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The sugar content of the syrup in canned peaches is normally distributed. A random sample of \(n=10\) cans yields a sample standard deviation of \(s=4.8\) milligrams. Calculate a \(95 \%\) two-sided confidence interval for \(\sigma .\)

Short Answer

Expert verified
The 95% confidence interval for \(\sigma\) is (3.29, 8.73).

Step by step solution

01

- Identify Parameters

Identify the key parameters: sample size \(n = 10\), sample standard deviation \(s = 4.8\), and confidence level \(95\%\). The degrees of freedom \(df = n-1 = 9\).
02

- Locate Chi-Square Values

To find the confidence interval for the population standard deviation \(\sigma\), use the Chi-Square distribution with \(df=9\). For a 95\% confidence interval, find the critical values \( \chi^2_{\alpha/2} \) and \( \chi^2_{1-\alpha/2} \), where \(\alpha = 0.05\).Using a Chi-Square table or calculator:- \( \chi^2_{0.025, 9} \approx 19.023\) - \( \chi^2_{0.975, 9} \approx 2.700\).
03

- Calculate Confidence Interval for Variance

Use the formula for the confidence interval for the variance:\[ \left( \frac{(n-1) \cdot s^2}{\chi^2_{\alpha/2}}, \frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha/2}} \right) \]Substitute the values:\\[ = \left( \frac{9 \cdot 4.8^2}{19.023}, \frac{9 \cdot 4.8^2}{2.700} \right) \]Calculate this to get the interval as \[(10.850, 76.292)\].
04

- Convert to Confidence Interval for Standard Deviation

Take the square root of the interval endpoints to convert from variance to standard deviation:\[ \left( \sqrt{10.850}, \sqrt{76.292} \right) \]Calculate this to yield the interval for \(\sigma\): \\((3.29, 8.73)\).

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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 vital concept in statistics and is often referred to as a bell curve due to its distinctive bell-like shape. This distribution is symmetrical around its mean, with most data points clustering around the central peak. It is characterized by two parameters: the mean (average) and the standard deviation, which measures how spread out the values are from the mean.

This distribution is important because it describes many natural phenomena and is the basis for statistical inference. In the context of our exercise, the sugar content of the syrup in canned peaches is normally distributed, which allows us to make inferences about the population standard deviation from the sample standard deviation. Normal distribution is assumed to be in place for many statistical techniques, allowing us to use different methodologies like confidence intervals to estimate parameter values with specified probability levels.
Chi-Square Distribution
The chi-square distribution is used extensively in statistics for hypothesis testing and constructing confidence intervals. It is particularly useful when dealing with sample variances. Unlike the normal distribution, the chi-square distribution is not symmetrical. It is skewed to the right, especially when the degrees of freedom are low.

When we calculate the confidence interval for the population variance and standard deviation, we employ the chi-square distribution because we assume our data comes from a normally distributed population. In our solution, the degrees of freedom are derived from the sample size minus one ( - 1 = 9), which was used to find the critical values of the chi-square distribution used to build our confidence interval. These critical values determine the range within which the true population variance is expected to lie with a certain level of confidence, such as the 95% used here.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation indicates that the values are spread out over a wider range.

In the exercise, the sample standard deviation is given as 4.8 milligrams, which serves as an estimate for the population standard deviation. Calculating a confidence interval for this parameter helps understand the precision of this estimate. By following the steps and using the chi-square distribution, we convert our findings on the variance into a range for the standard deviation. This process is critical in statistics as it provides insight into the data's spread in the population from which the sample was drawn.
Population Variance
Population variance is the average of the squared differences from the mean. It's an essential measure in statistics because it describes the extent to which individuals within a population differ from the population mean.

In the exercise, we aim to calculate the variance for the population of canned peaches' syrup content. Using the given sample standard deviation, the variance can be calculated and thereafter transformed into a confidence interval. This interval provides a range within which we expect the true population variance to fall 95% of the time, reflecting our confidence level. Understanding population variance not only helps in understanding the data's spread but also enables the assessment of how much sample data deviates from the overall population, which is crucial for making informed decisions based on statistical results.

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Most popular questions from this chapter

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\). (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \% \mathrm{CI}\) on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

Determine the values of the following percentiles: \(\chi_{0.05,10}^{2}, \chi_{0.025,15}^{2}, \chi_{0.01,12}^{2}, \chi_{0.95,20}^{2}, \chi_{0.99,18}^{2}, \chi_{0.995,16}^{2},\) and \(\chi_{0.005,25}^{2}\).

An article in Technometrics (1999, Vol. 41, pp. 202- 211 ) studied the capability of a gauge by measuring the weight of paper. The data for repeated measurements of one sheet of paper are in the following table. Construct a \(95 \%\) one-sided upper confidence interval for the standard deviation of these measurements. Check the assumption of normality of the data and comment on the assumptions for the confidence interval. $$ \begin{array}{lllll} \hline &&{\text { Observations }} \\ \hline 3.481 & 3.448 & 3.485 & 3.475 & 3.472 \\ 3.477 & 3.472 & 3.464 & 3.472 & 3.470 \\ 3.470 & 3.470 & 3.477 & 3.473 & 3.474 \\ \hline \end{array} $$

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.

A manufacturer of electronic calculators takes a random sample of 1200 calculators and finds 8 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?

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