Answer to Question #185693 in Statistics and Probability for kevin

Question

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.

 

Answer 1

H₀:p=0.686

H₁:p≠0.686 (claim)

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

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

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

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

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%.