STAT100 Fall 19 Final Exam Practice Problems Answers PDF

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Questions on different types of variables 1. Gallup conducted research to measure recent college graduates’ career and life outcomes. A 2014 study collected information about graduates’ annual income, whether they were living with their parents, employment status (employed or unemployed), zip code, their undergraduate grade point average (GPA), and whether the graduate was a member of a sorority/fraternity in college. a. What are the quantitative variables? Circle correct answer(s). There may be more than one correct answer. Annual income Living with Parents?

Employment Status

Zip code

Undergraduate GPA

Sorority/fraternity member

b. What are the categorical variables? Circle correct answer(s). There may be more than one correct answer. Annual income

Living with Parents?

Employment Status

Zip code

Undergraduate GPA

Sorority/fraternity member

Note that zip code is a categorical variable, even though it is represented by numbers. Quantitative variables must have a meaningful scale; i.e. the numbers measure something. For example, the sum of 2 zip codes or the average zip code for a group does not have a meaningful numeric interpretation.

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Questions on interpreting graphical displays 2. A histogram, boxplot, and summary statistics are given for levels of calcium concentration in drinking water for 172 locations. Summary statistics: Calcium (ppm) n Median Range Min 172 102.6 478.8 34.2

Max 513

Q1 68.4

Q3 205.2

a) Use the histogram to answer this question. What percent of locations had calcium concentrations of 250 ppm and above? Round your decimal to 3 decimal places, then report the corresponding percent (e.g., you would round .1248 to .125 and then write 12.5%). = (9+3+3+4+2)/172 = .122

b) Which of the following statements about the mean and median for this data is true? Circle the ONLY ONE correct statement. i)

The mean calcium (ppm) is greater than the median calcium (ppm) because the distribution is skewed left.

ii) The mean calcium (ppm) is greater than the median calcium (ppm) because the distribution is skewed right. iii) The median calcium (ppm) is greater than the mean calcium (ppm) because the distribution is skewed left. iv) The median calcium (ppm) is greater than the mean calcium (ppm) because the distribution is skewed right.

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Questions on two-way tables Scenario: Americans Without Health Insurance The United States federal government collects information on Americans who do not have health insurance. Data from 2004 are broken down into 4 regions of the country. These data are summarized in the table provided below. Using this table, answer the following questions.

1) Based on these figures, what percent of the Midwestern Americans who were surveyed are uninsured? Round your decimal to 3 decimal places, then report the corresponding percent (e.g., you would round .1248 to .125 and then write 12.5% = (# of Midwestern Americans who are uninsured)/(# of Midwestern Americans surveyed) = 7,757/64,892 = .1195 = .120 = 12.0% 2) Based on these figures, what percent of all the Americans surveyed are from the West? Round your decimal to 3 decimal places, then report the corresponding percent (e.g., you would round .1248 to .125 and then write 12.5%. = (# of Americans who are from the West)/(# of all Americans surveyed) = 67,103/290,710 = .2308 = .231 = 23.1% 3) Based on these figures, what percent of all Americans surveyed are from the South and are uninsured? Round your decimal to 3 decimal places, then report the corresponding percent (e.g., you would round .1248 to .125 and then write 12.5%. = (# of Americans who are from the South AND are uninsured)/(# of all Americans surveyed) = 19.090/290,710 = .0656 = .066 = 6.60%

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Questions on regression 1. Assume that you are given a regression equation of Y = 4.2 + (−2)X. Which of the following is the correct interpretation of the slope? a) For each decrease of one unit in the explanatory variable (X), we expect the response variable (Y) to decrease by two units. b) For each increase of one unit in the explanatory variable (X), we expect the response variable (Y) to increase by two units. c) For each increase of one unit in the explanatory variable (X), we expect the response variable (Y) to change by 4.2 + (−2) = 2.2 units. d) For each increase of one unit in the explanatory variable (X), we expect the response variable (Y) to decrease by two units.

Questions on discrete random variables Here is the probability distribution of Y, the number of people living in each apartment in a building:

1. What is the mean of Y? Show all work.

Use the formula for mean of a discrete random variable from the formula sheet. Mean of X = (0)(.05) + (1)(.05) + (2)(.10) + (3)(.75) + (4)(.05) = 2.70

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Questions on regression Results of a least-squares linear regression for 58 U.S. cities are shown below. This regression models whether a city’s mortality rate (deaths per 100,000 people) can be predicted by the city’s education level (average number of years in school). Correlation Coefficient: -.6405

Scatterplot of Mortality versus Education

Regression Equation: Mortality = 1493.3 - 49.92 Education

Mortality Rate

1050

1000

10.0

10.5

11.0

12.0

12.5

Education Level

a. What is the explanatory variable? Education level b. What is the response variable? Mortality rate c. Use the regression equation to predict the mortality rate for a city with an education level of 10.25 years. Round your answer to 3 decimal places.

= 1493.3 – 49.92 (10.25) = 981.62 d. The slope of the regression line is

-49.92

(enter a numeric value).

e. Interpret what the value of the correlation coefficient indicates about the strength and direction of the linear relationship between a city’s education level and mortality rate. r = -.6405; moderate to strong negative linear relationship

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Questions on binomial random variables 30% of female statistics students have tattoos. Suppose we take a random sample of 4 female statistics students. Let X represent the number of students with tattoos. This situation can be modeled with a binomial random variable X. The cumulative distribution function for X is shown below.

x 0 1 2 3 4

P(X ≤ x) 0.2401 0.6517 0.9163 0.9919 1.0000

1. What is the probability that exactly 2 students have tattoos? (HINT: see the formula sheet for help using Cumulative Probability Distributions) From formula sheet:

P(X = x) = P[X ≤ x] - P[X ≤ (x-1)] P(X = 2) = P(X≤2) – P(X≤1) = .9163 - .6517 = .2646 2. What is the probability that at least 3 students have tattoos? (HINT: see the formula sheet for help using Cumulative Probability Distributions) From formula sheet:

P(X ≥ x) = 1 – P[X ≤ (x-1)] P(X ≥ 3) = 1 – P(X≤2) = 1 – .9163= .0837 3. Determine the mean and standard deviation of X. Round all answers to 3 decimal places. X is binomial with n = 4 and p =.30, so the mean of X = n*p = 4*.3 = 1.2 X is binomial with n = 4 and p =.30, so the standard deviation of X = �n*p*(1-p) = .917

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As of October 2015, 22% of American families had their credit card information stolen by hackers. Suppose we take a random sample of 8 American families. This situation can be modeled with a binomial random variable X. The cumulative distribution function (CDF) for X is shown below.

0

P( X ≤ x ) 0.13701

1

0.44617

2

0.75136

3

0.92351

4 5

0.98421 0.99791

6

0.99984

7 8

0.99999 1.00000

1. What is the probability that exactly 3 families out of the random sample of 8 families had their credit card information stolen by hackers? (HINT: see the formula sheet for help using Cumulative Probability Distributions) P (exactly 3 families) =P(X = 3) Use the Formulas for Cumulative Probability Distributions of Discrete Random Variables from the formula sheet:

P(X = x) = P[X ≤ x] - P[X ≤ (x-1)] P(X = 3) = P(X≤3) – P(X≤2) = .92351 - .75136 = .17215

2. What is the probability that at least 3 families out of the random sample of 8 families had their credit card information stolen by hackers? (HINT: see the formula sheet for help using Cumulative Probability Distributions) P (at least 3 families) = P(X ≥ 3) Use the Formulas for Cumulative Probability Distributions of Discrete Random Variables from the formula sheet:

P(X ≥ x) = 1 – P[X ≤ (x-1)] P(X ≥ 3) = 1 – P(X≤2) = 1 –. 75136= .24864

Determine the mean and standard deviation of X. Round all answers to 3 decimal places. X is binomial with n = 8 and p =.22, so the mean of X = n*p = 8*.22 = 1.76 X is binomial with n = 8 and p =.22, so the standard deviation of X = �n*p*(1-p) = 1.1717 = 1.172

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Questions on appropriate notation and symbols a. According to national data on the sleeping habits of adults, the amount of sleep per night of all U.S. adults follows a normal distribution with a mean of 7.5 hours and a standard deviation of 1.2 hours. A study surveyed a random sample of 700 U.S. adults and found that their average amount of sleep per night was 6.85 hours with a standard deviation of 1.88 hours. Fill in each of the blanks below with the appropriate number based on the information listed above. n = _________700____________

σ = _________1.2___________

µ = _________7.5____________

s = __________1.88_________

฀฀ = ________6.85______________

b. According to data recorded over many years, 12% of the population in North America is left-handed. A random sample of 4,500 North Americans were surveyed and 500 of the survey responders reported that were lefthanded. Enter ฀฀ and p as decimals rounded to three decimal places.

n = ________4500_____________ ฀฀ = ________500/4,500 = .111____________ population proportion (p) = ______.12_______________

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Questions on normal random variables For the following three problems (1. – 3.), suppose the scores on an exam in a large class are normally distributed with a mean μ = 81 points (out of 100), and standard deviation σ = 5 points. 1) The instructor decides to set the "failing" score at 70, so anyone who scored below a 70 has a failing score. What proportion of exams will have failing scores? Show all work, use the provided Standard Normal Table (Z table), and show the proportion as a decimal rounded to four decimal places. We need to find the proportion of exams below 70, given that the mean exam score is μ = 81 and standard deviation σ = 5. Drawing a diagram can help. P (X < 70):

We convert from the given X value to a z-score. z-score for X = 70: z = (x - µ)/ σ = (70-81)/5 = -2.2 Round z-scores to 2 decimal places: z = -2.20 When we convert to a z-score, we are standardizing the x value. A Standard Normal, or Z, distribution is a normal distribution with µ = 0 and σ = 1. The shaded area below represents the probability we are looking for in a Standard Normal, or Z distribution. Note how it is the same area as the shaded area above for P (X < 70). The reason that we convert to z-scores is so that we can find the probability using a Z table. P (Z < -2.2)

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Then we use the Z-table to find the probability. P (Z < -2.2) = .0139

2) What is the exam score (in points out of 100) for an exam whose z-score is 2.25? Show all work and round your answer to two decimal places. z-scores are calculated by z = (x - µ)/ σ So if we know a z-score = 2.25, then 2.25 = (x-81)/5 x = 81 +2.25(5) = 92.25 3) What proportion of exams will have scores above 87? Show all work, use the provided Standard Normal Table (Z table), and show the proportion as a decimal rounded to four decimal places. We need to find the proportion of exams above 87, given that mean exam score is μ = 81 and standard deviation σ = 5. Drawing a diagram can help. The shaded area below represents the probability we are looking for: P (X > 87):

We convert from the given X value to a z-score. z-score for X = 87: z =(x - µ)/ σ = (87-81)/5 = 1.2 Round z-scores to 2 decimal places: z = 1.20 The shaded area below represents the probability we are looking for in a Standard Normal, or Z distribution. Note how it is the same area as the shaded area above for P (X > 87). P (Z > 1.2):

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Then we use the Z-table to find the probability. But remember that the Z table we use in STAT100 always gives us P (X < x). So we need to subtract the value in the Z-table from 1. P (Z > 1.2) = 1 – P (Z < 1.2) = 1 - .8849 = .1151 4) Which is the correct interpretation of a value x that has a z-score of -0.75 (negative 0.75)? (Highlight the correct answer) i.

There is 75% chance of getting a value smaller than x

ii.

There is 75% chance of getting a value larger than x

iii.

There is 75% chance of getting the value x

iv.

x is .75 standard deviations above the mean

v.

x is .75 standard deviations below the mean

5) The diameters of trees in a large forest are normally distributed with a mean diameter of 14.4 inches and a standard deviation of 4.6 inches. What is the probability that a randomly selected tree is between 10 and 13 inches in diameter? Show all work, use the provided Standard Normal Table (Z table) and give the probability as a decimal rounded to four decimal places. We need to find the probability that a randomly selected tree is between 10 and 13 inches in diameter, given that mean diameter of 14.4 inches and standard deviation is 4.6 inches. Drawing a diagram can help. The shaded area below represents the probability we are looking for: P (10 < X < 13):

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We convert from the given X values to z-scores. z-score for X = 10: z =(x - µ)/ σ = (10-14.4)/4.6 = -.9565 Round z-scores to 2 decimal places: z = -.96 z-score for X = 13: z = (x - µ)/ σ = (13-14.4)/4.6 = -.3034 Round z-scores to 2 decimal places: z = -.30 The shaded area below represents the probability we are looking for in a Standard Normal, or Z distribution. Note how it is the same area as the shaded area above for P (10 < X < 13). The reason that we convert to z-scores is so that we can find the probability using a Z table. P (-.96 < X < -.30):

We use the Z-table to find the probability by subtracting P (X < -.30) - P (X < -.96) A diagram may help to see why we do this: P (X < -.30) P (X < -.96)

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= P ( -.96 < X < -.30):

Then we use the Z-table to find the probability. P ( -.96 < X < -.30) = P (X < -.30) - P (X < -.96) = .3821 - .1685 = .2136 6) Suppose the scores on an exam in a statistics class are normally distributed with a mean μ = 78 points (out of 100), and a standard deviation of σ = 7 points. If a student received a 75 on the exam, what percentile would this test score be? (You can round the percentile to the nearest whole number). Show all work and use the provided Standard Normal Table (Z table).

We need to find the percentile for a test score of 75 if the mean exam score is μ = 78 and standard deviation σ = 7. So we basically need to find the proportion of exams less than 75. Drawing a diagram can help. The shaded area below represents the percentile for a test score of 75. P (X < 75):

We convert from the given X value to a z-score. z-score for X = 75: z = (x - µ)/ σ = (75-78)/7 = -4285 Round z-scores to 2 decimal places: z = -.43 The shaded area below represents the probability we are looking for in a Standard Normal, or Z distribution. Note how it is the same area as the shaded area above for P (X < 75). P (Z < -.43):

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Then we use the Z-table to find the probability. P (Z < -.43) = .3336 Then convert to a percent and round to the nearest whole number: = 33.36% = 33%. So a test score of 75 is in the 33rd percentile for this class.

7) Suppose high school GPAs of female students are normally distributed with a mean μ = 3.40 and a standard deviation σ = 0.50. High school GPAs of male students are normally distributed with a mean μ = 3.23 and a standard deviation σ = 0.54. Which would be more likely: a female student with a high school GPA of at least 4.0, or a male student with a high school GPA of at least 3.75? Show all work and use the Standard Normal Table. Probability of female GPA higher than 4.0: P(X>4.0) z=(4.0-3.40)/0.50=1.20 P(Z>1.20) = 1-0.8849=0.1151 Probability of male GPA higher than 3.75: P(X>3.75) z=(3.75-3.23)/0.54 = 0.9623 P(Z>0.96) = 1-0.8315=0.1685 Male GPA higher than 3.75 is more likely

8) Suppose the scores on an exam in a statistics class are normally distributed with a mean μ = 78 points (out of 100), and a standard deviation of σ = 7 points. Find the exam score that would be in the 67th percentile of exams scores in this statistics class. You can round the exam score to the nearest whole number. We need to find the exam score such that the probability of getting a score below it is 0.67. In other words we are solving for x such that P(X < x) = .67 Given the way the Z table is set up, we need to look in the body of the table for the table entry that is closest to 0.67 (which is 0.6700). Then we know that the exam score that we are looking for has a zscore of 0.44. So P (Z < 0.44) = .67. This means that the exam score that we are looking for is 0.44 standard deviations above the mean, and therefore x = 78 + .44 * 7 = 81.08 Page 14 of 35

Round to the nearest whole number. 81.08 -> 81 Thus an exam score of 81 is in the 67th percentile

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Questions on confidence intervals 1) A survey was given to a random sample of 283 American teens aged 16 to 17. Of the 283 surveyed teens,74 indicated that they had sent a text message while driving. Compute a 90% confidence interval for the population proportion of teens aged 16 to 17 who have sent a text message while driving. Show all work, and round all decimals that you use to three decimal places. 90% confidence interval for the population proportion p: ฀฀ ± 1.645�

฀฀�(1−฀฀�) ฀

฀

=(74/283) ± 1.645 �

(.2615(1−.2615) 283

= 0.2615 ± (.043)

= (.2185, .3045).

2) A random sample of U.S. college students found that 59 out of the 630 students surveyed were a member of a sorority or a fraternity. Compute a 95% confidence interval for the proportion of U.S. college students who are a member of a sorority or a fraternity. Show all work and round your answer to three decimal places. 95% confidence interval for the population proportion p: � + z* � ฀฀

�) ฀฀�(฀฀−฀฀ ฀

฀

(can use either z* = 2.0 or z* = 1.96)

฀฀ � = 59/630 = .094

.094 + 2 �

.฀฀฀฀฀฀(฀฀−.฀฀฀฀฀฀)) ฀ ฀

(.071, .117)

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= .071 and .117

3) The Law School Admission Test (LSAT) is an examination for prospective law school students. Scores on the LSAT are known to have a normal distribution and a population standard deviation of σ = 10. A random sample of 250 LSAT takers produced a sample mean of 502. Compute a 99% confidence interval for the population mean LSAT score. Show all work, and round all decimals that you use to three decimal places. 99% confidence interval for the population mean when σ is known: σ

฀฀ + 2.576*

√฀฀