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A precision instrument is guaranteed to read accurately to within 2 units. A sample of four instrument readings on the same object yielded the measurements \(353,351,351,\) and \(355 .\) Find a \(90 \%\) confidence interval for the population variance. What assumptions are necessary? Does the guarantee seem reasonable?

Short Answer

Expert verified
The guarantee does not seem reasonable as the 90% confidence interval for the variance is (1.41, 50.97).

Step by step solution

01

Find the Sample Mean

First, calculate the sample mean of the four instrument readings: \[ \bar{x} = \frac{353 + 351 + 351 + 355}{4} \]\[ \bar{x} = \frac{1410}{4} = 352.5 \]
02

Calculate Sample Variance

Next, compute the sample variance using the formula for variance:\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \]Calculate each squared deviation from the mean and divide by \(n-1\):\[ s^2 = \frac{(353 - 352.5)^2 + (351 - 352.5)^2 + (351 - 352.5)^2 + (355 - 352.5)^2}{3} \]\[ s^2 = \frac{0.25 + 2.25 + 2.25 + 6.25}{3} \]\[ s^2 = \frac{11}{3} = 3.67 \]
03

Determine Chi-Square Values

For constructing a confidence interval for the variance, use the chi-square distribution. Find the chi-square values for a 90% confidence interval with \(n-1\) degrees of freedom (\(n=4\), so \(df=3\)).\( \chi^2_{0.05} = 0.216 \) and \( \chi^2_{0.95} = 7.815 \) based on chi-square tables.
04

Construct the Confidence Interval

The confidence interval for the variance \(\sigma^2\) is given by:\[ \left(\frac{(n-1)s^2}{\chi^2_{0.95}}, \frac{(n-1)s^2}{\chi^2_{0.05}}\right) \]Substitute the known values:\[ \left(\frac{3(3.67)}{7.815}, \frac{3(3.67)}{0.216}\right) \]\[ \left(\frac{11.01}{7.815}, \frac{11.01}{0.216}\right) \]\[ (1.41, 50.97) \]
05

Assumptions and Reasonableness of the Guarantee

We assume that the instrument readings are normally distributed and that each reading is independent of others. With a variance interval ranging from 1.41 to 50.97, the precision guarantee of within 2 units does not appear reasonable since the interval is much larger than what 2 units suggest.

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Key Concepts

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

Sample Mean
The sample mean is one of the most fundamental statistics used in data analysis. It represents the average of a set of numbers. The sample mean is calculated by summing all the observations and dividing by the number of observations. In this exercise, four instrument readings are given as 353, 351, 351, and 355. By applying the formula for the sample mean:\[ \bar{x} = \frac{353 + 351 + 351 + 355}{4} = 352.5 \] This calculation gives us a better sense of the central tendency of the instrument readings. The sample mean acts as a reliable summary statistic when data is collected randomly and each observation is independent. It is crucial for further statistical calculations, such as variance and confidence intervals.
Sample Variance
Sample variance quantifies the extent of variability or spread in a collection of data points. It measures how much each reading differs from the sample mean. The formula for sample variance is:\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \] Using the example provided, the calculated sample variance is:\[ s^2 = \frac{(353 - 352.5)^2 + (351 - 352.5)^2 + (351 - 352.5)^2 + (355 - 352.5)^2}{3} = 3.67 \] These computations indicate how spread out the measurements are around the mean. A higher variance suggests more dispersion, while a lower variance indicates that data points are closer to the mean. Variance is critical in constructing confidence intervals, identifying the reliability of measurements, and understanding data variability.
Chi-Square Distribution
The Chi-Square Distribution is a key concept when dealing with statistical analyses involving variance. It is specifically used for inferential statistics when estimating the population variance based on sample variance. In this exercise, we need it to construct a confidence interval for variance. The Chi-Square Distribution depends on the degrees of freedom, which is the sample size minus one.For four readings (=4), the degrees of freedom are three. The chi-square critical values for a 90% confidence interval were identified as \( \chi^2_{0.05} = 0.216 \) and \( \chi^2_{0.95} = 7.815 \). These values are derived from chi-square distribution tables and are vital for calculating the confidence interval bounds of the variance. The knowledge of this distribution assists in determining how likely sample variances are to differ from the population variance.
Assumptions in Statistics
Assumptions in statistical analyses are conditions that are supposed to hold for the results to be valid. For confidence intervals on variance, several assumptions must be verified:
  • The data is assumed to be randomly sampled from the population.
  • Measurements must be independent of one another.
  • The population distribution is assumed to be normal.
In the present scenario, the assumption of normality is crucial as it affects the reliability of the confidence interval obtained using the Chi-Square Distribution. Additionally, the independence of readings ensures that one measurement does not influence another. These assumptions mean that the findings, such as the wide variance interval in this exercise, must be considered with a degree of scrutiny. If any assumptions are violated, it could question the applicability of the conclusions drawn, such as the unreasonableness of the instrument's precision guarantee.

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Most popular questions from this chapter

An increase in the rate of consumer savings frequently is tied to a lack of confidence in the economy and is said to be an indicator of a recessional tendency in the economy. A random sampling of \(n=200\) savings accounts in a local community showed the mean increase in savings account values to be \(7.2 \%\) over the past 12 months, with standard deviation \(5.6 \% .\) Estimate the mean percentage increase in savings account values over the past 12 months for depositors in the community. Place a bound on your error of estimation.

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