Chapter 8 Recitation Problems - More

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

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Solutions for More Recitation Problems from Chapter 8 From the Supplementary Exercises 44. Based on a survey of 54 discount brokers, the mean price charged for a trade of 100 shares at \$50 per share is \$33.77. It is known that the standard deviation of all such trades by all discount brokers is \$15.00. REMARKS: The objective after reading through the parts below is to confidently estimate the true mean price per 100 share trade 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 = \$33.77 σ = \$15.00 n = 54 a. What is the margin of error at 95% confidence? At 95% confidence, z = 1.960 Then E = z × σ /√ n = 1.960×\$15 /√54 = \$29.40/√54, so E = \$ 4.00 b. Calculate a 95 percent confidence interval for the actual mean household income. xbar – E μ ≤ xbar + E , with xbar given as \$33.77 and E = \$4.00 at 95% confidence from a, so \$33.77 – \$4.00 ≤ μ ≤ \$33.77 + \$4.00, so \$27.77 ≤ μ ≤ \$37.77 Chapter 8 More Recitation Problems 1 © 2009 Harvey A. Singer

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In a sketch New questions: 1. What is the point estimate for mean trade price? μ ~ xbar = \$33.77 2. What is the chance that the true mean trade price is contained within the interval calculated in part b? 95% 3. What is the chance that the true mean trade price is not contained within the interval calculated in part b? 5% 4. Based on this confidence interval, state a likely (and reasonable) upper bound for the mean trade price. Upper confidence bound = \$37.33 at 95% confidence. 5. What is the chance that the true mean trade price is more than the upper bound stated in part e? 2.5% (= 0.5×5%) Chapter 8 More Recitation Problems 2 © 2009 Harvey A. Singer x μ Sampling distribution of xbar (normal) ½×0.05 = 0.025 \$29.77 \$37.77 0.95
6. A broker executive claims that the mean price for such trades is \$28.50. Based on the above results, should this claim be doubted or believed? The claimed value of \$28.50 is less than the lower limit of the 95% confidence interval (= \$29.77). The chance of this happening in less than 2.5%. This small a chance is evidence that the executive’s claim should be doubted. 7. At 95 percent confidence, how big a sample size is needed to reduce the margin of error calculated in part b in half? At 95% confidence, E = \$4.00, so half the margin of error is \$4.00/2 = \$2.00. From E = z × σ /√ n , solving for the sample size n in terms of E results in n = ( z × σ / E ) 2 Substituting σ = \$15.00, z = 1.960 at 95% confidence, and E = \$2.00, n = (1.960×\$15 /\$2) 2 = 14.7 2 = 216.09 = 217 To reduce the margin of error by half to \$2.00, need a sample of at least n = 217 such trades. Chapter 8 More Recitation Problems 3 © 2009 Harvey A. Singer

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8.45. A survey conducted by the American Automobile Association showed that a family of four spends an average of \$215.60 per day while on vacation. Based on a random sample of 64 families of four vacationing at Niagara Falls, the mean amount spent per day is \$252.45 with a standard deviation of \$74.50 .
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