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A rivet is to be inserted into a hole. A random sample of \(n=15\) parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is \(s=0.008\) millimeters. Construct a \(99 \%\) lower confidence bound for \(\sigma^{2}\).

Short Answer

Expert verified
The 99% lower confidence bound for \( \sigma^2 \) is approximately 0.0000307 mm².

Step by step solution

01

Identify the given values

We are given that the sample size is \( n = 15 \), and the sample standard deviation is \( s = 0.008 \) millimeters. We are to construct a 99% lower confidence bound for \( \sigma^2 \), the population variance.
02

Find the appropriate chi-square distribution value

Since we're constructing a 99% lower confidence bound, we need the chi-square value corresponding to \( 1 - 0.99 = 0.01 \). The degrees of freedom \( df = n - 1 = 14 \). Using \( \chi^2 \) tables or a calculator, \( \chi^2_{0.01, 14} \approx 29.141 \).
03

Compute the lower bound for the variance

The formula for the lower confidence bound for variance is \((n - 1) \cdot \frac{s^2}{\chi^2_{\alpha, df}} \). Substituting the known values, we calculate: \( 14 \cdot \frac{(0.008)^2}{29.141} \approx 0.0000307 \) mm².
04

Interpret the result

This calculation indicates that we are 99% confident that the true variance of the hole diameter is greater than approximately \( 0.0000307 \) squared millimeters.

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Key Concepts

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

Chi-Square Distribution
The chi-square distribution is crucial when dealing with confidence intervals for variances. It is a special kind of distribution that describes how the variance of a sample is distributed under the assumption that the associated population is normally distributed. This distribution is characterized by its degrees of freedom (df), which are typically calculated as the sample size minus one:
  • When we have a small sample size, the chi-square distribution is more spread out.
  • As sample size increases, it resembles a normal distribution.
When constructing a confidence interval for the population variance, we use chi-square values to determine bounds. Since the interval depends on chi-square values, it is vital to use a chi-square table or calculator:
  • Find the location on the chi-square distribution table using the level of confidence (like 99%) and the corresponding degrees of freedom.
  • Apply it in the formula for the lower or upper limit based on the need.
Remember, the chi-square test is sensitive to non-normal data, so ensure your data fits the normal distribution closely before applying it.
Population Variance
Population variance (\( \sigma^2 \)) measures the spread of data points in a population. It's an average of how much each data point in a set differs from the mean of the entire dataset. Unlike the sample variance, which estimates this spread within a sample, population variance looks to capture the true variability:
  • It's important when assessing if a sample accurately represents a population.
  • Helps in constructing the confidence intervals to understand how dispersion behaves for various confidence levels.
In statistical analysis, the true population variance is rare to observe directly, so we estimate it based on sample variance.
Calculating the variance involves squaring deviations from the mean, making all differences positive. By building a confidence interval, one can say with certain confidence that it contains the true population variance. This helps us make informed decisions when we only have sample data.
Sample Standard Deviation
The sample standard deviation (\( s \)) is a measure that describes how spread out the values are in a sample. It helps us quantify the dispersion or spread of a data set:
  • Calculated by determining the square root of the sum of squared deviations from the sample mean.
  • It's related to variance and is essentially the square root of sample variance.
In practice, the sample standard deviation provides insight into the consistency of measurements:
  • If \( s \) is small, the measurements are close to the sample mean.
  • If larger, it indicates more spread out data.
When we have a sample statistic like the sample standard deviation, we often infer about the whole population. That includes estimating the population standard deviation and variance, crucial in statistics for constructing precise confidence intervals. It ensures we accurately reflect what's happening in the larger group based on smaller data points.

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Most popular questions from this chapter

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with \(\sigma^{2}=1000(\mathrm{psi})^{2}\). A random sample of 12 specimens has a mean compressive strength of \(\bar{x}=3250\) psi. (a) Construct a \(95 \%\) two-sided confidence interval on mean compressive strength. (b) Construct a \(99 \%\) two-sided confidence interval on mean compressive strength. Compare the width of this confidence interval with the width of the one found in part (a).

A manufacturer of electronic calculators takes a random sample of 1200 calculators and finds that there are eight defective units. (a) Construct a \(95 \%\) confidence interval on the population proportion. (b) Is there evidence to support a claim that the fraction of defective units produced is \(1 \%\) or less?

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (37.53,49.87) and (35.59,51.81) (a) What is the value of the sample mean? (b) One of these intervals is a \(99 \%\) CI and the other is a \(95 \%\) CI. Which one is the \(95 \%\) CI and why?

A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and on 18 of these helmets some damage was observed. (a) Find a \(95 \%\) two-sided confidence interval on the true proportion of helmets of this type that would show damage from this test. (b) Using the point estimate of \(p\) obtained from the preliminary sample of 50 helmets, how many helmets must be tested to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than \(0.02,\) regardless of the true value of \(p ?\)

An article in the Australian Journal of Agricultural Research ["Non-Starch Polysaccharides and Broiler Performance on Diets Containing Soyabean Meal as the Sole Protein Concentrate" (1993, Vol. 44, No. 8, pp. \(1483-1499)\) ] determined that the essential amino acid (Lysine) composition level of soybean meals is as shown below \((\mathrm{g} / \mathrm{kg})\): $$ \begin{array}{lllll} 22.2 & 24.7 & 20.9 & 26.0 & 27.0 \\ 24.8 & 26.5 & 23.8 & 25.6 & 23.9 \end{array} $$ (a) Construct a \(99 \%\) two-sided confidence interval for \(\sigma^{2}\). (b) Calculate a \(99 \%\) lower confidence bound for \(\sigma^{2}\). (c) Calculate a \(90 \%\) lower confidence bound for \(\sigma\). (d) Compare the intervals that you have computed.

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