document.write( "Question 868431: Although male polar bears weigh only about one pound at birth, the mean weight of a random sample of 10 adult male polar bears is 1200 pounds with a standard deviation of 100 pounds. Construct and interpret a 98% confidence interval for the mean weight of all adult male polar bears. Assume that the population is evenly distributed. \n" ); document.write( "
Algebra.Com's Answer #523872 by jim_thompson5910(35256)\"\" \"About 
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Use a table to find that the critical value is \"t=2.821\". The degrees of freedom are \"df+=+n-1+=+10-1+=+9\". Look in the row that starts with \"9\" and look above the 98%. The value you'll see in this spot is \"2.821\"\r
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\n" ); document.write( "\n" ); document.write( "So we're given\r
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\n" ); document.write( "\n" ); document.write( "xbar = 1200 (sample mean)
\n" ); document.write( "t = 2.821 (see above)
\n" ); document.write( "s = 100 (sample standard deviation)
\n" ); document.write( "n = 10 (sample size)\r
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\n" ); document.write( "\n" ); document.write( "Now compute the lower bound (L) and the upper bound (U) of the confidence interval.\r
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\n" ); document.write( "\n" ); document.write( "Lower Bound:\r
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\n" ); document.write( "\n" ); document.write( "L = xbar - t*s/sqrt(n)
\n" ); document.write( "L = 1200 - 2.821*100/sqrt(10)
\n" ); document.write( "L = 1,110.79214720666\r
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\n" ); document.write( "\n" ); document.write( "Upper Bound:\r
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\n" ); document.write( "\n" ); document.write( "U = xbar + t*s/sqrt(n)
\n" ); document.write( "U = 1200 + 2.821*100/sqrt(10)
\n" ); document.write( "U = 1,289.20785279334\r
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\n" ); document.write( "\n" ); document.write( "The 98% confidence interval is approximately (1110.79214720666, 1289.20785279334)\r
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\n" ); document.write( "\n" ); document.write( "So if we construct 100 confidence intervals, then about 98 of them will contain the true population mean.
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