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Ch. 8 - Hypothesis Testing with Two Samples
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoLarson 8th EditionCh. 8 - Hypothesis Testing with Two SamplesProblema 8.CR.12a
Capitolo 8, Problema 8.CR.12a

The mean room rate for two adults for a random sample of 26 three-star hotels in Cincinnati has a sample standard deviation of \$31. Assume the population is normally distributed. (Adapted from Expedia)


Construct a 99% confidence interval for the population variance.

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Step 1: Understand the problem. We are tasked with constructing a 99% confidence interval for the population variance based on a sample of 26 hotels. The sample standard deviation is given as \$31, and the population is assumed to be normally distributed. This means we can use the chi-square distribution to calculate the confidence interval.
Step 2: Recall the formula for the confidence interval for the population variance. The formula is: \[ \left( \frac{(n-1)s^2}{\chi^2_{\text{upper}}}, \frac{(n-1)s^2}{\chi^2_{\text{lower}}} \right) \] where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \chi^2_{\text{upper}} \) and \( \chi^2_{\text{lower}} \) are the critical values of the chi-square distribution corresponding to the desired confidence level.
Step 3: Calculate the sample variance \( s^2 \). The sample variance is the square of the sample standard deviation: \( s^2 = 31^2 \). This will be used in the formula.
Step 4: Determine the degrees of freedom \( df \). The degrees of freedom for the chi-square distribution is \( n-1 \), where \( n \) is the sample size. Here, \( df = 26 - 1 = 25 \).
Step 5: Find the critical chi-square values for a 99% confidence interval. Using a chi-square table or statistical software, locate \( \chi^2_{\text{upper}} \) and \( \chi^2_{\text{lower}} \) for \( df = 25 \) and a 99% confidence level. Plug these values, along with \( (n-1)s^2 \), into the formula to compute the confidence interval for the population variance.

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Confidence Interval

A confidence interval is a range of values, derived from a sample, that is likely to contain the population parameter with a specified level of confidence. In this case, a 99% confidence interval means that if we were to take many samples and construct intervals in the same way, approximately 99% of those intervals would contain the true population variance.
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Percorso guidato
06:33
Introduction to Confidence Intervals

Sample Standard Deviation

The sample standard deviation is a measure of the amount of variation or dispersion in a set of sample data. It quantifies how much the individual data points deviate from the sample mean. In this scenario, the sample standard deviation of $31 will be used to estimate the population variance, which is the square of the standard deviation.
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Percorso guidato
08:45
Calculating Standard Deviation

Chi-Squared Distribution

The chi-squared distribution is a statistical distribution that is used to estimate the variance of a population based on sample data. When constructing confidence intervals for variance, the chi-squared distribution is applied, particularly when the population is normally distributed, as is the case here with the three-star hotels.
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Percorso guidato
07:01
Intro to Least Squares Regression
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In Exercises 3–6, construct the indicated confidence interval for the population mean . Which distribution did you use to create the confidence interval?


c=0.90, x̅=8.21, σ=0.62, n=8

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[APPLET] The annual earnings (in dollars) for 30 randomly selected locksmiths are shown below. Assume the population is normally distributed. (Adapted from Salary.com)


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In Exercises 3–6, construct the indicated confidence interval for the population mean . Which distribution did you use to create the confidence interval?


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In a survey of 4860 U.S. adults, 77% said they would date or have already dated someone whose religion was different from theirs. (Source: Pew Research Center)


Construct a 95% confidence interval for the proportion of U.S. adults who say they would date or have already dated someone whose religion was different from theirs.

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[APPLET] The annual earnings (in dollars) for 30 randomly selected locksmiths are shown below. Assume the population is normally distributed. (Adapted from Salary.com)


48,69446,85642,91261,67271,11254,861


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66,36048,16465,27235,25061,12765,397


58,92558,91659,01753,07045,19969,941


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A researcher claims that the mean annual earnings for locksmiths is \$55,000. At α=0.05, can you reject the researcher’s claim? Interpret the decision in the context of the original claim.

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