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Unformatted text preview: Solutions for Recitation Problems from Chapter 8 8.9. Given: xbar = 3.37 = 0.28 n = 120 1 = 0.95 Is known? Yes, so use the normal distribution. At 95% confidence, 1 = 0.95, = 0.05 so /2 = 0.025. From normal table, z /2 = z 0.025 = 1.96. Then at 95% confidence, E = 1.96 (0.28/120) = 0.050 In general xbar E xbar + E , so at 95% confidence 3.37 0.050 3.37 + 0.050 3.37 1.96 (0.28/120) 3.37 0.05 or 3.32 to 3.42 Chapter 8 Recitation Problems 1 2009 Harvey A. Singer 8.10 ALTERNATIVE LANGUAGE TO TEXTBOOK. A magazine has reported that, on the basis of a random sample of 80 households, the mean annual household income of its readers is $119,155. It is known from previous research that the standard deviation of mean annual income of all households subscribing to the magazine is $30,000. REMARKS: The objective after reading through the parts below is to confidently estimate the true mean annual household income at the 95% confidence level. Note the keywords known and all (referring to the population of households.) As a result, sigma is known: is known: use z Given: xbar = $119,155 = $30,000 n = 80 a. What is the point estimate for mean annual household income? ~ xbar = $119,155 b. What is the margin of error at 90% confidence? At 90% confidence, z = 1.645 Then E = z / n = 1.645$30,000/80 = $49350/80, so E = $5517.50 c. Calculate a 90 percent confidence interval for the actual mean household income. xbar E xbar + E , with xbar given as $119,155 and E = $5517.50 from b, so $119,155 $5517.50 $119,155 + $5517.50, so Chapter 8 Recitation Problems 2 2009 Harvey A. Singer $113,637.50 $113,637....
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This note was uploaded on 01/26/2011 for the course OM 210 taught by Professor Singer during the Fall '08 term at George Mason.
 Fall '08
 SINGER

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