document.write( "Question 437869: A study was done to determine the average number of homes that a homeowner owns in his or her lifetime. For the 50 homeowners surveyed, the sample average was 3.8 amd the sample standard deviation was 1.5. calculate the 95% confidence interval for the true average number of homes that a person owns in his or her lifetime. \r
\n" ); document.write( "\n" ); document.write( "Am i on the right track- > n= 50 standard dev = 1.5 Xbar= 3.8\r
\n" ); document.write( "\n" ); document.write( "My computation- (E= Z a/2 * st dev / sq rt of n)\r
\n" ); document.write( "\n" ); document.write( "1.67 (z score) * 1.5/ sq rt 50 = 0.354260497..... lost from here = (
\n" ); document.write( "Thanks for ANY help \n" ); document.write( "

Algebra.Com's Answer #302893 by stanbon(75887) $\"\"$ $\"About$

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A study was done to determine the average number of homes that a homeowner owns in his or her lifetime. For the 50 homeowners surveyed, the sample average was 3.8 amd the sample standard deviation was 1.5. calculate the 95% confidence interval for the true average number of homes that a person owns in his or her lifetime.
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\n" ); document.write( "x-bar = 3.8
\n" ); document.write( "ME = zs/sqrt(n) = 1.96 * [1.5/sqrt(50)] = 0.4158
\n" ); document.write( "----
\n" ); document.write( "95% CI: 3.8-0.4158 < u < 3.8+0.4158
\n" ); document.write( "95% CI: 3.3842 < u < 4.2158
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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