Chapter Additional: Standard Deviation and Chi Square


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1 Chapter Additional: Standard Deviation and Chi Square Chapter Outline: 6.4 Confidence Intervals for the Standard Deviation 7.5 Hypothesis testing for Standard Deviation
2 Section 6.4 Objectives Interpret the chisquare distribution and use a chisquare distribution table Use the chisquare distribution to construct a confidence interval for the variance and standard deviation
3 The ChiSquare Distribution The point estimate for σ is s The point estimate for σ is s s is the most unbiased estimate for σ Estimate Population Parameter with Sample Statistic Variance: σ s Standard deviation: σ s
4 The ChiSquare Distribution You can use the chisquare distribution to construct a confidence interval for the variance and standard deviation. If the random variable x has a normal distribution, then the distribution of ( n1) s σ forms a chisquare distribution for samples of any size n > 1.
5 Properties of The ChiSquare Distribution 1. All chisquare values χ are greater than or equal to zero.. The chisquare distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ, use the χ distribution with degrees of freedom equal to one less than the sample size. d.f. = n 1 Degrees of freedom 3. The area under each curve of the chisquare distribution equals one.
6 Properties of The ChiSquare Distribution 4. Chisquare distributions are positively skewed. Chisquare Distributions
7 Critical Values for χ There are two critical values for each level of confidence. The value χ R represents the righttail critical value The value χ L represents the lefttail critical value. 1 c c 1 c The area between the left and right critical values is c. L R χ
8 Example: Finding Critical Values for χ Find the critical values R and L for a 95% confidence interval when the sample size is 18. Solution: d.f. = n 1 = 18 1 = 17 d.f. Each area in the table represents the region under the chisquare curve to the right of the critical value. Area to the right of χ R = Area to the right of χ L = 1 c c
9 Solution: Finding Critical Values for χ Table 6: χ Distribution L R % of the area under the curve lies between and
10 Confidence Intervals for σ and σ Confidence Interval for σ : ( n 1) s ( n 1) s σ R L Confidence Interval for σ: ( n 1) s ( n 1) s σ R L The probability that the confidence intervals contain σ or σ is c.
11 Confidence Intervals for σ and σ In Words 1. Verify that the population has a normal distribution.. Identify the sample statistic n and the degrees of freedom. 3. Find the point estimate s. 4. Find the critical values χ R and χ L that correspond to the given level of confidence c. In Symbols s d.f. = n 1 ( xx) n 1 Use Table 6 in Appendix B.
12 Confidence Intervals for and In Words 5. Find the left and right endpoints and form the confidence interval for the population variance. 6. Find the confidence interval for the population standard deviation by taking the square root of each endpoint. In Symbols σ R L ( n 1) s ( n 1) s ( n 1) s ( n 1) s σ R L
13 Example: Constructing a Confidence Interval You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.0 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. Solution: d.f. = n 1 = 30 1 = 9 d.f.
14 Solution: Constructing a Confidence Interval Area to the right of χ R = 1 c Area to the right of χ L = 1 c The critical values are χ R = and χ L = 13.11
15 Solution: Constructing a Confidence Interval Confidence Interval for σ : Left endpoint: ( n1) s R (30 1)(1.0) Right endpoint: ( n1) s L (30 1)(1.0) < σ < 3.18 With 99% confidence, you can say that the population variance is between 0.80 and 3.18.
16 Solution: Constructing a Confidence Interval Confidence Interval for σ : ( n 1) s ( n 1) s σ R L (30 1)(1.0) (30 1)(1.0) < σ < 1.78 With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams.
17 Section 6.4 Summary Interpreted the chisquare distribution and used a chisquare distribution table Used the chisquare distribution to construct a confidence interval for the variance and standard deviation
18 Section 7.5 Objectives Find critical values for a χ test Use the χ test to test a variance or a standard deviation
19 Finding Critical Values for the χ Test a. Specify the level of significance α. b. Determine the degrees of freedom d.f. = n 1. c. The critical values for the χ distribution are found in Table 6 in Appendix B. To find the critical value(s) for a a. righttailed test, use the value that corresponds to d.f. and α. b. lefttailed test, use the value that corresponds to d.f. and 1 α. c. twotailed test, use the values that corresponds to d.f. and ½α, and d.f. and 1 ½α.
20 Finding Critical Values for the χ Test Righttailed Lefttailed Twotailed L R
21 Example: Finding Critical Values for χ Find the critical χ value for a lefttailed test when n = 11 and α = Solution: Degrees of freedom: n 1 = 11 1 = 10 d.f. The area to the right of the critical value is 1 α = = χ 0 =.558 From Table 6, the critical value is
22 Example: Finding Critical Values for χ Find the critical χ value for a twotailed test when n = 9 and α = Solution: Degrees of freedom: n 1 = 9 1 = 8 d.f. The areas to the right of the critical values are From Table 6, the critical values are L.180 and R
23 The ChiSquare Test χ Test for a Variance or Standard Deviation A statistical test for a population variance or standard deviation. Can be used when the population is normal. The test statistic is s. The standardized test statistic ( n1) s follows a chisquare distribution with degrees of freedom d.f. = n 1.
24 Using the χ Test for a Variance or Standard Deviation In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.. Specify the level of significance. 3. Determine the degrees of freedom. 4. Determine the critical value(s). In Symbols State H 0 and H a. Identify α. d.f. = n 1 Use Table 6 in Appendix B.
25 Using the χ Test for a Variance or Standard Deviation In Words 5. Determine the rejection region(s). 6. Find the standardized test statistic and sketch the sampling distribution. In Symbols ( n1) s 7. Make a decision to reject or fail to reject the null hypothesis. 8. Interpret the decision in the context of the original claim. If χ is in the rejection region, reject H 0. Otherwise, fail to reject H 0.
26 Example: Hypothesis Test for the Population Variance A dairy processing company claims that the variance of the amount of fat in the whole milk processed by the company is no more than 0.5. You suspect this is wrong and find that a random sample of 41 milk containers has a variance of 0.7. At α = 0.05, is there enough evidence to reject the company s claim? Assume the population is normally distributed.
27 Solution: Hypothesis Test for the Population Variance H 0 : σ 0.5 (Claim) H a : σ > 0.5 α = 0.05 df = 41 1 = 40 Rejection Region: Test Statistic: ( n1) s (411)(0.7) Decision: Fail to Reject H 0. At the 5% level of significance, there is not enough evidence to reject the company s claim that the variance of the amount of fat in the whole milk is no more than 0.5.
28 Example: Hypothesis Test for the Standard Deviation A company claims that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes. A random sample of 5 incoming telephone calls has a standard deviation of 1.1 minutes. At α = 0.10, is there enough evidence to support the company s claim? Assume the population is normally distributed.
29 Solution: Hypothesis Test for the Standard Deviation H 0 : H a : α = σ 1.4 min. σ < 1.4 min. (Claim) 0.10 Test Statistic: ( n1) s (5 1)(1.1) 1.4 df = 5 1 = Rejection Region: Decision: Reject H 0. At the 10% level of significance, there is enough evidence to support the claim that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes.
30 Example: Hypothesis Test for the Population Variance A sporting goods manufacturer claims that the variance of the strengths of a certain fishing line is A random sample of 15 fishing line spools has a variance of 1.8. At α = 0.05, is there enough evidence to reject the manufacturer s claim? Assume the population is normally distributed.
31 Solution: Hypothesis Test for the Population Variance H 0 : σ = 15.9 (Claim) H a : σ 15.9 α = 0.05 df = 15 1 = 14 Rejection Region: L R Test Statistic: (n 1)s (15 1)(1.8) 15.9 Decision: Fail to Reject H 0 At the 5% level of significance, there is not enough evidence to reject the claim that the variance in the strengths of the fishing line is 15.9.
32 Section 7.5 Summary Found critical values for a χ test Used the χ test to test a variance or a standard deviation
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