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5 Step Hypothesis Variance - 2 For a sample of 25 items of...

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Describe a 5-step hypothesis test for two variances for a particular claim related to your work or  life environment.  You need to state the claim, define the null and alternate hypothesis, identify  the test significance and the test statistic, and state the decision rule that will be used.   The  decision rule may be based on critical values or on the p-value approach.  You do not need to  collect data and actually conduct the test calculations unless you wish to do so. Pitcher Granite, LLC recently ran quantitative statistical tests on its receivables to determine the  level of significance in variance. The goal was to ascertain the effect of increasing fuel costs on  setting price level for freight and sell price per unit: 1. The firm discovered that a sample of 25 items purchased on July 25,2011 has a mean of  $5.29 with a standard deviation of $3.02.  2. For a sample of 25 items of the same category purchased today the mean is $5.12 with a 
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Unformatted text preview: 2. For a sample of 25 items of the same category purchased today the mean is $5.12 with a standard deviation of $2.14. 3. The firm decided to ascertain if at = .05, is there a difference in α variance ? Pitcher Granite, LLC used the following parameters: a. Hypotheses: H : 12 = 22 versus H σ σ a : 12 ≠ 22 σ σ b. Level of Significance : = 5% = 0.05 α c. Decision Rule : Reject the null hypothesis if p-value < 0.05. d. Test characteristic: F = s12 ÷ s22 = 3.022 ÷ 2.142 = 1.99 e. For Degrees of Freedom(numerator) = Degrees of Freedom(denominator) = e.i. = 24 and corresponding to F-value = 1.99, the p-value = 0.049 f. Conclusion: Since 0.049 is just less than 0.05, we reject H . There is just about enough statistical evidence to conclude at 5% level of significance that there is a difference in the variances ....
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