The cost of weddings in the United States has skyrocketed in recent years. As a

Question

The cost of weddings in the United States has skyrocketed in recent years. As a result many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than $10,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings. At the .05 significance level is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised?

Explanation / Answer

assuming you have a large enough sample such that the central limit theorem holds and the mean is normally distributed then to test the null hypothesis H0: µ = ?? find the test statistic z = (xBar - ?) / (sx / sqrt(n)) where xbar is the sample average sx is the sample standard deviation n is the sample size The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis. H1: µ > ?; p-value is the area to the right of z H1: µ < ?; p-value is the area to the left of z H1: µ ? ?; p-value is the area in the tails greater than |z| for a small sample test for the mean every thing is the same save the test statistic is a t statistic with n - 1 degrees of freedom. However, in this case the underlying distribution must be normal for this test to be valid. The sample mean is 9.78750 the sample standard deviation is 0.91720 Here we have a small sample and will use the student t statistic. The null and alternative hypothesis are: H0: µ = 10 vs. H1 µ < 10 the test statistic is: t = (xbar - 10) / (s / sqrt(8)) = -0.66 the p-value is P(t_7 < -0.66) = 0.267 With such a large p-value we fail to reject the null hypothesis. There is insufficient evidence to support the claim the average price is less than $10,000

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