document.write( "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.\r
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\n" ); document.write( "The level of significance is a = _____, and the critical region is z = ____.
\n" ); document.write( "Compute for the value of one sample z-test.
\n" ); document.write( "Decision rule:
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\n" ); document.write( "Compute for the confidence interval.
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Algebra.Com's Answer #795576 by Boreal(15235)\"\" \"About 
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Ho: NW=590K
\n" ); document.write( "HA: NW NE 590K
\n" ); document.write( "Level of significance is 0.01
\n" ); document.write( "z=2.576 critical value (+/-); reject if |z|>2.576
\n" ); document.write( "z=(x-mean-/sigma/sqrt(50)
\n" ); document.write( "=130*sqrt(50)/65=14.14, highly significant.\r
\n" ); document.write( "\n" ); document.write( "99% Ci: half interval is 2.576*65/sqrt(50)=23.68
\n" ); document.write( "the CI is (696.32, 743.68). It does not contain 590 K so again highly unlikely the true mean its 590.\r
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