document.write( "Question 1196688: ou want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a simple random sample of 32 apartments advertised in the local newspaper is $1 400. Assume that the standard deviation is known to be $220.\r
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document.write( "Find a 99% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community. \n" );
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Algebra.Com's Answer #829662 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "At 99% confidence, the z critical value is roughly z = 2.576 \n" ); document.write( "Use a table like this \n" ); document.write( "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf \n" ); document.write( "to get that value. Look at the bottom row labeled \"Z\" and above the 99% confidence level.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We use a Z distribution instead of a T distribution since \n" ); document.write( "a) sigma is given \n" ); document.write( "b) n > 30\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Given info: \n" ); document.write( "n = 32 = sample size \n" ); document.write( "xbar = 1400 = sample mean \n" ); document.write( "sigma = 220 = population standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Side note: We often won't know what sigma is since it's a population parameter. The goal of statistics is to estimate parameter. Use the sample standard deviation (variable 's') to estimate sigma, if sigma is unknown.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "E = margin of error \n" ); document.write( "E = z*sigma/sqrt(n) \n" ); document.write( "E = 2.576*220/sqrt(32) \n" ); document.write( "E = 100.18288875851\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "L = lower boundary of confidence interval \n" ); document.write( "L = xbar - E \n" ); document.write( "L = 1400 - 100.18288875851 \n" ); document.write( "L = 1299.8171112415 \n" ); document.write( "L = 1299.82\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "U = upper boundary of confidence interval \n" ); document.write( "U = xbar + E \n" ); document.write( "U = 1400 + 100.18288875851 \n" ); document.write( "U = 1500.1828887585 \n" ); document.write( "U = 1500.18\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: (1299.82, 1500.18) which is in the format (L, U) \n" ); document.write( "This is equivalent to saying 1299.82 < mu < 1500.18 in the format L < mu < U. \n" ); document.write( "Your answers may slightly vary depending on how you round the decimal values. \n" ); document.write( " \n" ); document.write( " |