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 ```A supervisor of White Industries is considering a new method of assembling its cart product. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, with a standard deviation of 2.7 minutes. Using the 0.05 level of significance, can we conclude that the assembly time using the new method is faster?``` Request for Question Clarification by scriptor-ga on 25 Aug 2005 15:59 PDT ```Google Answers discourages and may remove questions that are homework or exam assignments. Scriptor```
 ```Lola5 ? What we?re seeking here is the confidence interval with p = 0.05. Restated, it?s a 95% confidence that the same fall within the range. It's calculated as: mean +/- T-value * SD BMJ.com ?Statements of probability and confidence intervals? (undated) http://bmj.bmjjournals.com/collections/statsbk/4.shtml The small sample size here calls for the use of the Student?s T-distribution, which narrows the range of values depending on the sample size. It's important to remember that the degrees of freedom are one less than the sample size: Statsoft Electronic Textbook ?Student?s T Table? (2003) http://www.statsoft.com/textbook/sttable.html Thus at p < 0.05, the mean is 40.6 +/- (1.713872) * 2.7 So, anything in that range is statistically expected and we can?t reject the hypothesis H0 (also called the null hypothesis) ? that this new assembly technique is the same as the 42.3 minutes of the old technique. Only those values outside the range would support hypothesis H1 ? the alternate hypothesis -- Statistics Glossary ?Hypothesis Testing? http://www.cas.lancs.ac.uk/glossary_v1.1/hyptest.html So, values between 40.6 +/- 4.63 minutes or in the range of 35.97 ? 45.23 are indistinguishable from our sample. Of course 42.3 minutes is well within that range, so we can?t conclude that we?re doing any better. Google search strategy: Probablility + ?confidence interval? T distribution Best regards, Omnivorous-GA```