MATH534 Final Exam Solution’s Study Guide-100% verified solutions 2024-2025 .docx, Exams of Mathematics

Download MATH534 Final Exam Solution’s Study Guide-100% verified solutions 2024-2025 .docx and more Exams Mathematics in PDF only on Docsity! MATH534 Final Exam Solution’s Study Guide-comprehensive You should work each of the following on your own, then review the solution’s guide. DO NOT look at the solution’s guide first. 1. The following numbers represent the weights in pounds of six 7-year old children in Mrs. Jones' second grade class (25, 60, 51, 47, 49, 45). Find the mean, median, mode, variance, and standard deviation. Solution: This would be a sample from the class mean = 46.166 (=AVERAGE) median = 48 (=MEDIAN) mode does not exist (looking at the data) variance = 134.5667 (=VARIANCE.S) standard deviation =11.60029 (=STDEV.S) These can also be found using the MATH533 spreadsheet. 2. If the variance is 846, what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 3. If we have the following data: 34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66. Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 | 2 1 9 2 0 0 3 | 4 8 7 1 4 4 | 0 1 9 7 5 | 6 | 6 This distribution is right skewed (positively skewed) because the “tail” extends to the right. 4. For two events, C and D, P(C) = 0.6, P(D)=0.3, and P(C|D) = 0.2. Find P(C𝗇D). Solution: P(C𝗇D) = P(D) * P(C|D) = 0.3*0.2 = 0.06 a. Find the braking distance that represents the 91st percentile. b. Find the probability that the braking distance is less than or equal to 45 m. c. Find the probability that the braking distance is greater than 46.8 m. d. Find the probability that the braking distance is between 45 m and 46.8 m. Solution: You can use normal probability spreadsheet for each of these answers. To do these by hand in Excel, here are the directions. (a) =NORM.INV(0.91, 45.1, 0.5) to get 45.77m. (b) =NORM.DIST(45, 45.1, 0.5, TRUE) to get 0.4207. (c) Find the probability of less than 46.8 first and then subtract that probability from 1. So it would be =NORM.DIST(46.8, 45.1, 0.5, TRUE), which is 0.99966. More than 46.8 would be 1 - 0.99966 = 0.00034. (d) Find the probability of each, and subtract the larger from the smaller. In one cell, =NORM.DIST(46.8, 45.1, 0.5, TRUE) – NORM.DIST(45, 45.1, 0.5, TRUE) to get 0.579. 10.A sample of 58 days of gas prices has a mean of $2.75 and a standard deviation of $0.86. Construct the 95% confidence interval. Solution: You can use the Excel spreadsheet for confidence intervals. In Excel, find E and then add and subtract that from the mean of 2.75. In Excel, type =CONFIDENCE.NORM(0.05, 0.86, 58), which works out to 0.22. The interval would be (2.75-0.22, 2.75 + 0.22) or (2.53, 2.97). 11.A drug manufacturer wants to estimate the mean heart rate for patients with a certain heart condition. Because the condition is rare, the manufacturer can only find 14 people with the condition currently untreated. From this small sample, the mean heart rate is 101 beats per minute with a standard deviation of 8. a. Find a 99% confidence interval for the true mean heart rate of all people with this untreated condition. Show your calculations. b. Interpret this confidence interval, and write a sentence that explains it. Solution: You can use the Excel spreadsheet for confidence intervals. Because this is a small sample, need to first find tc and then find the maximum error. Tc = T.INV((1-0.99)/2, 13) = -3.01 Maximum error = -3.01(8/sqrt(14)) = -6.44 Then find the confidence interval by adding and subtracting from the mean. (101 – 6.44 , 101 + 6.44) (94.56, 107.44) This means that we are 99% confident that the true population mean falls between 94.56 and 107.44. 12.Determine the minimum required sample size if you want to be 80% confident that the sample mean is within 2 units of the population mean given sigma = 9.4. Assume the population is normally distributed. Solution: n = (Zc*sigma/E)^2 = [(1.28 * 9.4)/ 2]^2 = (6.016)^2 = 36.192256 = 37 rounded up. (We always round up to the next whole number.) 13.A social service worker wants to estimate the true proportion of pregnant teenagers who miss at least one day of school per week on average. The social worker wants to be within 5% of the true proportion when using a 90% confidence interval. A previous study estimated the population proportion at 0.21. a. Using this previous study as an estimate for p, what sample size should be used? b. If the previous study was not available, what estimate for p should be used? Solution: You can use the Excel spreadsheet for proportion confidence intervals. (a). The critical z-value for a 90% confidence interval is 1.645. Because a previous study is known, we can use it to estimate p = 0.21. The maximum error is 0.05. Sample size = n = p*(1-p)*( z / error )^2 = 0.21*(0.79)*(1.645/0.05)^2 = 179.5718 = 180 rounded up Thus, at least 180 pregnant teenagers must be sampled. (b). If no estimate of p is known, we must use p = 0.5 to have a large enough sample size to meet the desired maximum error. = 0.50*(0.50)*(1.645/0.05)^2 = 270.6 which would be 271 when rounded up. 14. A restaurant claims that its speed of service time is less than 15 minutes. A random selection of 49 service times was collected, and their mean was calculated to be 14.5 minutes. To find the area to the left: =NORM.DIST((0.266667-0.3)/sqrt(0.3*0.7/420),0,1,TRUE) = 0.068 Because it is two-tailed, we multiply this by 2 for the p-value, 2*0.068 = 0.136. Because p-value = .136 > .06 = alpha, we do not reject Ho. The implication is that the doctor can claim that 30% of patients have a cold. 17.To test if there was a positive relationship between the dependent and independent variables, a hypothesis test found a t-statistic of 1.43 based on a sample size of 20. Based on this information, what could be concluded about the relationship between these variables using an alpha of 0.05? Solution: The null hypothesis would be that beta-1 is 0 while the alternative hypothesis is that beta-1 is greater than 0, or positive. Using the MATH533 Excel sheet, a t-statistic of 1.43 and 18 degrees of freedom would lead to a p value of =1 – 0.9151 = 0.0849. Because this is more than 5%, there is insufficient evidence to reject the null hypothesis that there is no positive linear relationship. 18. In a multiple regression with four independent variables and 39 in the sample size, a beta is estimated to be 3.98. Using a standard deviation of this beta of 0.87, find the 95% confidence interval for the beta. Solution: The alpha is 1 – c, so would be 1-0.95 = 0.05 The degrees of freedom would be 39-(4+1) = 34 Using Excel, find the t value using half of alpha: =T.INV(0.025, 34) = -2.032 The confidence interval is based on the positive of this value, the original beta value and its standard deviation. (3.98 – 2.032*0.87, 3.98 + 2.032*0.87) (2.212, 5.748) 19.Find the regression equation of the following data. X 6 5 7 6 5 6 8 9 4 y 14 33 43 54 21 33 43 24 28