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Statistics Ch 7 Quiz 2 of 4

Statistics Ch 7 Quiz 2 of 4 - right hand tail of the...

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Unformatted text preview: right hand tail of the distribution which is the “reject the null” region. So we reject the null and accept the alternative that those polled were not representative of the actual population. So what happened here? Well, we have a couple of possibilities: one that the sample was bad. This is doubtful since we were told it was a random sample. A second possibility is that we are making an error in rejecting a true null, but this is, indeed, a small probability. We are left with the possibility that people identify themselves with a winner. So the population “changed” after the results of the election become known. 2. The Orange County Bureau of Weights and Measures were receiving complaints that the Windsor Bottling Company was cheating consumers by putting less than 12 oz of root beer in its 12 oz cans. When 24 cans were randomly selected and measured, the amounts were found to have a mean of l 1.4 oz per can and a standard deviation of 0.62 oz. The company president claims that the sample is too small to be meaningful. Use the sample data to test the claim that consumers are being cheated at a 0.05 significance level. Does the company president’s argument have any validity? Solution: Here again we have a binomial distribution which can be approximated by a normal. However we are concerned about population means and we do not know the population standard deviation, so the t test using the Student t distribution is the most appropriate. Be careful. In the problem we have the sample standard deviation and not the population standard deviation. So a 2 statistic and the normal distribution is ONLY appropriate if the population standard deviation is known. Another way of saying the same thing is that we can always get information from the null hypothesis. Here, however, the null concerns itself with the mean and not the standard deviation, so we can assume the population mean from the null hypothesis, we can say nothing about the standard deviation. The claim: ,1: <12 02. So the counter claim is that ,u 2 12. b. The null comes from the claim involving the equality, so the null is HD : ,u =12 , The alternate hypothesis comes from the claim which does 9‘ not involve equality, so H1 : ,u <12. c. The hypothesis shows us we need a one tailed left handed test. d. The degrees of freedom are one less than the sample size, so DOF=23. . . x— - 11.4wl2 —0.6 e. We now compute the t stat1st1c as r = 'u‘ = —— - — = —4.74 i 0.62 _ 0.127 J3 J2? f. From the table at a 5% level of significance and a dof of 23, we find that t- crit is -1.7l4. g. Comparing the two, we find that the observed sample is again in the red region, and thus suggests we should reject the null and accept the alternative, i.e. that indeed the company is cheating the consumer. The ...
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