SOLUTION: In a sample of 1000 home​ buyers, you find that 459 home buyers found their real estate agent through a friend. At αe=0.08​, can you reject the claim that 46​
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Question 1076466: In a sample of 1000 home buyers, you find that 459 home buyers found their real estate agent through a friend. At αe=0.08, can you reject the claim that 46% of home buyers find their real estate agent through a friend?
what is the standardized test statistic with steps in getting the answer
Thank you! Answer by Theo(13342) (Show Source):
459 of those 1000 found their real estate agent through a friend.
alpha = .08
claim is that 46% of home buyers find their real estate agent through a friend.
46% is equal to .46.
your null hypothesis is that the probability that prospective home buyers find their real estate agent through a friend is .46
your alternate hypothesis is that the probability that prospective h ome buyers find their real estate agent through a friend is something different than .46
this indicates a two tailed alpha since the claim is that the real percent could be more than or less than 46%.
since alpha is .08, then you would divide it by 2 to get alpha/2 = .04
the critical z-score for an alpha of .04 would be found by looking in the z-score table or using a z-score calculator to find a z-score that has an area of .96 to the left of it.
i used the z-score calculator by statrek to find that the critical z-score would be plus or minus 1.751 which i then rounded to 1.75.
using my ti-84 plus calculator, the critical z-score was found to be plus or minus 1.750686071.
you may round as your instructor requires.
my answer is based on the critical z-score of plus or minus 1.75.
your sample p is equal to .459
your sample q is equal to 1 - p which makes it equal to .541.
your sample standard error is equal to sqrt((p*q/n) = sqrt(.459*.541/1000) which is equal to .01567581408 which i rounded to .0156758.
your sample z-score would be equal to (x-m)/s which translates to (.459 - .46) / .0156758 which is equal to -.0637925975 which i rounded to -.06.
your sample z-score of -.06 is less than your critical z-score of -1.75.
this indicates that the probability of a difference between your sample mean and the state mean is due largely to chance variations in sample means of samples with a sample size of 1000.
you therefore can't reject the null hypothesis.
that's my take on this problem based on the information you provided.
a reference for determining the mean and standard error of a proportion can be found here:
