7.4 Confidence Intervals for Variance and Standard Deviation Statistics.
Lesson 9 - 4 Confidence Intervals about a Population Standard Deviation.
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Transcript of Lesson 9 - 4 Confidence Intervals about a Population Standard Deviation.
![Page 1: Lesson 9 - 4 Confidence Intervals about a Population Standard Deviation.](https://reader036.fdocuments.net/reader036/viewer/2022083007/56649e4f5503460f94b47042/html5/thumbnails/1.jpg)
Lesson 9 - 4
Confidence Intervals about a Population Standard Deviation
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Objectives• Find critical values for the chi-square distribution
• Construct and interpret confidence intervals about the population variance and standard deviation
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Vocabulary• Chi-Square distribution
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Characteristics of the Chi-Square Distribution
• It is not symmetric
• The shape of the chi-square distribution depends on the degrees of freedom (just like t-distribution)
• As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric
• The values of χ² are nonnegative; that is, values of χ² are always greater than or equal to zero (0)
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Chi-Square DistributionIf a simple random sample of size n is obtained from a normally distributed population with mean μ and standard deviation σ, then
has a chi-squared distribution with n-1 degrees of freedom
(n – 1) s²χ² = ----------- σ²
n=5 degrees of freedom
n=10 degrees of freedom
n=30 degrees of freedom
α/2 α/2
1 – α
χ² 1-α/2 χ² α/2
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A (1 – α) * 100% Confidence Interval about σ²
If a simple random sample of size n is obtained from a normal population with mean μ and standard deviation σ, then a (1 – α) * 100% confidence interval about σ² is given by
(n – 1) s²Lower bound = ----------- χ²α/2
(n – 1) s²Upper bound = ----------- χ²1-α/2
![Page 7: Lesson 9 - 4 Confidence Intervals about a Population Standard Deviation.](https://reader036.fdocuments.net/reader036/viewer/2022083007/56649e4f5503460f94b47042/html5/thumbnails/7.jpg)
Example 1
We have measured a sample standard deviation of s = 8.3 from a sample of size n = 12. Compute a 90% confidence interval for the standard deviation.
n = 12, so there are 11 degrees of freedom90% confidence means that α = 0.05χ2
0.05 = 19.68 and χ20.95 = 4.57
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Example 2
We have measured a sample standard deviation of s = 6.1 from a sample of size n = 15. Compute a 95% confidence interval for the variance.
n = 15, so there are 14 degrees of freedom95% confidence means that α = 0.025χ2
0.025 = and χ20.975 =
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Summary and Homework
• Summary– We can construct confidence intervals for population
variances and standard deviations in much the same way as for population means and proportions
– We use the chi-square distribution to obtain critical values
– We divide the sample variances and standard deviations by the critical values to obtain the confidence intervals
• Homework– pg 491 – 492; 2, 4, 7, 10, 15b, c, d