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Problem Identification: Hypothesis Test of the Variance using the Chi-Square Distribution Fill in the Yellow areas. The green areas will be automatically updated based on your inputs. 1. The population parameter of interest is the mean (describe the application) Describe the parameter here 2. The Null Hypothesis makes this claim about the population Variance: σ2 (sigma, <= 0.0004 units squared) ↑↑↑ ↑↑↑ ↑units of measure (put <= or = or >= in this box) (put a number value in this box) (use '= to get Excel to accept =) ok ok 3. The Alternative Hypothesis makes this claim about the population Mean: σ2 > 0.0004 units 5. We are using the Chi-Square Distribution We are assuming that the population is generally normally distributed. 6. We choose the level of significance for this test to be (alpha) α = 0.05 ok This is a one-tailed test. on the right side The Alternative Hypothesis occupies probability of 0.0500 (5%) in the right tail 7. We have collected sample data from the population: The sample variance is: s2 = 0.0007 and the sample size is: n = 28 therefore the degrees of freedom is: df = n - 1 = 27 ok All Inputs Ok? = TRUE ok 8. The calculated value of the Test Statistic is χ2 = 47.2500 =(df*SampleVariance)/(NullHypothesisValue) Area to the left of this is: 0.990701 (n 1) s 2 Area to the right of this is: 0.009299 2* 2 9. p-Value approach: The probability that the Test Statistic is this value, or more extremely, TestStat > 47.2500 , is pValue: 0.009299 <= 0.05 and this pValue is LESS THAN α or 9. Critical Value approach The Critical Value is =CHIINV(alpha or alpha/2, df) 40.1133 The Test Statistic, 47.2500 is INSIDE the Critical Region. Excel formulas for critical values: =CHIINV(IF(Tailed="one-tailed",alpha,alpha/2),df) p-Value formula: =IF(Tailed="two-tailed",CHIINV(1-alpha/2,df),"") =IF(Tail="left",1-CHIDIST(TestStatistic,df),CHIDIST(TestStatistic,df)) 10. Conclusion: REJECT There IS sufficient evidence at the α = 0.05 level of significance to reject the null hypothesis. Problem Identification: Hypothesis Test of the Variance using the Chi-Square Distribution Fill in the Yellow areas. The green areas will be automatically updated based on your inputs. 1. The population parameter of interest is the mean (describe the application) Describe the parameter here 2. The Null Hypothesis makes this claim about the population Variance: σ2 (sigma, squared) ↑↑↑ ↑↑↑ ↑units of measure (put <= or = or >= in this box) (put a number value in this box) 3. The Alternative Hypothesis makes this claim about the population Mean: σ2 5. We are using the Chi-Square Distribution We are assuming that the population is generally normally distributed. 6. We choose the level of significance for this test to be (alpha) α = This is a test. The Alternative Hypothesis occupies probability of 7. We have collected sample data from the population: The sample variance is: 2 s = and the sample size is: n = therefore the degrees of freedom is: df = n - 1 = 8. The calculated value of the Test Statistic is 2 χ = Area to the left of this is: Area to the right of this is: 9. p-Value approach: The probability that the Test Statistic is this value, or more extremely, TestStat , is pValue: and this pValue is 10. Conclusion: 2* (n 1) s 2 2 or 9. Critical Value approach The Critical Value is =CHIINV(alpha or alpha/2, df) The Test Statistic, is There seems to be no built-in Excel function. Use formulas like the "Automatic" sheet to accomplish this kind of a test. There is the Analysis add-in's "F-Test Two-Sample for Variances". There seems to be no built-in STATS > TEST feature for this kind of a problem. ≠