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
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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- Spring '10