document.write( "Question 1063468: (5 points) For the following data set
\n" ); document.write( "136 141 137 145
\n" ); document.write( "131 135 138 136
\n" ); document.write( "130 132 136 134
\n" ); document.write( "137 133 141 139
\n" ); document.write( "Table 2: Data set Exercise 2
\n" ); document.write( "Construct a\r
\n" ); document.write( "\n" ); document.write( "1. 90% confidence interval for the population mean;
\n" ); document.write( "2. 95% confidence interval for the population mean;
\n" ); document.write( "3. 99% confidence interval for the population mean;
\n" ); document.write( "4. What assumption do you need to make about the population of interest to construct the confidence
\n" ); document.write( "intervals
\n" ); document.write( "

Algebra.Com's Answer #678667 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
The mean of x is 136.31 to two decimal places
\n" ); document.write( "s=3.96
\n" ); document.write( "The general form of a CI is
\n" ); document.write( "for ci of 1-alpha, x bar+/- t(df=15,alpha/2)s/sqrt (n)
\n" ); document.write( "so for 90%, where t=1.753, 136.31+/-1.753*(3.96/4)=136.31 +/-1.74 or (134.57, 138.05)
\n" ); document.write( "for 95%, where t=2.1314, 136.31+/- 2.1314*0.99=136.31+/-2.11 or (134.20, 138.42)
\n" ); document.write( "for 99%, where t=2.9467, 136.31+/- 2.9467*0.99=136.31+/-2.92 or (133.39, 139.23)
\n" ); document.write( "Assumptions: population is normally distributed and standard deviation of the sample is an unbiased estimator of the standard deviation of the population.
\n" ); document.write( "
\n" );