solutions hw 8

solutions hw 8 - COMPREHENSIVE HOMEWORK PROBLEMS 1) a) b)...

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Unformatted text preview: COMPREHENSIVE HOMEWORK PROBLEMS 1) a) b) c) d) In thinking about doing statistical analysis, the sample mean should be interpreted as: a constant value that is equal to the population mean. a constant value that is approximately equal to the population mean. a random variable that is approximately equal to the population mean when sampling is done without replacement. e)* a random variable that is approximately equal to the population mean if n > 30 and when sampling is done without replacement. a random variable that when averaged across many samples is approximately equal to the population mean. 2) a) b)* c) d) e) Which of the following are random? x̄ after a sample is taken x̄ before a sample is taken µ after a sample is taken µ before a sample is taken More than one answer is correct. 3) a)* b) c) d) e) 4) a) b)* c) d) e) 5) a) b) c) d)* e) 6) The monthly earnings of teachers is normally distributed with a mean of $3,000 and the standard deviation of $250. We select a sample of 87 teachers. The sampling distribution of the sample mean has an expected value and standard deviation of: 3,000 and 26.8 E(x) = μ = $ 3,000 ̄ 3,000 and 1.69 se(x) = σ ∕ √n = 26.80 ̄ 3,000 and 250 3,000 and 2.87 3,000 and 321.6 T he following data was collected by taking a simple random sample of a population 13 15 14 16 12 From this we know that, T he population mean is 14. T he point estimate of the population mean is 14. T he population mean must be 14 since the sample mean is 14. Both a. and b. are correct. Both a., b., and c. are correct. x̄ = 14 A direct mail company wishes to estimate the proportion of persons on a large mailing list that will purchase a product. Suppose the true proportion is 0.07. If 486 are sampled, what is the probability that the sample proportion will differ from the population proportion by less than 0.03? 0.0240 π = 0.07 P(0.04 < p̄ < 0.10) 0.1428 n = 486 se(p) = √π(1 - π) ∕ n ̄ se(p) = 0.0116 ̄ 0.4952 z = (p̄ − π) ∕se(p) ̄ 0.9904 P(-2.59 < z < 2.59) = 0.9904 z = ± 2.59 None of the above answers is correct. a) b) c) d) e)* A quality control expert wants to test car engines. The production manager claims they have an average life of 92 months with a standard deviation of 8. If the claim is true, what is the probability that the mean engine life would be greater than 90.8 months in a sample of 93 engines? 0.0596 μ = 92 P(x̄ > 90.8) 0.0735 σ=8 se(x) = 0.830 ̄ 0.4265 n = 93 z = (x̄ − μ) ∕se(x) ̄ 0.5596 z = -1.45 0.9265 P(z > -1.45) = 0.9265 7) a) b) c)* d) e) Increasing the size of a sample from 100 to 200 will reduce the standard error of the mean to one-half its original value. have no effect on the standard error of the mean. reduce the standard error of the mean to approximately 70% of its current value. double the standard error of the mean. None of the above answers is correct. let σ = 10 n = 100 n = 200 se(x) = 1 ̄ se(x) = 0.71 ̄ NEXT T WO ARE RELAT ED QUEST IONS ABOUT SAMPLING DIST RIBUT IONS. One hundred samples of size 85 each are drawn from an unknown population distribution of x and a sample mean is calculated for each sample. 8) If the number of samples stays at 100, but the size of each sample is increased from 85 to 125, then one would expect the variation in sample means observed across samples to: a) increase. n₁ = 85 b)* decrease. n₂ = 125 c) remain approximately the same. se(x) = σ ∕ √n ̄ σ ∕ √125 < σ ∕ √85 d) change a lot, but not necessary increase or decrease. e) be similar to the variation of x values in the population. 9) a) b) c) d) e)* 10) a) b) c) d) e)* 11) a) b) c) d) e) If the number of samples stays at 100, but the size of each sample is increased from 85 to 125, then one would expect the distribution of sample means observed across samples to: remain unknown. to depend upon the population distribution of x. approximate the normal distribution, but not more closely than when 100 samples were drawn. less closely approximate the normal distribution. more closely approximate the normal distribution. per CENTRAL LIMIT THEOREM. Annual part-time earnings in the U.S. average $15,000 and have a standard deviation $3,000. A sample of 62 part-time earners is selected. The standard error of the sample mean is: $5 $7 se(x) = σ ∕ √n ̄ se(x) = 381.00 ̄ $48 $242 $381 A speedboat engine company makes engines with the following specifications: the engine delivers an average power of 220 horsepower with a standard deviation of 16. Assuming that horsepower is normally distributed, if a randomly selected single engine is tested, what is the probability that the horsepower will exceed 224? 0.4013 0.3783 P(x > 224) 0.352 z = (x - μ) ∕σ 0.33 z = 0.25 0.2643 P(z > 0.25) = 0.4013 12) a) b) c) d)* e) In the previous question, if a sample of 80 engines are tested. What is probability that the sample mean will exceed 222 horsepower. 0.4483 μ = 220 P(x̄ > 222) 0.3446 σ = 16 se(x) = 1.789 ̄ 0.2148 n = 80 z = (x̄ − μ) ∕se(x) ̄ 0.1314 z = 1.12 0.1056 P(z > 1.12) = 0.1314 13) a) Consider the horsepower average and standard deviation in the previous question: µ = 220 and σ = 16. If the sample size is n = 100, in the sampling distribution of x̄ what interval of values would contain 95% of all sample means? MOE = z0.025 se(x) ̄ 205.98 to 234.02 MOE = 3.136 x̄L = μ − MOE = 216.86 x̄U = μ + MOE = 223.14 b) 210.08 to 229.92 c) d) e)* 212.99 to 227.01 215.57 to 224.43 216.86 to 223.14 14) a) b) c) d)* e) According to the central limit theorem, as the sample size increases, the expected value of x̄ approaches 0. the expected value of standard error of x̄ approaches 1. the standard error of x̄ approaches the population standard deviation. the distribution of approaches the normal. the distribution of s, the sample standard deviation, approaches 0. 15) a) b) c) d) e)* The actual proportion of defective jeans in a large warehouse is 0.20. What is the probability that in a random sample of 500 jeans the sample proportion p̄ will be within a margin of ±0.04 of the population proportion of defective jeans? 0.8198 P(0.16 < p̄ < 0.24) 0.8415 se(p) = √π(1 - π) ∕ n ̄ se(p) = 0.0179 ̄ 0.8859 z = (p̄ − π) ∕se(p) ̄ 0.9164 z = ± 2.23 P(-2.23 < z < 2.23) = 0.9743 0.9743 b) In the previous question, what interval of p ̄ values contains 90% of all p ̄ ’s? MOE = z0.05 se(p) ̄ 0.171 to 0.229 MOE = 0.029 pL = π − MOE = 0.171 ̄ 0.165 to 0.235 c) d) e) 0.154 to 0.246 0.145 to 0.255 0.122 to 0.278 16) a)* p̄U = π + MOE = 0.229 17) a) b) c)* d) e) 18) a) b) c) d) e)* A sample of 64 patients in a walk-in clinic showed that they had to wait an average of 48 minutes before they could see a doctor. The sample standard deviation was 20 minutes. What is the 95% confidence Interval for the population average waiting time? 45.2 50.8 L, U = x̄ ± MOE 1 - α = 0.95 MOE = zα/2 se(x) ̄ 44.6 51.4 se(x) = 2.5 ̄ 43.1 52.9 42.2 53.8 41.8 54.2 MOE = 4.9 L = 43.1 U = 52.9 What is the minimum sample size to estimate the 95% confidence interval in the previous question to within plus or minus 3 minutes. Use a planning value of 20 minutes for the standard deviation. 131 n = (zα/2 ŝ ∕ MOE)² 141 n = 170.74 151 161 171 19) b)* Here is a problem similar to the previous question, but now the sample of patients is only 9. In this sample the patients had to wait an average of 45 minutes and the sample standard deviation was 12 minutes. What is the margin of error for a 95% confidence Interval for the population average waiting time? (Assume that the population of the waiting times in normal). 7.8 1 - α = 0.95 MOE = tα/2,(n -1) se(x) ̄ 9.2 se(x) = 4 ̄ c) d) e) 10.6 12 13.4 a) t0.025,8 = 2.306 MOE = 9.2 20) a) b)* c) d) e) Imagine 526 statisticians each took a different random sample of the population of patients visiting the walk-in clinic in Question 17 (each took a sample size of 64). About how many would produce confidence intervals that contained the population mean? 100 526 × 0.95 = 499.7 500 Almost all of them All of them. Cannot tell based on the information provided. 21) b) c) d)* e) To build a confidence interval for the average age of the civilian labor force, a sample of 100 people was selected. The sample mean was 38.5 years and the standard deviation was 13.2 years. The lower and upper boundaries of a 95% confidence interval are: 37.8 to 39.2 L, U = x̄ ± MOE 1 - α = 0.95 MOE = zα/2 se(x) ̄ 37.4 to 39.6 se(x) = 1.32 ̄ 36.3 to 40.7 MOE = 2.6 35.9 to 41.1 L = 35.9 34.4 to 42.6 U = 41.1 22) a) b) c) d)* e) We can make a confidence interval more precise (narrower) by, increasing the sample size. reducing the confidence level (or confidence coefficient). increasing the confidence level Both (a) and (b) are correct. Both (a) and (c) are correct. a) 23) a)* b) c) d) e) 24) a)* b) c) d) e) 25) To estimate the population average age of the civilian labor force to within a margin of error of 0.5 years at 95% level of confidence, what is the minimum sample size? Assume the population standard deviation is known to be 12.1 years. n = (zα/2 ŝ ∕ MOE)² 2,250 n = 2249.79 2,145 1,972 1,576 1,255 Using the standard deviation 2.61 as a planning value, what sample size should be used in order to obtain a 95% confidence interval with a margin of error of ±0.5? 105 n = (zα/2 ŝ ∕ MOE)² 94 n = 104.68 85 64 56 To build an interval estimate of commuting time from Fishers to downtown Indianapolis in a midweek rush hour period five trial runs were made, obtaining the following results (in minutes). b) 55 45 43 34 38 Assuming the population commuting time is normally distributed, build a 95% confidence interval for the population mean commuting time. T he interval is: 28.6 to 57.4 x̄ = 43 L, U = x̄ ± MOE 1 - α = 0.95 MOE = tα/2,(n -1) se(x) ̄ 29.9 to 56.1 s = 7.969 se(x) = 3.564 ̄ c) d)* e) 31.2 to 54.8 33.1 to 52.9 36.0 to 50.0 26) For another interval estimate of the commuting time a sample of 100 trial runs were made. The lower and upper bounds of the interval were: L = 43.67 and U = 48.33 minutes. The sample standard deviation was s = 10 minutes. What is the confidence level for this interval estimate? 98 percent. MOE = (48.33 - 43.67) / 2 = 2.33 MOE = zα/2 se(x) ̄ 96 percent. se(x) = 1 ̄ a) a)* b) c) zα/2 = MOE ∕ se(x) ̄ 94 percent. d) e) t0.025,4 = 2.776 MOE = 9.9 L = 33.1 U = 52.9 92 percent. 90 percent. zα/2 = 2.33 T he tail area for z = 2.33 is P(z > 2.33) = 0.01 α/2 = 0.01 1 - α = 0.98 27) a) b) c)* d) e) A survey of 200 individuals who completed four years of college showed that 36 smoked regularly. Using this survey result what is the 95% confidence interval for the proportion of all individuals with four years of college education who smoke? 0.097 to 0.263 L, U = p̄ ± MOE 1 - α = 0.95 p̄ = 36 / 200 = 0.18 MOE = zα/2 se(p) ̄ 0.107 to 0.253 se(p) = 0.0272 ̄ se(p) = √p̄ (1 − p) ∕ n ̄ ̄ 0.127 to 0.233 MOE = 0.053 0.137 to 0.223 L = 0.127 0.147 to 0.213 U = 0.233 28) a) b) c)* d) e) In the previous question, what is the minimum sample size to estimate the population proportion of all individuals with four years of college education who smoke to within a margin of error of ±0.03. We expect 19 out of every 20 such interval estimates to contain the population proportion. Use the sample proportion in the previous question as the planning value. n = p(1 − p)(zα/2 ∕ MOE)² ̂ ̂ 227 1 - α = 0.95 p̂ = 0.18 355 631 993 1418 n = 630.02 29) a) b) c)* d) e) 30) a)* b) c) d) e) The director of admission at a large state university advises parents of incoming students about the cost of textbooks during a typical semester. A sample of 100 students enrolled in the university indicates a sample mean cost of $315.40 with a sample standard deviation of $69. The sample is used to test the hypothesis that the population mean is at most $300. Which of the following is the correct statement of the null and alternative hypotheses? H0 : µ ≥ 300 H1 : µ < 300 H0 : µ > 300 H1 : µ ≤ 300 H0 : µ ≤ 300 H1 : µ > 300 H0 : µ < 300 H1 : µ ≥ 300 H0 : µ = 300 H1 : µ ≠ 300 Regardless how you answered Question 29, which of the following statements is correct? If the mean cost of text books is in fact greater than $300 and the hypothesis test leads you to conclude that it is at most $300, the you have committed a Type II error. If the mean cost of text books is in fact greater than $300 and the hypothesis test leads you to conclude that it is at most $300, the you have committed a Type I error. If the mean cost of text books is at most $300 and the hypothesis test leads you to conclude that it is greater than $300, the you have committed a Type II error. If the mean cost of text books is in less than $300 and the hypothesis test leads you to conclude that it is greater than $300, the you have committed a Type II error. If the mean cost of text books is in less than $300 and the hypothesis test leads you to conclude that it is at least $300, the you have committed a Type II error. You commit a T ypt II error when you do not reject a false null hypothesis. 31) a) b) c) d) e)* 32) a) b) c) d) e) Test H0 : µ ≤ 5,000 versus H1 : µ > 5,000 when a sample of size 100 yields a mean of 5,315.4 and a standard deviation of 1400. Conduct the test with a probability of type I error = 0.10. Also compute the probability value. Which of the following is the correct decision: T he probability value is 0.02. Do not reject the null hypothesis that the mean is less than or equal to 5000. T he probability value is 0.10. Do not reject the null hypothesis that the mean is less than or equal to 5000. T he probability value is 0.02. Reject the null hypothesis that the mean is less than or equal to 5000. T he probability value is 0.01. Conclude that the mean is less than 5000. T he probability value is 0.01. Conclude that the mean is greater than 5000. test stat: z = (x̄ − μ₀) ∕ se(x) ̄ z = 2.25 p-value: P(z > 2.25) = 0.0122 Reject H₀ : µ ≤ 5,000 The automobile manufacturer Toyonda substitutes a different engine in cars of a model that were known to have an average miles per gallon (mpg) rating of 30 on the highway. To test whether the new engine changes the average mpg a random sample of 100 trial runs gives x̄ = 28.3 mpg and s = 6.6 mpg. At α = 0.05 level of significance, is the average highway mpg rating for new engines different from the rating for the old engines? The standardized test statistic exceeds the critical value. The average highway mpg rating for new engines is different from that of the old engines. The probability value is less than the level of significance. The average highway mpg rating for new engines is different from that of the old engines. The standardized test statistic exceeds the critical value. The average highway mpg rating for new engines is NOT different from that of the old engines. The probability value is greater than the level of significance. The average highway mpg rating for new engines is NOT different from that of the old engines. Both (a) and (b) are correct. H₀: μ = 30 H₁: μ ≠ 30 test stat: z = (x̄ − μ₀) ∕ se(x) ̄ z = 2.58 p-value: P(z > 2.58) = 0.004 9 Reject H₀ : µ = 30 cv = 1.64 33) a) b) c) d)* e) 34) The engineering team of Honota Motors has designed a new engine for cars of Model X240 which it claims will improve (increase) the gas mileage while maintaining the same horsepower. The current average highway mileage is 25 mpg. A sample of 40 trial runs gives x̄ = 26.4 mpg. Based on the average trial-run mpg the management rejects the engineering team’s claim and does not adopt the new design. Suppose the new design does in fact improve the gas mileage. Which of the following correctly describes the management’s decision: T he management has rejected a true null hypothesis. T herefore, it has committed a T ype I Error. T he management has rejected a true null hypothesis. T herefore, it has committed a T ype II Error. T he management has not rejected a false alternative hypothesis. T herefore, it has committed a T ype I Error. T he management has not rejected a false null hypothesis. T herefore, it has committed a T ype II Error. Both (a) and (c) are correct. H₀: μ ≤ 25 H₁: μ > 25 Use the sample of commuting times from Fishers to downtown Indianapolis in a midweek rush hour period: 55 45 43 34 38 Perform a test of hypothesis that the average time exceeds 36 minutes, using α = 0.05. Based on the sample data, a) b) The test statistic is 1.964 and the critical value is 1.64. The sample mean is significantly greater than 36. Reject the null hypothesis. The test statistic is 1.64 and the critical value is 1.964. The sample mean is not significantly greater than 36. Do not reject the null hypothesis. c) The test statistic is 1.074 and the critical value is 1.64. The sample mean is not significantly less than 50. Do not reject the null hypothesis. d) The test statistic is 1.074 and the critical value is 2.132. The sample mean is not significantly above 35. Do not reject the null hypothesis. e)* The test statistic is 1.964 and the critical value is 2.132. The sample mean is not significantly above 36. Do not reject the null hypothesis. s = 7.969 H₀: μ ≤ 36 H₁: μ > 36 x̄ = 4 3 t0.05,4 = 2.132 test stat: t = (x̄ − μ₀) ∕ se(x) ̄ t = 1.964 35) a) b)* c) d) e) 36) a) b) c)* d) e) To test the hypothesis that the percentage of individuals with four years of college education who smoke has decreased from 21% a decade ago, a random sample of 1200 such individuals revealed that 222 smoked. Use α = 0.05. Based on the sample result, The sample proportion is not significantly less than 21%. Do not reject the null hypothesis. Conclude that the proportion of college educated individuals who smoke has not decreased compared to a decade ago. The sample proportion is significantly less than 21%. Reject the null hypothesis. Conclude that the proportion of college educated individuals who smoke has decreased compared to a decade ago. The sample proportion is significantly less than 21%. Do not reject the null hypothesis. Conclude that the proportion of college educated individuals who smoke has decreased compared to a decade ago. The sample proportion is not significantly less than 21%. Reject the null hypothesis. Conclude that the proportion of college educated individuals who smoke has not decreased compared to a decade ago. The test statistic is less than the critical value. Do not reject the null hypothesis. Conclude that the proportion of college educated individuals who smoke has not decreased compared to a decade ago. H₀: π ≥ 0.21 H₁: μ < 0.21 test stat: z = (p̄ − π₀) ∕ se(p) ̄ π₀ = 0.21 se(p) = 0.0118 ̄ p̄ = 0.185 z0.05 = 1.64 z = 2.12 In the previous question, the probability value is: 0.095 P(z > 2.12) =0.0170 0.032 0.017 0.01 0.004 NEXT T HREE QUEST IONS ARE BASED ON T HE FOLLOWING SCENARIO: The professors at Budget University make $75,000 on average. The professors want to convince the Budget administrators that professors from comparable universities make higher salaries. The Budget professors collect sample data on salaries from comparable universities to provide a test of their hypothesis. 37) T he general form of the test should be a) H₀: μ = $75,000 H₁: μ.≠ $75,000 b) H₀: μ ≠ $75,000 H₁: μ = $75,000 c) H₀: μ ≤ $75,000 H₁: μ ≠ $75,000 d)* H₀: μ ≤ $75,000 H₁: μ > $75,000 e) H₀: μ ≥ $75,000 H₁: μ< $75,000 38) b)* If the Budget professors economists to use a 1 percent significance level instead of a 5 percent significance level, the critical value (s) will be _______ in absolute value and it is _______ likely that the null hypothesis will be rejected. larger, more z0.01 = 2.33 larger, less c) d) e) smaller, more smaller, less unaffected, equally 39) The economists decide to use a 1 percent significance level. They collect sample data on salaries from 20 comparable universities. The sample mean is $81,000 and the sample standard deviation is $10,000. The test statistic is ________, which causes them to ________ the null hypothesis. z = 2.68, reject t = 2.68, fail to reject test stat: t = (x̄ − μ₀) ∕ se(x) ̄ t0.01,19 = 2.539 z = 0.81, reject t = 2.683 t = 0.81, fail to reject t = 2.68, reject a) a) b) c) d) e)* z0.05 = 1.64 NEXT FIVE QUEST IONS ARE BASED ON T HE FOLLOWING REGRESSION OUT PUT : T he following data for a sample of 10 individuals shows the hourly earnings and years of schooling. Hourly Earnings 17.24 15.00 14.91 4.50 18.00 6.29 19.23 18.69 7.21 42.06 Years of Schooling 15 16 8 6 15 12 12 18 12 20 T he following regression Summary Output is used to study the relationship between hourly earnings and years of schooling: SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.7311 0.5344868664 0.4763 7.656124999 10 ANOVA SS 538.41 468.93 1007.34 MS 538.40905 58.61625 Coefficients Standard Error -7.791 8.3134 1.799 0.5935 df t Stat -0.9371 3.0312 Regression Residual Total Intercept X Variable 1 1 8 9 F Significance F 9.1853643 0.016291153 P-value 0.3761 0.0163 Lower 95% -26.962 Upper 95% 11.38 Answer the next FIVE questions using the information in the Summary Output. 40) What is the predicted hourly earnings for 12 years of schooling? a) 3 b) 6.6 ŷ = -7.791 + 1.799(12) ŷ = 13.797 c)* 13.8 d) 19.19 e) 24.59 41) a) b) c) d)* e) What percentage of hourly earnings is explained by years of schooling? 74.60% 73.10% R² = SSR / SST = 0.5345 69.30% 53.40% 47.60% 42) a) c) d) e) What is the 95% confidence interval for the population slope parameter β1? 0.8 2.8 L, U = b₁ ± MOE MOE = tα/2,(n -2) se(b₁) 0.43 3.17 s e(b₁) = 0.5935 t0.025,8 = 2.306 0.23 3.37 0.13 3.47 MOE = 1.369 0.03 3.57 L = 0.430 U = 3.17 43) a)* b) c) d) e) To perform a test of hypothesis that the population slope parameter β ₁ is zero, the test statistic is: 3.031 2.306 t = b₁ ∕ se(b₁) t = 3.0312 2.262 2.228 0.33 44) a) b) Given the P-value of 0.0163, we can conclude, at 5% level of significance, that: T he population slope parameter is zero. T here is NO relationship between hourly earnings and years of schooling. b)* T he population slope parameter is different than zero. T here is NO relationship between hourly earnings and years of schooling. c)* d) e) T he population slope parameter is different than zero. T here is a relationship between hourly earnings and years of schooling. T he population slope parameter is zero. T here is a relationship between hourly earnings and years of schooling. There is a small probability of a Type II error, accepting the hypothesis that the slope parameter is not equal to zero, when in fact it is. To study the relationship between manufacturers’ market share and the quality of product. The following data on market share (in percentage) and product quality (ratings on the scale of 0 to 100) are available. The question is, are the variations in market share explained by the quality of the product? y x Market share Product Quality (%) (Scale: 0 to 100) 2 27 3 39 10 73 9 66 4 33 6 43 5 47 8 55 7 60 9 68 Using the following calculations complete the relevant parts (the shaded cells) of the Excel regression output below and answer FOUR questions. ȳ = 6.3 x̄ = 51.1 See the lease squares formuals for b₁ and b₀ in your notes. ∑xy = 3592 ∑x² = 28331 ∑(x − x)(y − y) = 372.7 ̄ ̄ ∑(x − x)² = 2218.9 ̄ SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.9193 0.8291 10 ANOVA df Regression Residual Total Intercept X Variable 1 8 SS 62.601 5.499 68.1 Coefficients Standard Error -2.2831 0.1680 0.01760 MS F Significance F P-value Lower 95% 0.687375 t Stat Upper 95% 9.543191996 45) a) b)* c) d) e) T he predicted market share for a product quality rating of 90 is: 11.2 12.8 ŷ = -2.2831 + 0.168(90) = 12.83 13.4 13.9 14.4 46) a) b)* c) d) e) T he proportion of the variations in market share explained by product quality rating is: 0.96 0.92 R² = 0.919 0.86 0.82 0.78 47) a) c) d)* e) T he upper boundary of the 95% confidence interval for the population slope parameter is: 0.241 L, U = b₁ ± MOE MOE = tα/2,(n -2) se(b₁) 0.235 s e(b₁) = 0.01760 t0.025,8 = 2.306 0.228 0.209 MOE = 0.041 0.117 L = 0.127 U = 0.209 48) a) b) c)* d) e) T he t Stat for the test of hypothesis that the population slope parameter is zero is: 2.95 7.65 t = b₁ ∕ se(b₁) 9.54 10.11 10.98 b) Next THREE questions use the following data describing the median annual family income (in $1000s) and the median sale price of a house (in $1000s) for a sample of 12 housing markets. The data are used to regress the median price in a housing market on the median income in that market. The regression output follows the data. x y Income Price Market ($1000s) ($1000s) Syracuse, NY 4 1.80 76 Springfield, IL 4 7.70 91 Lima, OH 4 0.00 65 Dayton, OH 4 4.30 88 Beaumont, T X 37.30 70 Lakeland, FL 35.90 73 Baton Rouge, LA 39.30 85 Nashua, NH 56.90 118 Racine, WI 4 6.70 81 Des Moines, IA 48.30 89 Minneapolis 54.60 110 Wilmington, DE-MD 55.50 110 Average 45.692 SUMMARY OUTPUT Regression Statistics Multiple R R Square 0.8606269791 Adjusted R Square Standard Error 6.6343 Observations 12 ANOVA df Regression Residual Total Intercept X Variable 1 10 SS 2717.86 440.14 3158 Coefficients Standard Error -11.802 2.1843 0.2780 MS F Significance F 44.014 t Stat P-value 7.8583 49) a) b)* c) d) e) T he point estimate of the median price in a housing market with a median family income of $50,270 per annum is: $97,000 $98,000 x = 50.27 $97,270 ŷ = -11.802 + 2.1843(50.27) = 98.003 $98,990 $99,000 50) a) b) c) d)* e) What percent of the total variation in the median sale price of houses is explained by the estimated regression line? 93 84 R² = 0.861 63 86 80 51) The sum of squared deviations x’s is: ∑(x − x)² = 569.669. Calculate the t statistic for testing the null hypothesis of no linear relation (i.e. ̄ the slope parameter is zero) at a 10% level of significance (assume that the errors are normal, so there is no problem using the t distribution). The conclusion would be: reject the null hypothesis; there is a significant linear relationship. do not reject the null hypothesis; there is no significant linear relationship. reject the null hypothesis; there is no significant linear relationship. do not reject the null hypothesis; there is a significant linear relationship. reject the null hypothesis at 10%; but you would not reject at 1%. se(b₁) = se(e) ∕ √∑(x − x)² ̄ se(b₁) = 0.2780 t = b₁ ∕ se(b ₁) = 7.858 a)* b) c) d) e) 52) a) b)* c) d) e) 53) a) b) c) d)* e) Below are some data on weekly sales (in $1000s). If you are using a four-week moving average to forecast the next week’s sales, what would your forecast have been for week 11? 23 19.5 sales 19.3 week ($1000s) α = 0.4 18.6 1 17 15 2 21 17 3 19 18.6 4 23 5 18 6 16 7 20 8 16 9 22 10 20 11 23 19.5 12 19 Using the data from Question 52, what is the exponential smoothing forecast for week 3 sales when α = 0.4? 23 19.5 19.3 18.6 20 NEXT T WO QUEST IONS ARE BASED ON T HE FOLLOWING SALES DAT A FROM YOUR ICE CREAM BUSINESS: (sales in $100,000s) January February March April 54) a) b) c)* d) e) 12 14 12 9 13.6 10 9.92 11 9.984 What is your three-month moving average forecast for May? 9 9.98 11 11.38 11.67 55) a) b)* c) d) e) What is your forecast for May using exponential smoothing and a smoothing parameter α = 0.8? 9 9.98 11 11.38 11.67 The following data shows a state’s building contracts (in millions of dollars) for a 12-month period. Next two questions are based on this table. Period 1 2 3 4 5 6 7 8 9 10 11 12 13 Contracts 260 380 250 280 300 350 24 0 34 0 260 34 0 260 250 Moving Average Exponential Smoothing 283.3333 289.3 273.58 56) a) b)* c) d) e) Using a three-month moving averages, what is the forecast for period 13? 286.7 283.3 313.3 280 310 57) a)* b) c) d) e) Using a smoothing constant α = 0.4, what is the forecast for period 13? 273.6 308.8 287.9 306.6 284.3 ...
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