Midterm exam 5 October 2018, questions and answers PDF

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Midterm 1 questions and answers of business statistics worth 25% of course grade...

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ACCT 1230 Midterm 1 Answer Key

Section 1 1. (15 points total) ABC news has phoned 3000 Pennsylvania voters in order to estimate which presidential candidate will be selected by Pennsylvania in the upcoming U.S. elections. ABC's goal is thus to estimate the voting preferences of Pennsylvania voters in the upcoming election. ABC asked respondents about the following five topics: their income, their race, the size of their family, their job and who they plan to vote for in the presidential election. a. What is the population in this study? (1 point) The population is all registered voters in Pennsylvania. b. What is the sample in this study? (1 point) The 3000 voters who took part in the study. c. What are the variables in this study? (1 point) Income, race, size of family, job, who they plan to vote for

d. What type of data is each variable? (2.5 points) Quantitative; Categorical; Quantitative; Categorical; Categorical e. What is each variable’s scale of measurement? (2.5 points) Ratio; Nominal; Ratio; Nominal, Nominal f. Describe the difference between inferential and descriptive statistics. Is the ABC study primarily an example of inferential or descriptive statistics? Why? (2 points) Inferential statistics examines a sample of data in order to draw conclusions about the larger population. Descriptive statistics is the presentation and analysis of a data set. The ABC study is an example of inferential statistics. g. If you’ve said the study is an example of descriptive statistics, how could you modify the study to make it an example of inferential statistics? If, on the contrary, you’ve said the study is an example of inferential statistics, how could you modify the study to make it an example of descriptive statistics? (5 points)

There were a variety of acceptable answers for this. Essentially, if ABC had not attempted to draw conclusions about the larger population, but had rather simply analyzed or presented the data drawn from the sample it would have been a purely descriptive study. The key thing to understand is that inferential statistics implies moving from sample data to draw conclusions about a larger population. 2. (10 points total) Six friends compare their scores on the statistics midterm. They find they were given the following marks: 89 74 95 68 55 74. The average grade for the class was 75 and the standard deviation for the class was 10. a. Without using your computer (but please feel free to use your calculator), calculate the mean, median, mode, sample variance and standard deviation for the group of friends. (If necessary consult the attached formula pages.) Show your work. (5 points) Mean = 75.8 Median= 74 Mode = 74 Variance = 208.57 Standard Deviation = 14.44 (Provided you showed your work, I awarded full credit even if there was a mistake in the mathematical calculation. If you didn’t show your work and you had a mistake (this happened a lot with the variances and standard deviations), I deducted points.) Also, some people didn’t understand the following: the standard deviation for the class was ten, but I was asking you to find the standard deviation for the group of friends. b. Comment on the amount of variation you found among the friends. How does it compare to the amount of variation in the class as a whole? Cite the appropriate statistics to back up your claim. (2 points) There was more variation among the friends. (14.4 versus 10 standard deviation) c. What’s the range of the scores among the six friends? Assuming test scores are normally distributed, what percent of the class would you expect scored within that range? (3 points) Range = 95-55

Around 95%, as the range is 2 standard deviations from the mean.

3. (3 points) Recently, a Kwantlen instructor chose three students from a class of 10. How many different groups of 3 could the instructor have potentially selected? Applying the “Combination formula” we find the answer is 120. C=N!/n!(N-n)! C=10!/3!*7!=120

4. (3 points) A recent study concluded that 90 percent of Kwantlen students have a cell phone. If there are 5 Kwantlen students in a study group, what is the chance that three of them don’t have cell phones? Applying the binomial probability function:

Binomial Probability Function=

[ [

] ]

x

n! p (1− p)(n −x ) x !(n−x) ! .3

5! (5 − 3 ) .1 (1−.1) Binomial Probability Function= 3 ! (5−3)! (*Note – we use a probability of .1 as p, rather than .9, b/c the question asks the probability that three of the students won’t have a cell phone.)

=10*.001*.81 =.0081 There is a .81% chance that in a randomly selected group of 5 Kwantlen students, 3 of them don’t have a cell phone.

Section 2 Please Make Sure You’ve Handed in Section 1 Before Starting the Excel Portion of This Exam.

Student Name: ___________________________

5. (25 points) Recently, several Kwantlen instructors have documented the heights of the students in their classes. On the M-drive (in the ACCT 1230 directory) you will find a data set from one of these classes (Class heights_midterm). Use the exploratory tools of your choice to begin to make sense of the data set. Potential exploratory tools include: a stem and leaf diagram, a box and whiskers diagram, and a five number summary. (Use any one or two of these – you need not use all three.) Comment on the distribution of data in this sample. (7 points) Using Excel, continue your exploration of the data. Use all the tools at your disposal to describe the dataset. You will be rewarded points in proportion to the follow terms and tools that you use correctly: ordered array, frequency distribution, percent frequencies, cumulative frequencies, bar chart, histogram, ogive, quartiles, mean, mode, median, standard deviation, variance, range, and interquartile range. Briefly explain the measures you choose to use. (18 points) (*Note* Not all of these can be used given the nature of the dataset – i.e. remember that some graphs are only appropriate for categorical data while some are only appropriate for quantitative data.) Arrange your conclusions neatly and print your answer, making sure that your name is on your printout. This problem is graded exactly as it sounds – you needed to use all the appropriate tools. This means, that you need to know that height is a quantitative variable, so you would not use a bar chart or pie chart, but you would use a histogram. Beyond that, it’s simply a question of using Excel and the various exploratory techniques correctly.

6. (11 points) You are given the following information about the heights of Kwantlen students: the mean height is 6 feet (72 inches), and the standard deviation is 4 inches. Assume that heights are normally distributed (as, in fact, height tends to be). Use Excel’s tools for finding the normal distribution to answer the following questions. (The formula begins with NORMDIST and NORMINV.) a. Sketch a graph (pen and paper is fine) of the distribution of heights, including any information labels that you can. (3 points)

99.72%

INCHES b. Between what heights would you expect to find 95% of Kwantlen students? In addition to writing the number here, show this on the graph you’ve sketched above. (2 points) 64 inches and 80 inches c. Between what two heights would you expect to find the middle 50% of Kwantlen students? In addition to writing the number here, show this on the graph you’ve sketched above. (3 points) Using the excel NORMINV function you can find the heights below which you’d expect to find 25% and 75% of the heights of the class. You’d expect to find 50% of the class between these two heights. The Excel arguments would go as follows: NORMINV(.25,72,4) & NORMINV (.75,72,4) and you’d find that you’d expect to find 50% of the class between 69.3 inches and 74.7 inches. d. What’s the probability that a randomly selected individual will be taller than 6’9’’ (81 inches) tall? In addition to writing the number here, show this on the graph you’ve sketched above. (3 points)

Using the excel NORMDIST function you can find the cumulative probability that a randomly selected individual will be shorter than 81 inches. You can subtract this probability from 1 to find the probability that a randomly selected individual will be taller than 81 inches. The Excel argument would go as follows: NORMDIST(81,72,4,TRUE), and you’d find that there’s a 1.2% probability that a randomly selected individual will be taller than 6’9” (81 inches.)

General Guidelines: -Try every question. -Show all mathematical calculations (where practical). Showing me the right formula – even if you don’t apply it correctly – will get you almost full credit....