/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 In a study of speed dating condu... [FREE SOLUTION] | 91Ó°ÊÓ

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In a study of speed dating conducted at Columbia University, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below \((1=\) not attractive; \(10=\) extremely attractive). Construct a \(95 \%\) confidence interval estimate of the standard deviation of the population from which the sample was obtained.

Short Answer

Expert verified
Calculate the sample standard deviation, find the critical chi-square values, and use the confidence interval formula for the standard deviation.

Step by step solution

01

- List the Given Data

Write down the sample data given in the exercise. Suppose the sample data is: [3, 5, 6, 7, 8, 4, 7, 6, 5, 7].
02

- Find the Sample Standard Deviation

Calculate the sample standard deviation using the formula: \[ s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n}{(x_i - \bar{x})^2} } \]where \(n\) is the number of samples, \(x_i\) are the sample values, and \(\bar{x}\) is the sample mean.
03

- Determine the Degrees of Freedom

The degrees of freedom (df) is equal to the sample size minus 1. For our sample, if \(n = 10\), then \(df = n - 1 = 9\).
04

- Find Critical Chi-Square Values

Use a Chi-square distribution table to find the critical values for \(\chi^2\) at \(95\%\) confidence level with \(df = 9\). These are typically denoted as \(\chi^2_{\alpha/2,df}\) and \(\chi^2_{1-\alpha/2,df}\).
05

- Use the Formula for Confidence Interval

Apply the formula to find the confidence interval for the population standard deviation:\[ \sqrt{ \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2), df}} } \leq \sigma \leq \sqrt{ \frac{(n-1)s^2}{\chi^2_{\alpha/2, df}} } \]where \((1-\alpha/2)\) and \(\alpha/2\) correspond to the upper and lower critical values respectively.
06

- Calculate the Interval

Substitute the given values into the formula and solve to get the confidence interval for the population standard deviation.

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Key Concepts

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

Sample Standard Deviation
In statistics, the sample standard deviation is a measure of the amount of variation or dispersion in a set of values. When you have a sample from a larger population, calculating the sample standard deviation helps you understand the spread of the data within that sample. The formula for the sample standard deviation is:
\( s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n}{(x_i - \bar{x})^2} } \)
Here:
  • \(s\): Sample standard deviation
  • \(n\): Number of sample values
  • \(x_i\): Each individual sample value
  • \(\bar{x}\): Mean of the sample values
Basically, you:
  • Find the mean of the sample values.
  • Subtract the mean from each sample value to find the deviation for each.
  • Square each of these deviations.
  • Sum these squared deviations.
  • Divide by the number of samples minus one.
  • Take the square root of that result.
This gives you a numerical value that provides insight on how spread out the sample data is.
Degrees of Freedom
Degrees of freedom, often abbreviated as df, is a concept referring to the number of values in a calculation that are free to vary. When calculating the sample standard deviation or creating confidence intervals, the degrees of freedom is important because it adjusts for the fact that these calculations are based on sample data and not the entire population.
For a sample where the size is \(n\), degrees of freedom is calculated as: \( df = n - 1 \).
  • For instance, if your sample size is 10 (\(n = 10\)), then the degrees of freedom would be 9 (\( df = 9 \)).
  • Degrees of freedom help make our sample estimate more representative of the population parameter by compensating for the fact that we are using a sample mean to compute the variance.
This adjustment ensures that our estimates for various statistical parameters remain unbiased.
Chi-Square Distribution
The Chi-square distribution is a widely used probability distribution in statistics, particularly when analyzing variance and categorical data. It is especially useful for constructing confidence intervals for variance and standard deviation. The Chi-square distribution depends on the degrees of freedom (df).

If you want to find a confidence interval for the population standard deviation using sample data, use the critical values from the Chi-square distribution table corresponding to your specified confidence level and degrees of freedom.
  • For a \(95\)% confidence interval and \( df = 9\), you'll find two critical values: one for \(\alpha/2\) and one for \((1-\alpha/2)\).
  • These values help define the range in which we expect the population parameter to lie, by acting as cutoff points in our calculations.
Understanding Chi-square distribution is key to correctly interpreting statistical data, especially when dealing with variances and standard deviations.
Population Standard Deviation
The population standard deviation (\(\sigma\)) is a measure of the spread of all the values in a population. Unlike the sample standard deviation, which is calculated from a sample, the population standard deviation considers every value in the entire population.
Calculating the population standard deviation provides a precise measure of variability or dispersion for the whole population. When sample data is used, constructing a confidence interval gives a range of possible values within which the population standard deviation likely lies.
The formula for the confidence interval for the population standard deviation is:
\[ \sqrt{ \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2), df}} } \leq \sigma \leq \sqrt{ \frac{(n-1)s^2}{\chi^2_{\alpha/2, df}} } \]
  • Here, \(n\) represents the sample size, and \(s\) is the sample standard deviation.
  • \( \chi^2_{(1-\alpha/2)}\) and \( \chi^2_{\alpha/2}\) are the critical values from the Chi-square distribution.
By understanding these values, you can better estimate the true variability in your population, based on your sample.

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Most popular questions from this chapter

The drug OxyContin (oxycodone) is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 227 subjects were treated with OxyContin and 52 of them developed nausea (based on data from Purdue Pharma L.P.). a. Construct a \(95 \%\) confidence interval estimate of the percentage of OxyContin users who develop nausea.b. Compare the result from part (a) to this \(95 \%\) confidence interval for 5 subjects who developed nausea among the 45 subjects given a placebo instead of OxyContin: \(1.93 \%

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In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) had a mean of \(0.4\) and a standard deviation of \(21.0\) (based on data from "Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia," by Gardner et al., Archives of Internal Medicine, Vol. 167). Construct a \(98 \%\) confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

In a study of speed dating conducted at Columbia University, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below \((1=\) not attractive; \(10=\) extremely attractive). Use a \(99 \%\) confidence level. Can the result be used to estimate the mean amount of attractiveness of the population of all adult males?

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