Question #185693

A recent survey found that 68.6% of people own their homes. In a random sample of 150 heads of household, 92 responded that they own their homes. At a=0.01, does that suggest a difference for the proportion? Find the CV and compute the test value (TV) , and state your conclusion.

 


1
Expert's answer
2021-05-07T09:28:28-0400

H0:p=0.686

H1:p≠0.686 (claim)


p=0.686,n=150,p^=92/150=0.613p=0.686, n=150, \hat{p}=92 /150=0.613


CV at α=0.01 is equal to 2.58CV \text{ at } \alpha=0.01 \text{ is equal to }2.58


z=(p^p)p(1p)Nz=\dfrac{(\hat{p}-p)}{\sqrt{\frac{p(1-p)}{N}}} =(0.6130.686)0.001436=1.918=\dfrac{(0.613-0.686)}{\sqrt{0.001436}} ​ =-1.918


p-value=(2)P(z<1.918)=(2)(0.0276)=0.0552( two-tailed test )= (2)P(z < -1.918) = (2)(0.0276) = 0.0552 (\text{ two-tailed test })


Since p-value > 0.01, do NOT reject the null hypothesis. There is significant evidence that the mean proportion of homeowners is not different from 68.6%.


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