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A simple random sample of size \(n\) is drawn from a population that is known to be normally distributed. The sample variance, \(s^{2}\) is determined to be 12.6 . (a) Construct a \(90 \%\) confidence interval for \(\sigma^{2}\) if the sample size, \(n,\) is 20 (b) Construct a \(90 \%\) confidence interval for \(\sigma^{2}\) if the sample size, \(n\), is \(30 .\) How does increasing the sample size affect the width of the interval? (c) Construct a \(98 \%\) confidence interval for \(\sigma^{2}\) if the sample size, \(n\), is 20. Compare the results with those obtained in part (a). How does increasing the level of confidence affect the width of the confidence interval?

Short Answer

Expert verified
90% CI for \sigma^2 with n=20: (7.94, 23.70); n=30: (8.59, 22.77). 98% CI for \sigma^2 with n=20: (6.62, 26.89). Increasing sample size narrows the interval; higher confidence level widens it.

Step by step solution

01

Calculate Chi-Square Critical Values for 90% CI, n=20

Find the chi-square critical values \(χ^2_{0.05, 19}\) and \(χ^2_{0.95, 19}\) using chi-square distribution table. For a 90% confidence interval with 19 degrees of freedom (df = n-1): \(χ^2_{0.05, 19} = 10.117\) and \(χ^2_{0.95, 19} = 30.144\).
02

Calculate Confidence Interval for \sigma^2, n=20

Using the formula \[\left( \frac{(n-1) s^2}{χ^2_{0.95, n-1}}, \frac{(n-1) s^2}{χ^2_{0.05, n-1}} \right) \], substitute s^2 = 12.6, n = 20: \[\left( \frac{19 \times 12.6}{30.144}, \frac{19 \times 12.6}{10.117} \right) = (7.94, 23.70) \].
03

Calculate Chi-Square Critical Values for 90% CI, n=30

Find the chi-square critical values \(χ^2_{0.05, 29}\) and \(χ^2_{0.95, 29}\) using chi-square distribution table. For a 90% confidence interval with 29 degrees of freedom (df = n-1): \(χ^2_{0.05, 29} = 16.047\) and \(χ^2_{0.95, 29} = 42.557\).
04

Calculate Confidence Interval for \sigma^2, n=30

Substitute s^2 = 12.6 and n = 30: \[\left( \frac{29 \times 12.6}{42.557}, \frac{29 \times 12.6}{16.047} \right) = (8.59, 22.77) \]. The interval is narrower than when n = 20.
05

Calculate Chi-Square Critical Values for 98% CI, n=20

Find chi-square critical values for a 98% confidence interval with 19 degrees of freedom: \(χ^2_{0.01, 19} = 8.907\) and \(χ^2_{0.99, 19} = 36.191\).
06

Calculate Confidence Interval for \sigma^2, 98% CI, n=20

Substitute s^2 = 12.6 and n = 20 in the same formula: \[\left( \frac{19 \times 12.6}{36.191}, \frac{19 \times 12.6}{8.907} \right) = (6.62, 26.89) \]. This interval is wider than the 90% confidence interval with the same sample size.
07

Compare the Intervals

Increasing the sample size reduces the width of the confidence interval. A higher confidence level results in a wider interval.

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Key Concepts

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

Simple Random Sample
A simple random sample is a basic sampling technique where each member of a population has an equal chance of being chosen. It ensures that the sample is unbiased and accurately represents the population. This method is foundational in statistics because it enables researchers to draw general conclusions about a population based on a subset of collected data. To select a simple random sample:
- Define the population clearly.
- Assign a number to each member of the population.
- Use a random number generator to pick members from the population.
This technique is essential for maintaining randomness and reducing selection bias in studies.
Chi-Square Distribution
A chi-square distribution is a statistical method that helps assess the variance in a data set. It's commonly used when constructing confidence intervals for variance because it handles sample variances derived from normally distributed populations. In our exercise, we use the chi-square distribution to calculate confidence intervals for the population variance. The steps involve:
- Determining degrees of freedom (df), which is typically the sample size minus one (n-1).
- Using chi-square tables or statistical software to find critical values for the desired confidence level (e.g., 90%, 98%).
- Applying these critical values in the formula \(\frac{(n-1)s^2}{\chi^2_{upper}}, \frac{(n-1)s^2}{\chi^2_{lower}}\) to get the confidence interval.
The chi-square distribution is pivotal for understanding how data variability might differ from the expected distribution in hypothesis testing and interval estimation.
Sample Size Effect
The sample size directly influences the confidence interval's width. When sample size increases, the interval tends to get narrower, providing a more precise estimate.
Key effects of sample size include:
- **Larger Sample Sizes:** These yield more reliable and accurate estimates as they reduce the margin of error. Consequently, the confidence interval for the population variance becomes narrower.
- **Smaller Sample Sizes:** These can lead to wider confidence intervals, indicating less precision in the estimate of the population parameter.
In our exercise, increasing the sample size from 20 to 30 resulted in narrower confidence intervals for variance (7.94, 23.70 to 8.59, 22.77). Hence, it's crucial to consider sample size appropriately to ensure the statistical reliability of your results.

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Most popular questions from this chapter

Construct the appropriate confidence interval. A simple random sample of size \(n=12\) is drawn from a population that is normally distributed. The sample variance is found to be \(s^{2}=23.7\). Construct a \(90 \%\) confidence interval for the population variance.

Find the critical values \(\chi_{1-\alpha / 2}^{2}\) and \(\chi_{\alpha / 2}^{2}\) for the given level of confidence and sample size. \(98 \%\) confidence, \(n=23\)

A random sample of 1003 adult Americans was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without (a) Obtain a point estimate for the population proportion of adult Americans who believe that televisions are a luxury they could do without. (b) Verify that the requirements for constructing a confidence interval about \(p\) are satisfied. (c) Construct and interpret a \(95 \%\) confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without. (d) Is it possible that a supermajority (more than \(60 \%\) ) of adult Americans believe that television is a luxury they could do without? Is it likely? (e) Use the results of part (c) to construct a \(95 \%\) confidence interval for the population proportion of adult Americans who believe that televisions are a necessity.

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Blood Alcohol Concentration A random sample of 51 fatal crashes in 2013 in which the driver had a positive blood alcohol concentration (BAC) from the National Highway Traffic Safety Administration results in a mean BAC of 0.167 gram per deciliter \((\mathrm{g} / \mathrm{dL})\) with a standard deviation of \(0.010 \mathrm{~g} / \mathrm{dL}\) (a) A histogram of blood alcohol concentrations in fatal accidents shows that BACs are highly skewed right. Explain why a large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive \(\mathrm{BAC}\) (b) In \(2013,\) there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(90 \%\) confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. (d) All 50 states and the District of Columbia use a BAC of \(0.08 \mathrm{~g} / \mathrm{dL}\) as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication level? Explain.

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