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Section 8.5-1 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Lecture Slides Elementary Statistics Twelfth Edition and the Triola Statistics Series by Mario F. Triola

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Section 8.5-2 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Chapter 8 Hypothesis Testing 8-1 Review and Preview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim about a Proportion 8-4 Testing a Claim About a Mean 8-5 Testing a Claim About a Standard Deviation or Variance

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Section 8.5-3 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Key Concept This section introduces methods for testing a claim made about a population standard deviation σ or population variance σ 2. The methods of this section use the chi-square distribution that was first introduced in Section 7-4.

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Section 8.5-4 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Requirements for Testing Claims About σ or σ 2 = sample size = sample standard deviation = sample variance = claimed value of the population standard deviation = claimed value of the population variance

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Section 8.5-5 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Requirements 1. The sample is a simple random sample. 2. The population has a normal distribution. (This is a much stricter requirement than the requirement of a normal distribution when testing claims about means.)

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Section 8.5-6 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Chi-Square Distribution Test Statistic

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Section 8.5-7 Copyright © 2014, 2012, 2010 Pearson Education, Inc. P-Values and Critical Values for Chi-Square Distribution P-values: Use technology or Table A-4. Critical Values: Use Table A-4. In either case, the degrees of freedom = n –1.

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Section 8.5-8 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Caution The χ 2 test of this section is not robust against a departure from normality, meaning that the test does not work well if the population has a distribution that is far from normal. The condition of a normally distributed population is therefore a much stricter requirement in this section than it was in Section 8-4.

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Section 8.5-9 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Properties of Chi-Square Distribution All values of χ 2 are nonnegative, and the distribution is not symmetric (see the Figure on the next slide). There is a different distribution for each number of degrees of freedom. The critical values are found in Table A-4 using n – 1 degrees of freedom.

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Section 8.5-10 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Properties of Chi-Square Distribution Properties of the Chi-Square Distribution Different distribution for each number of df. Chi-Square Distribution for 10 and 20 df

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Section 8.5-11 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example Listed below are the heights (inches) for a simple random sample of ten supermodels. Consider the claim that supermodels have heights that have much less variation than the heights of women in the general population. We will use a 0.01 significance level to test the claim that supermodels have heights with a standard deviation that is less than 2.6 inches. Summary Statistics: 707169.2568.569707170 69.5

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Section 8.5-12 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Requirement Check: 1. The sample is a simple random sample. 2. We check for normality, which seems reasonable based on the normal quantile plot.

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Section 8.5-13 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Step 1: The claim that “the standard deviation is less than 2.6 inches” is expressed as σ < 2.6 inches. Step 2: If the original claim is false, then σ ≥ 2.6 inches. Step 3: The hypotheses are:

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Section 8.5-14 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Step 4: The significance level is α = 0.01. Step 5: Because the claim is made about σ, we use the chi-square distribution.

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Section 8.5-15 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Step 6: The test statistic is calculated as follows: with 9 degrees of freedom.

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Section 8.5-16 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Step 6: The critical value of χ 2 = 2.088 is found from Table A-4, and it corresponds to 9 degrees of freedom and an “area to the right” of 0.99.

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Section 8.5-17 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Step 7: Because the test statistic is in the critical region, we reject the null hypothesis. There is sufficient evidence to support the claim that supermodels have heights with a standard deviation that is less than 2.6 inches. Heights of supermodels have much less variation than heights of women in the general population.

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Section 8.5-18 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued P-Value Method: P-values are generally found using technology, but Table A-4 can be used if technology is not available. Using a TI-83/84 Plus, the P-value is 0.0002897.

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Section 8.5-19 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued P-Value Method: Since the P-value = 0.0002897, we can reject the null hypothesis (it is under the 0.01 significance level). We reach the same exact conclusion as before regarding the variation in the heights of supermodels as compared to the heights of women from the general population.

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Section 8.5-20 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Confidence Interval Method: Since the hypothesis test is left-tailed using a 0.01 level of significance, we can run the test by constructing an interval with 98% confidence. Using the methods of Section 7-4, and the critical values found in Table A-4, we can construct the following interval:

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Section 8.5-21 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Example - Continued Based on this interval, we can support the claim that σ < 2.6 inches, reaching the same conclusion as using the P-value method and the critical value method.

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