Question 523155
Of 150 adults who tried a new rum-flavored peppermint patty, 87 rated it excellent.:: pa-hat = 87/150 = 0.580 
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Of 200 children sampled, 123 rated it excellent.:: pc=hat = 123/200 = 0.615
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Using the .10 level of significance, can we conclude that there is a significant difference in the proportion of adults and the proportion of children who rate the new flaŽvor excellent?
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(a) State the null hypothesis and the alternate hypothesis.
Ho: pa-pc = 0
Ha: pa-pc is not 0
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(b) What is the probability of a Type I error?::P(Type I) = alpha = 10%
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(c) Is this a one-tailed or a two-tailed test?:::: 2-tail
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(d) What is the decision rule?:: Reject Ho is test stat <-1.645 or >+1.645
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(e) What is the value of the test statistic?:::
ts = (0.580-0.615)/sqrt[(.58*0.42/150)+(0.615*0.385/200)] = -0.6605
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(f) What is your decision regarding the null hypothesis?
Since the ts is not in a reject interval, fail to reject Ho.
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(g) What is the p-value? Explain what it means in terms of this problem.
p-valu = 2*P(z < -0.6605) = 2*0.25 = 0.50
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There is a 50% probability that the test results could have provided
stronger evidence for rejecting Ho.
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Cheers,
Stan H.
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