
The margin of error represents the range of values above and below the sample statistic in a confidence interval. It is calculated using the formula:\[ ME = z^* \times \frac{\sigma}{\sqrt{n}} \]where:
  • \( z^* \) is the critical value.
  • \( \sigma \) is the population standard deviation.
  • \( n \) is the sample size.
The margin of error accounts for random sampling variability and increases with higher confidence levels and smaller sample sizes.
In our example, the margin of error was calculated to be approximately 0.0081, reflecting how much the sample mean could vary from the actual population mean.
Population Standard Deviation
Population standard deviation \( \sigma \) measures the amount of variation or dispersion of a set of values.
Key characteristics include:
  • A smaller standard deviation means the data points are closer to the mean.
  • A larger standard deviation indicates a wider spread in data values.
In the given exercise, knowing the population standard deviation \( \sigma = 0.01 \) inch is important because it allows the use of a z-distribution to calculate the confidence interval.
Having a known standard deviation simplifies the calculation process and enhances the accuracy of the confidence interval for estimating the true mean.

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Most popular questions from this chapter

Following are two confidence interval estimates of the mean \(\mathrm{m}\) of the cycles to failure of an automotive door latch mechanism (the test was conducted at an elevated stress level to accelerate the failure). $$ 3124.9 \leq \mu \leq 3215.7 \quad 3110.5 \leq \mu \leq 3230.1 $$ (a) What is the value of the sample mean cycles to failure? (b) The confidence level for one of these CIs is \(95 \%\) and for the other is \(99 \%\). Both CIs are calculated from the same sample data. Which is the \(95 \%\) CI? Explain why.

The brightness of a television picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in \(\bar{x}=317.2\) and \(s=15.7\). Find (in microamps) a \(99 \%\) confidence interval on mean current required. State any necessary assumptions about the underlying distribution of the data.

An article in the Journal of Agricultural Science ["The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects" (1997, Vol. 128, pp. \(135-142\) ) ] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the United Kingdom. The analysis used a variety of nitrogen fertilizer applications \((\mathrm{kg} \mathrm{N} / \mathrm{ha}),\) temperature \(\left({ }^{\circ} \mathrm{C}\right),\) and total monthly rainfall \((\mathrm{mm})\). The following data below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and \(1993 .\) The temperatures measured in June were obtained as follows: $$ \begin{array}{llllll} 15.2 & 14.2 & 14.0 & 12.2 & 14.4 & 12.5 \\ 14.3 & 14.2 & 13.5 & 11.8 & 15.2 & \end{array} $$ Assume that the standard deviation is known to be \(\sigma=0.5\) (a) Construct a \(99 \%\) two-sided confidence interval on the mean temperature. (b) Construct a \(95 \%\) lower-confidence bound on the mean temperature. (c) Suppose that you wanted to be \(95 \%\) confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used? (d) Suppose that you wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at \(95 \%\) confidence. What sample size should be used?

A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with \(\sigma=0.001\) millimeters. A random sample of 15 rings has a mean diameter of \(\bar{x}=74.036\) millimeters. (a) Construct a \(99 \%\) two-sided confidence interval on the mean piston ring diameter. (b) Construct a \(99 \%\) lower-confidence bound on the mean piston ring diameter. Compare the lower bound of this confidence interval with the one in part (a).

An article in Nuclear Engineering International (February \(1988,\) p. 33 ) describes several characteristics of fuel rods used in a reactor owned by an electric utility in Norway. Measurements on the percentage of enrichment of 12 rods were reported as follows: \(\begin{array}{llllll}2.94 & 3.00 & 2.90 & 2.75 & 3.00 & 2.95 \\ 2.90 & 2.75 & 2.95 & 2.82 & 2.81 & 3.05\end{array}\) (a) Use a normal probability plot to check the normality assumption. (b) Find a \(99 \%\) two-sided confidence interval on the mean percentage of enrichment. Are you comfortable with the statement that the mean percentage of enrichment is \(2.95 \%\) ? Why?
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