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Finding Confidence Intervals. In Exercises \(9-16,\) assume that each sample is a simple random sample obtained from a population with a normal distribution. Speed Dating 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 \((1=\text { not attractive; } 10=\text { extremely attractive). Construct a } 95 \%\) confidence interval estimate of the standard deviation of the population from which the sample was obtained. $$ \begin{array}{rrrrrrr} 7 & 8 & 2 & 10 & 6 & 5 & 7 & 8 & 8 & 9 & 5 & 9 \end{array} $$

Short Answer

Expert verified
The 95% confidence interval for the standard deviation is approximately 2.46 to 24.54.

Step by step solution

01

- Write down the sample data

List the sample data provided: 7, 8, 2, 10, 6, 5, 7, 8, 8, 9, 5, 9.
02

- Calculate the sample mean

The sample mean \( \bar{x} \) is calculated by summing all the sample values and dividing by the total number of samples. \[ \bar{x} = \frac{7+8+2+10+6+5+7+8+8+9+5+9}{12} = \frac{84}{12} = 7 \]
03

- Calculate the sample variance

The sample variance \( s^2 \) is calculated using the formula: \[ s^2 = \frac{\frac{1}{n-1} \bigsqcup\big( x_i - \bar{x} \big)^2}{n-1} \] Here, \[ (7-7)^2 + (8-7)^2 + (2-7)^2 + (10-7)^2 + (6-7)^2 + (5-7)^2 + (7-7)^2 + (8-7)^2 + (8-7)^2 + (9-7)^2 + (5-7)^2 + (9-7)^2 \] = 0 + 1 + 25 + 9 + 1 + 4 + 0 + 1 + 1 + 4 + 4 + 4 = 54 So, \[ s^2 = \frac{54}{11} = 4.91\text{ (approximately) } \]
04

- Calculate the sample standard deviation

The sample standard deviation \( s \) is the square root of the sample variance. \[ s = \frac{\bigsqcup 4.91} = 2.22 \text{(approximately)}. \]
05

- Identify the degrees of freedom

The degrees of freedom \( df \) is equal to the sample size minus one \( n-1 \). Here, \[ df = 12 - 1 = 11. \]
06

- Determine chi-squared critical values

For a 95% confidence interval with 11 degrees of freedom, consult the chi-squared distribution table. The critical values for \( \frac{\bigsqcup Chi_{ 0.025}}^2 \) = 2.20 and \( \frac{\bigsqcup Chi_{0.975}}^2 \) = 21.92.
07

- Calculate the confidence interval

Use the chi-squared critical values along with the sample standard deviation to calculate the confidence interval for the population standard deviation. The formula is: \(\text{lower bound} = \frac{(n-1)s^2}{chi_{0.025}}) = ) \frac{11*4.91}{21.92}=2.46\) \( \text{upper bound} = [\frac(n-1)*s^2}{chi_{0.975}=\frac{11*4.91}{2.20}=24.54 \) The 95% confidence interval for \( \theta \) is approximately 2.46 to 24.54.

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Key Concepts

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

Sample Variance
The sample variance measures how much the individual data points in a sample differ from the sample mean. It gives an estimation of the spread or dispersion of the sample data. To calculate the sample variance, follow these steps:
  • First, find the mean (\bar{x}) of the sample. This is the sum of all sample values divided by the number of samples.
  • Next, subtract the mean from each data point to obtain the deviation of each point from the mean.
  • Square each deviation.
  • Sum up all the squared deviations.
  • Finally, divide this sum by the number of data points minus one (n-1). This gives you the sample variance (s^2).
Using our example data (7, 8, 2, 10, 6, 5, 7, 8, 8, 9, 5, 9), the mean is 7. Subtract the mean from each data point, square the result, sum these squared deviations, and divide by 11 (since there are 12 data points):
\[ s^2 = \frac{1}{11} \sum (x_i - \bar{x})^2 \approx 4.91. \]
Sample Standard Deviation
The sample standard deviation is a measure of the dispersion of the sample data points around the sample mean, representing a degree of variability within the sample. It is the square root of the sample variance. It is calculated as follows:
  • First, find the sample variance (s^2).
  • Then take the square root of the sample variance.
For our example, since the sample variance is approximately 4.91, the sample standard deviation will be:
\[ s = \sqrt{4.91} \approx 2.22. \]
Chi-Squared Distribution
The chi-squared distribution is a probability distribution that is used to estimate the variance of a population based on a sample. This distribution is particularly useful in hypothesis testing and constructing confidence intervals. Its shape depends on the degrees of freedom. When calculating a confidence interval for the standard deviation of a population, critical values from the chi-squared distribution are used to determine the bounds.
In our example, we want a 95% confidence interval for the standard deviation. We use the chi-squared distribution with 11 degrees of freedom to find critical values for chi-squared at 0.025 and 0.975.
These critical values help establish the lower and upper bounds of the confidence interval:
\( \text{Lower bound} = \frac{(n-1) s^2}{\chi^2_{0.975}} \text{ and } \text{Upper bound = } \frac{(n-1) s^2}{\chi^2_{0.025}}.\)
Degrees of Freedom
Degrees of freedom (df) refers to the number of independent values or quantities that can be assigned to a statistical distribution. In the context of calculating the sample variance and standard deviation, the degrees of freedom are equal to the sample size minus one (n-1).
Degrees of freedom are critical in determining the appropriate distribution (such as the chi-squared distribution) to use for statistical inferences.
In our example, the sample size is 12, so the degrees of freedom is:
\[ df = 12 - 1 = 11. \]
Sample Mean
The sample mean (\bar{x}) represents the average value of the sample data. It is calculated by summing up all the sample values and then dividing by the total number of samples. The sample mean is a primary measure of central tendency in statistics.
Calculating the sample mean involves:
  • Adding up all sample values.
  • Dividing this sum by the number of samples (n).
Using our sample data (7, 8, 2, 10, 6, 5, 7, 8, 8, 9, 5, 9), the sum of the samples is 84 and the number of samples is 12. Therefore, the sample mean is:
\[ \bar{x} = \frac{84}{12} = 7. \]

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Most popular questions from this chapter

Finding Critical Values and Confidence Intervals. In Exercises \(5-8,\) use the given information to find the number of degrees of freedom, the critical values \(\mathcal{X}_{L}^{2}\) and \(\mathcal{X}_{R}^{2},\) and the confidence interval estimate of \(\boldsymbol{\sigma} .\) The samples are from Appendix \(\boldsymbol{B}\) and it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Heights of Men \(99 \%\) confidence; \(n=153, s=7.10 \mathrm{cm}\)

Use technology to create the large number of bootstrap samples. Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands. Using the bootstrap method with 1000 bootstrap samples, construct a \(99 \%\) confidence interval estimate of \(\mu\). How does the result compare to the confidence interval found in Exercise 22 "Caffeine in Soft Drinks" in Section \(7-2\) on page \(330 ?\) $$\begin{array}{rrrrrrrrrrrr}0 & 0 & 34 & 34 & 34 & 45 & 41 & 51 & 55 & 36 & 47 & 41 & 0 & 0 & 53 & 54 & 38 & 0 & 41 & 47\end{array}$$

Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes pulse rates of 153 randomly selected adult males, and those pulse rates vary from a low of 40 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want \(99 \%\) confidence that the sample mean is within 2 bpm of the population mean. a. Find the sample size using the range rule of thumb to estimate \(\sigma .\) b. Assume that \(\sigma=11.3\) bpm, based on the value of \(s=11.3\) bpm for the sample of 153 male pulse rates. c. Compare the results from parts (a) and (b). Which result is likely to be better?

Construct the confidence interval estimate of the mean. Data Set 3 "Body Temperatures" in Appendix B includes a sample of 106 body temperatures having a mean of \(98.20^{\circ} \mathrm{F}\) and a standard deviation of \(0.62^{\circ} \mathrm{F}\). Construct a \(95 \%\) confidence interval estimate of the mean body temperature for the entire population. What does the result suggest about the common belief that \(98.6^{\circ} \mathrm{F}\) is the mean body temperature?

Find the sample size required to estimate the population mean. The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of college professors. We want to be 99\% confident that our sample mean is within 4 IQ points of the true mean. The mean for this population is clearly greater than \(100 .\) The standard deviation for this population is less than 15 because it is a group with less variation than a group randomly selected from the general population; therefore, if we use \(\sigma=15\) we are being conservative by using a value that will make the sample size at least as large as necessary. Assume then that \(\sigma=15\) and determine the required sample size. Does the sample size appear to be practical?

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