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Chapter 8 Recitation Problems

# Chapter 8 Recitation Problems - Solutions for Recitation...

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Solutions for Recitation Problems from Chapter 8 8.9. Given: xbar = 3.37 σ = 0.28 n = 120 1 – α = 0.95 Is σ known? Yes, so use the normal distribution. At 95% confidence, 1 – α = 0.95, α = 0.05 so α /2 = 0.025. From normal table, z α /2 = z 0.025 = 1.96. Then at 95% confidence, E = 1.96 × (0.28/√120) = 0.050 In general xbar – E μ ≤ xbar + E , so at 95% confidence 3.37 – 0.050 ≤ μ ≤ 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

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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: σ is known: use z Given: xbar = \$119,155 σ = \$30,000 n = 80 a. What is the point estimate for mean annual household income? μ ~ xbar = \$119,155 b. What is the margin of error at 90% confidence? At 90% confidence, z = 1.645 Then E = z × σ /√ 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 μ ≤ xbar + E , with xbar given as \$119,155 and E = \$5517.50 from b, so \$119,155 – \$5517.50 ≤ μ ≤ \$119,155 + \$5517.50, so Chapter 8 Recitation Problems 2 © 2009 Harvey A. Singer
\$113,637.50 ≤ μ ≤ \$124,672.50 In a sketch d.

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