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HW 9 Chap 8 Exemplars

# HW 9 Chap 8 Exemplars - Exemplars for the Estimation...

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Exemplars for the Estimation Problems Required for Homework #10 Interval estimation with sigma known (from section 8.1). As a result of having σ , can use the normal distribution to find the z values z α /2 . 8.5. Given: σ = \$5.00 n = 49 Is σ known? Yes, so use the normal distribution for z . a. Given: 1 – α = 0.95 At 1 – α = 0.95, z = 1.96. Then: E = 1.96 / 1.96(5/ 49) 1.40 n σ = = b. Given: xbar = \$24.80 1 – α = 0.95 Already know from a that E = \$1.40. Then at 95% confidence, \$24.80 ± \$1.40 or \$23.40 to \$26.20 New question: How a big a sample is needed to reduce the margin of error to \$1.00 at 95% confidence? In other words for what n is E = \$1.00? In n = ( z σ / E ) 2 substitute z = 1.96, σ = 5.00, and E = 1.00, so n = (1.96 ×5.00/1.00) 2 = 9.8 2 = 96.04 Rounding up to the next integer, n = 97. HW #10 estimation exemplars 1 © 2010 Harvey Singer

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8.6. Given: xbar = 8.5 hours per week σ = 3.5 hours per week n = 300 1 – α = 0.95 Is σ known? Yes, so use the normal distribution for z = 1.96 at 1 – α = 0.95. In general at 95 % confidence, .025 ( / ) x z n σ ± with z = 1.96 at 1 – α = 0.95. Then 8.5 ± 1.96(3.5/ 300 ) so that 8.5 ± .4 or 8.1 to 8.9 8.6 ALTERNATIVE LANGUAGE TO TEXTBOOK. A major research organization has studied the weekly viewing time during nightly prime time hours (8:00 to 11:00 PM). They reported that, on the basis of a random sample of 300 households, the mean viewing time is 8.5 hours per week. It is known from previous research that the standard deviation of mean weekly viewing time of all households is 3.5 hours. 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 = 8.5 hours/week σ = 3.5 hours/week n = 300 a. What is the point estimate for mean annual household income? μ ~ xbar = 8.5 hours/week b. What is the margin of error at 95% confidence? HW #10 estimation exemplars 2 © 2010 Harvey Singer
At 95% confidence, z = 1.960 Then E = z × σ /√ n = 1.960×3.5/√300 E = 0.396 hours/week c. Calculate a 95 percent confidence interval for the actual mean household income. xbar – E μ ≤ xbar + E , with xbar given as 8.5 and E = 0.396 from b, so 8.5 – 0.396 ≤ μ ≤ 8.5 + 0.396, so 8.104 ≤ μ ≤ 8.896 hours/week d. What is the chance that the true mean annual household income is contained within the interval calculated in part b? 95% e. What is the chance that the true mean annual household income is not contained within the interval calculated in part b? 5% f. Based on this confidence interval, state a likely (and reasonable) upper bound for the mean annual household income. Upper confidence bound is 8.896 hours/week at 95% confidence g. What is the chance that the true mean annual household income is more than the upper bound stated in part f?

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