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Statistics Ch 7 Quiz 3 of 4

Statistics Ch 7 Quiz 3 of 4 - size of the sample is not a...

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Unformatted text preview: size of the sample is not a matter at issue here since all requirements for use of the Student t have been met. 3. This real example is very similar to one that was performed in in-patient pharmacy at Tripler several years ago. So, here goes: The Medassist Pharmaceutical Company uses a machine to pour cold medicine into bottles in such a way that the standard deviation of the weights is 0.15 oz. A new machine is tested on '1’] bottles, and the standard deviation for this sample is 0.12 oz. The Dayton Machine Company, which manufactures the new machine, claims that it fills bottles with lower variation. At the 0.05 significance level, test the claim made by the Dayton Machine Company. If Dayton’s machine is being used on a trial basis, should its purchase be considered? Solution: Here we want to test variances, so we must use the Chi-squared statistic and the Chi-squared distribution. a. The claim is that the Dayton machine has a fill variance less than 0.15, so the counter claim is that the Dayton machine has a fill variance of greater than or equal to 0.15. The null hypothesis is then H0 :0“ = 0.15. The alternative hypothesis comes in this case from the original claim (the other has the equality) so HI :0' < 0.15 . The hypothesis shows we need a left sided one tail test. The distribution for variances is the Chi-squared distribution. The dof is one less than the sample size or 70. 2 2 The Chi square statistic is given by 12 = ("—1215— = (19(0—122) or (0.15) For 70 degrees of freedom, and using the Chi-Square table for a 5% level of significance we have a bit of a problem- While the normal and the Student t are symmetric, the Chi-square is not, and the table values are read from the right, not the left. So since we want a left handed test, we much leave 95% of the area to the right of the critical value. We look up the value, then for 95%, not 5%: This critical value is 51.739. Now since 44.8 is to the left of 5 1 .739, our test statistic is in the rejection area. Hence we reject the null and accept the alternative: they do better than a standard deviation of less than 0.15. So this supports the procurement of the new machine. = 44.3 ...
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