Statistics Homework 1: Data Analysis and Problem Solving

MGMT 30500: Assignment 1 – Fall 2023 Due Date: September 3, 11:59 pm  To find the assignment problems and datafiles in the Textbook: Exercises: Methods, Exercises: Application, Supplementary Exercises, or Case Problems.  To submit all homework assignments: Brightspace -> Course Tools -> Assignments. PART A. From the Textbook Chapter 1: 21, 25 Chapter 2: 39 Chapter 3: 43 (Use =percentile.exc(data array, percent) to find percentiles), 53, 59 Chapter 6: 11, 41 (Show EXCEL functions for both exercise problems.) PART B. For each problem, briefly explain why you choose your answer. Variable Count Minimum Q1 Median Q3 Maximum Incomes 100 1.63 19.60 21.60 31.73 77.80 1. To evaluate a possible business expansion, we have compiled the above 5-number summary for a random sample of 100 household incomes (in US$1000) in Berlin, Germany. A boxplot of this data will show: A. No outliers B. Exactly 1 high outlier C. At least 1 high outlier D. At least 1 low outlier E. Both high and low outliers There is a maximum outlier because the maximum is above Q3+(1.5)IQR but there isn’t a low outlier because the minimum value is less than Q1-(1.5)IQR. We cannot say how many outliers there are exactly because we do not know every data value within the dataset. 2. Referring again to the 5-number summary from Question 1, approximately how many of the household incomes are between US$19,600 and US$31,730? A. 25 B. 50 C. 75 D. 90 50% of the data fall between Q1 (25%) and Q3 (75%). 1

3. Referring again to the 5-number summary from Question 1, a histogram of these household incomes would most likely be: A. Slightly skewed left. B. Slightly skewed right. The tail of the data would fall more to the right of Q3 than to the left of Q1 4. To study human IQs, we use Excel to randomly sample 200 numbers from the IQ distribution, N(100, 15). According to the empirical rule, approximately how many of these numbers do we expect to be greater than 120? A. 0 B. 3 C. 10 D. 32 E. None of the above. P(120<X) = (X-MU)/ = (120-100)/15= 1.333 Ỽ P(Z>1.333) = 1 – P(Z<1.333) = 1- .90824 = 0.09176 0.09176 * 100 = ~10 _____________________________________________ 5. Enduro Foods, Inc. makes long-lasting survival foods. To test a new preservative, 100 survival meals are left exposed at room temperature. Their time to spoilage in weeks is recorded and a histogram of the results is shown above. For this sample, the median time to spoilage is: A. Between 0 and 10. B. Between 10 and 20. C. Between 20 and 30. 2

D. Between 30 and 40. Majority of the values ~65% fall within 0 – 10 weeks. 6. Given that T is a random variable following a t-distribution with 59 degrees of freedom, P(T ≤ - 6 or T ≥ 6) is approximately A. 0. B. 0.34. C. 0.02. D. 0.003. E. Cannot be calculated. T distribution chart? Homework 1 – Problems in the Textbook Seen Below 3

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Homework 1 – Problems in the Textbook 1.21 A. The two populations studied are mothers that took DES during pregnancy and mothers that did not take DES during pregnancy. B. The data was most likely collected through a survey because the scientist did not manipulate any of the variables. They simply observed an outcome. C. We can use the sample proportion of women that developed abnormalities and use it gauge how many were impacted out of 1,000. (63/3980) * 1000 = ~15.83 women D. The study concluded that mothers that took the drug are 2x as likely to develop abnormalities so to find the likelihood of developing an abnormality while not taking the drug with: (15.8/2) = ~7.9 E. They use relatively large samples to improve the reliability of their conclusions. 4

1.25 A. There are 7 variables within the dataset. B. The Categorical variables are: Make, Model, Recommended, and Owner Satisfaction The quantitative variables are: Overall Score, Overall Miles per Gallon, and Acceleration 5

C. 7 vehicles are recommended out of 15 total vehicles which means we can find the percentage with (7/15) = 0.47 * 100 = 47% D. The average can be found by adding up all miles per gallon data points and dividing by the number of data points overall. (366/15) = 24.4 miles per gallon E. F. Class Intervals Class Boundaries Tally Bars Frequency 7.0-7.9 6.95-7.95 l 1 8.0-8.9 7.95-8.95 lllll 5 9.0-9.9 8.95-9.95 llll 4 10.0-10.9 9.95-10.95 lllll 5 Total 15 6 Yes No 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8 8.2 Owner Satisfaction

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2.39 A. B. There is an apparent negative relationship between the two variables where the MPG decreases as the Speed increases. 7 20 25 30 35 40 45 50 55 60 65 0 5 10 15 20 25 30 35 40 Fuel Efficiency as Speed Increases Speed (Miles/Hour) MPG (Miles/Gallon)

3.43 A. The mean is $72,342.94 and the median is $66,743.00. B. The mean is close to the median which means that the data has a symmetrical data. C. The range is $100,132.00 and the standard deviation is $23,711.80. D. The 1 st quartile is $56,799.50 and the 3 rd quartile is $81,533.25. E. The upper limit is $97,076.69 and the lower limit is $47,609.19. There are 2 datapoints above the upper limit which are the outliers within this data. The implications are that these points might be the cause of the data being positively skewed. 8

3.53 A. The median return is 13.9. B. 19 companies had a higher return than the average / 50 total companies = 0.38 *100 = 38% C. -13.9, 3.7, 13.9, 30, 117.1 D. The upper limit is 69.45 and the lower limit is -35.75. There is no data below the lower limit but there are 3 outliers above the upper limit. E. 9

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3.59 A. I expect a negative relationship between smoke detectors and deaths from fires due to smoke detectors warning occupants that there is a fire. B. The correlation between the two is – 0.90596. This is a negative correlation as expected. C. 10 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.0 5.0 10.0 15.0 20.0 25.0 Smoke Detector Usage and Deaths from Fire Smoke detectors used % Deaths from fire

6.11 A. p(z<=-1) = .15866 B. p(z>=-1) = 1- .15866 = 0.84134 C. p(z>=-1.5) = 1- .06681 = .93319 D. p(z<=-2.5) = .00621 E. p(0>=z>-3) = 0.49865 11

6.41 A. Z = 65,000 – 53901 / 15,000 = 0.7399 P( Z >= 65,000) = 1 - 0.77035 = 0.2297 B. Z = 65,000 – 51541 / 11,000 = 1.2235 P( Z >= 65,000) = 1 - 0.88877 = 0.11123 C. Z = 40,000 – 51541 / 11,000 = -1.05 P( Z < 40,000) = 0.14686 D. P( Z < X) = 0.01 X = 2.33 * 11000 + 51541 = 77,171 12

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Homework 1 word problems

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