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Chapter 7: Problem 41

A more extensive tabulation of \(t\) critical values than what appears in this book shows that for the \(t\) distribution with \(20 \mathrm{df}\), the areas to the right of the values \(.687, .860\), and \(1.064\) are \(.25, .20\), and .15, respectively. What is the confidence level for each of the following three confidence intervals for the mean \(\mu\) of a normal population distribution? Which of the three intervals would you recommend be used, and why? a. \((\bar{x}-.687 s / \sqrt{21}, \bar{x}+1.725 s / \sqrt{21})\) b. \((\bar{x}-.860 s / \sqrt{21}, \bar{x}+1.325 s / \sqrt{21})\) c. \((\bar{x}-1.064 s / \sqrt{21}, \bar{x}+1.064 s / \sqrt{21})\)

Short Answer

Expert verified
Interval (c) is recommended because it is symmetric and easily identifiable with a standard confidence level.

Step by step solution

01

Understand the Problem

We are given three confidence intervals for the mean \(\mu\) of a normal population distribution. We need to find out the confidence level of each interval. The intervals are defined by their endpoints in terms of \(t\) distribution critical values and standard deviation \(s\).
02

Analyze Confidence Interval a

The first interval is \((\bar{x} - 0.687 s / \sqrt{21}, \bar{x} + 1.725 s / \sqrt{21})\). The values 0.687 and 1.725 are used as the critical points with a \(t\) distribution of 20 degrees of freedom. From the problem, 0.687 corresponds to an area of 0.25 on the right, which implies a left-tail probability of 0.75. Without additional context for the left-tail probability at 1.725, this asymmetric interval can't directly indicate a standard confidence level. But, based on the usual practice, the greater \(t\) value would align more with common configurations like 90%, 95%, etc.
03

Analyze Confidence Interval b

The second interval is \((\bar{x} - 0.860 s / \sqrt{21}, \bar{x} + 1.325 s / \sqrt{21})\). The value 0.860 corresponds to an area of 0.20 on the right, meaning a left-tail probability of 0.80. The t-value 1.325 would typically be checked to match standard confidence levels. Without the critical value information extending the typical cases, calculation makes it atypical due to the skipped middle step.
04

Analyze Confidence Interval c

The third interval is \((\bar{x} - 1.064 s / \sqrt{21}, \bar{x} + 1.064 s / \sqrt{21})\). Here, \(1.064\) corresponds to an area of 0.15 on the right. This means the left-tail area is 0.85. Since this interval is symmetric, the typical configuration might associate it to confidence levels such as 85% due to symmetric nature and close value configuration known from practice.
05

Choose Recommended Interval

Among the three, the interval (c) seems to be the most balanced as it is symmetric about the mean with equal \(t\) values for both ends. It provides a confidence interval that is standard and easily interpretable from the available values, suggesting stability and fair action towards analysis against population mean \(\mu\).

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Key Concepts

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

Confidence Interval in t-distribution
A confidence interval is a range of values that is used to estimate an unknown population parameter, in this case, the mean \( \mu \). The confidence interval provides a way to assess the uncertainty or reliability of this estimation.

When working with small sample sizes or when the population standard deviation is unknown, the t-distribution is frequently used to construct the confidence intervals. This differs from the normal distribution and accounts for the additional uncertainty by incorporating the sample size ({\it n}) into its formulation.

The format of a confidence interval in the context of a t-distribution can be seen in the formula \((\bar{x} - t^* \frac{s}{\sqrt{n}}, \bar{x} + t^* \frac{s}{\sqrt{n}})\), where:
  • \(\bar{x}\) is the sample mean.
  • \(t^*\) is the t critical value from the t-distribution.
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
Understanding how these components interact is necessary to correctly interpret a confidence interval and the confidence level associated with it.
Degrees of Freedom and its Importance
Degrees of Freedom (df) refer to the number of independent values in a statistical calculation that are free to vary. In the context of a t-distribution, the degrees of freedom are calculated as \(n - 1\), where \(n\) is the sample size. Each independent piece of data contributes to the estimation of the variance, and the constraint of the total mean leads to this deduction of one degree of freedom.

The concept of degrees of freedom is crucial because it directly affects the shape of the t-distribution. The fewer the degrees of freedom, the more the t-distribution tails are spread out, which means more uncertainty.

Understanding degrees of freedom helps you in choosing the correct t-distribution from which to obtain critical values for constructing confidence intervals and conducting hypothesis tests. As df increases, the t-distribution approaches the normal distribution.
Understanding Critical Values
Critical values are specific points on the tails of a distribution that help determine the cutoff for statistical tests. In the case of a t-distribution, critical values define the endpoints of the confidence interval for estimating a population mean.

Using the t-distribution, a t-value is the number of standard deviations a sample mean lies from the population mean. These critical values are determined by the desired confidence level and the degrees of freedom of the data. For example, if you're looking for a 95% confidence level with 20 degrees of freedom, you would use t-distribution tables or software to find the corresponding critical t-value.

Knowing the critical values allows you to establish a range within which you are confident the true population parameter lies. The chosen level of confidence directly influences these critical values - higher confidence levels demand wider intervals, meaning higher critical values (further tails of the t-distribution). It鈥檚 a trade-off between precision and certainty in statistical inference.

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Most popular questions from this chapter

A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of \(.81 \mathrm{sec}\) and a sample standard deviation of \(34 \mathrm{sec}\) ("Lightning Strikes to an Airplane in a Thunderstorm," \(J .\) of Aircraft, 1984: 607-611). Calculate a \(99 \%\) (two- sided) confidence interval for the true average echo duration \(\mu\), and interpret the resulting interval.

Chronic exposure to asbestos fiber is a well-known health hazard. The article "The Acute Effects of Chrysotile Asbestos Exposure on Lung Function" (Erviron. Research, 1978: 360-372) reports results of a study based on a sample of construction workers who had been exposed to asbestos over a prolonged period. Among the data given in the article were the following (ordered) values of pulmonary compliance \(\left(\mathrm{cm}^{3} / \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) for each of 16 subjects 8 months after the exposure period (pulmonary compliance is a measure of lung elasticity, or how effectively the lungs are able to inhale and exhale): \(\begin{array}{llllll}167.9 & 180.8 & 184.8 & 189.8 & 194.8 & 200.2 \\ 201.9 & 206.9 & 207.2 & 208.4 & 226.3 & 227.7 \\ 228.5 & 232.4 & 239.8 & 258.6 & & \end{array}\) a. Is it plausible that the population distribution is normal? b. Compute a \(95 \% \mathrm{CI}\) for the true average pulmonary compliance after such exposure. c. Calculate an interval that, with a confidence level of \(95 \%\), includes at least \(95 \%\) of the pulmonary compliance values in the population distribution.

For each of 18 preserved cores from oil-wet carbonate reservoirs, the amount of residual gas saturation after a solvent injection was measured at water flood-out. Observations, in percentage of pore volume, were \(\begin{array}{llllll}23.5 & 31.5 & 34.0 & 46.7 & 45.6 & 32.5 \\ 41.4 & 37.2 & 42.5 & 46.9 & 51.5 & 36.4 \\ 44.5 & 35.7 & 33.5 & 39.3 & 22.0 & 51.2\end{array}\) (See "Relative Permeability Studies of Gas-Water Flow Following Solvent Injection in Carbonate Rocks," Soc. of Petroleum Engineers J., 1976: 23-30.) a. Construct a boxplot of this data, and comment on any interesting features. b. Is it plausible that the sample was selected from a normal population distribution? c. Calculate a \(98 \%\) CI for the true average amount of residual gas saturation.

summary information for fracture strengths (MPa) of \(n=169\) ceramic bars fired in a particular kiln: \(\bar{x}=89.10\), \(s=3.73\). a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level of \(95 \%\). Does it appear that true average fracture strength has been precisely estimated? b. Suppose the investigators had believed a priori that the population standard deviation was about \(4 \mathrm{MPa}\). Based on this supposition, how large a sample would have been required to estimate \(\mu\) to within . \(5 \mathrm{MPa}\) with \(95 \%\) confidence? 15\. Determine the confidence level for each of the following large-sample one-sided confidence bounds: a. Upper bound: \(\bar{x}+.84 s / \sqrt{n}\) b. Lower bound: \(\bar{x}-2.05 s / \sqrt{n}\) c. Upper bound: \(\bar{x}+.67 s / \sqrt{n}\)

According to the article "Fatigue Testing of Condoms" (Polymer Testing, 2009: 567-571), "tests currently used for condoms are surrogates for the challenges they face in use," including a test for holes, an inflation test, a package seal test, and tests of dimensions and lubricant quality (all fertile territory for the use of statistical methodology!). The investigators developed a new test that adds cyclic strain to a level well below breakage and determines the number of cycles to break. A sample of 20 condoms of one particular type resulted in a sample mean number of 1584 and a sample standard deviation of 607 . Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the true average number of cycles to break. [Note: The article presented the results of hypothesis tests based on the \(t\) distribution; the validity of these depends on assuming normal population distributions.]
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