/*! 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 Finding Confidence Intervals. In... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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:
\[ \bar{x} = \frac{84}{12} = 7. \]

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the critical value \(z_{a / 2}\) that corresponds to the given confidence level. $$99 \%$$

Find the critical value \(z_{a / 2}\) that corresponds to the given confidence level. $$99.5 \%$$

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. See the preceding exercise, in which we can assume that \(\sigma=15\) for the IQ scores. Attorneys are a group with IQ scores that vary less than the IQ scores of the general population. Find the sample size needed to estimate the mean IQ of attorneys, given that we want \(98 \%\) confidence that the sample mean is within 3 IQ points of the population mean. Does the sample size appear to be practical?

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.05 margin of error, and use a confidence level of \(95 \% .\) a. Assume that nothing is known about the percentage to be estimated. b. Assume that prior studies have shown that about \(40 \%\) of full-time students earn bachelor's degrees in four years or less. c. Does the added knowledge in part (b) have much of an effect on the sample size?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.