document.write( "Question 771919: Question 1:
\n" ); document.write( "From past record, 75% of the retirees stated that they preferred living in an apartment to living in a one-family home. A random sample of 30 retirees was taken, and they are asked whether they prefer living in an apartment. 21 of these retirees are responded yes. Construct a 99% confidence interval for the proportion of all retirees who prefer living in an apartment.
\n" ); document.write( "Question 2:
\n" ); document.write( "A survey is being planned to determine the mean amount of time corporation executives watch TV. A survey indicated that the mean time per week is 12 hours with a standard deviation of 3 hours. It is desired to estimate the mean viewing time within one-quarter hour. The 95% level of confidence is to be used. How many executives should be surveyed?
\n" ); document.write( "Question 3:
\n" ); document.write( "A population is estimated to have a standard deviation of 10. We want to estimate the population mean within 2, with a 95% level of confidence. How large a sample is required?
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Algebra.Com's Answer #470543 by oscargut(2103)\"\" \"About 
You can put this solution on YOUR website!
Hi,\r
\n" ); document.write( "\n" ); document.write( "I do question 1) here\r
\n" ); document.write( "\n" ); document.write( "CI at 99% is:\r
\n" ); document.write( "\n" ); document.write( "21/30 +- z(0.005)sqrt((21/30(1-21/30))/30)\r
\n" ); document.write( "\n" ); document.write( "= (0.484,0.916)\r
\n" ); document.write( "\n" ); document.write( "I can help you with more questions at: mthman@gmail.com\r
\n" ); document.write( "\n" ); document.write( "Thanks
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