Question 1170481: The treasurer of a municipality claims that the average net worth of families living in this municipality is P590,000. A random sample of 50 families selected from this area produced a mean net worth of P720,000 with a standard deviation of P65,000. Using 1% significance level, can we conclude that the claim is true? Also, find the 99% confidence interval of the true mean.
Hypothesis:
H_0:
H_1:
The level of significance is a = _____, and the critical region is z = ____.
Compute for the value of one sample z-test.
Decision rule:
Conclusion:
Compute for the confidence interval.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho: NW=590K
HA: NW NE 590K
Level of significance is 0.01
z=2.576 critical value (+/-); reject if |z|>2.576
z=(x-mean-/sigma/sqrt(50)
=130*sqrt(50)/65=14.14, highly significant.
99% Ci: half interval is 2.576*65/sqrt(50)=23.68
the CI is (696.32, 743.68). It does not contain 590 K so again highly unlikely the true mean its 590.
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