of the column at the right area with the row at df Keep in mind that \u03b1 1 The

Of the column at the right area with the row at df

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of the column at the right area with the row atdf.Keep in mind thatα=1-The confidence level.A. Bourhim (Syracuse University)Review-Test4MAT121, Spring 201517 / 21
FormulaIf the requirements are satisfied, then aconfidence interval (CI)for thepopulation varianceσ2is given by(n-1)s2χ2R<σ2<(n-1)s2χ2L,wherenis the sample size.sis the sample standard deviation.χ2Randχ2Lare the critical values.Note that aconfidence interval (CI)forσis given byvuut(n-1)s2χ2R<σ<vuut(n-1)s2χ2L.A. Bourhim (Syracuse University)Review-Test4MAT121, Spring 201518 / 21
15. Construct a 95% confidenceinterval estimate of the variance of the population.
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ExampleIf the nicotine content of cigarettes has a normal distribution, find the 95%confidence interval for the variance and standard deviation of the nicotinecontent of cigarettes manufactured if a simple random sample of 20 cigaretteshas a standard deviation of 1.6 milligrams.Requirements1The sample is a simple random sample (From the text).2The population must have normally distributed values (From the text).Thus the requirements are satisfied. From the text, we also have1The sample size isn=20.2The sample standard deviation is 1.6.So, we only needχ2Landχ2Rfrom the table A4 of Chi-square distribution. Wehaveα=1-95%=5%=.05 andα2=.052=.025.On the other hand, df=n-1=20-1=19.A. Bourhim (Syracuse University)Review-Test4MAT121, Spring 201520 / 21
Example (Continued)Therightcritical valueχ2Rhas area to therightofα2=.025 with df=19. ByTable A4, row at df=19 with column at .025 givesχ2R=32.852.Theleftcritical valueχ2Lhas area to theleftofα2=.025 and so is its rightarea is 1-.025=.975. By Table A4, row at df=19 with column at .975χ2L=8.907.The interval forσ2is(n-1)s2χ2R=(20-1)1.6232.852=1.5.(n-1)s2χ2R=(20-1)1.628.907=5.5.Thus1.5<σ2<5.5.The interval forσisp1.

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