Download Test 3 Practice Key

Elementary Statistics
Practice Test 3
Chapters 5, 6, 7, 8
Chapter 5
1. According to a study by Dr. John McDougall of his live-in weight loss program at St.
Helena Hospital, the people who follow his program lose between six and 15 pounds
a month until they approach trim body weight. Let’s suppose that the weight loss
is uniformly distributed. We are interested in the weight loss of a randomly selected
individual following the program for one month. Chapter 5 Practice 5 (edited).*
(a) X =
Solution: X = the amount of weight lost (in pounds) in one month by one of
the participants
(b) X ∼
Solution: X ∼ U (6, 15)
(c) Sketch the probability density curve for X. Be sure to label relevant values on the
x− and y−axes.
Solution:
(d) Find the 90th percentile for weight loss in a month and interpret it in a sentence.
Solution:
Shade an area of 0.9 on the left.
Area
= Length × Height
1
0.9
=
x
×
9
x = 0.9 ∗ 9 = 8.1 pounds over 6 pounds
6 + x = P90 = 14.1 pounds
Interpretation: 90% of participants lose less than 14.1 pounds in one month;
10% lose more than 14.1 pounds in one month.
(e) Find the probability that a person following the program loses more than 10 pounds
in a month.
Elementary Statistics
Practice Test 3 - Page 2
Solution:
Area
= Length ×
P (x > 10) = (15 − 10) ×
5
P (x > 10) =
9
Chapters 5, 6, 7, 8
Height
1
9
(f) Find the probability that a person following the program loses more than 10 pounds
in a month, assuming they lost at least 8 pounds.
Solution: Since we know the person lost at least 8 pounds, the new distribution
is X ∼ U (8, 15), which has a height of 17 .
P (x > 10) = length × height = (15 − 10) × 17 = 57 .
Chapter 6
2. The heights of the 430 National Basketball Association players were listed on team
rosters at the start of the 2005–2006 season. The heights of the basketball players have
an approximate normal distribution with mean of 79 inches (6 ft 7 in) and a standard
deviation of 3.89 inches. For each of the following heights, calculate the z-score and
interpret it using a complete sentence. Chapter 6 Homework 4 (edited).*
(a) 77 inches
≈ −0.51
Solution: z = 77−79
3.89
77 inches (6 ft 5 in) is about half a standard deviation shorter than the average
height of an NBA player.
Use 2 decimal places for z-scores.
(b) 85 inches
Solution: z = 85−79
≈ 1.54
3.89
85 inches (7 ft 1 in) is about one and a half standard deviations taller than the
average height of an NBA player.
(c) If an NBA player reported that his height had a z-score of 3.5, would you believe
him? Explain your answer by calculating an appropriate probability. How tall
would he be?
Solution: No. The probability that a data value from a normal distribution is at least 3.5 standard deviations above the mean is P (z > 3.5) =
normalcdf(3.5, 109 , 0, 1) ≈ 0.0002, which is very, very unlikely (2 in 10,000).
The player would be 79 + (3.5)(3.89) ≈ 93 inches = 7 ft 9 in tall.
(d) According to the Empirical Rule (or the 68-95-99.7 Rule), 99.7% of NBA players
are between
and
. Convert your answer to feet and inches. Round
Elementary Statistics
Practice Test 3 - Page 3
Chapters 5, 6, 7, 8
to the nearest inch.
Solution: According to the Empirical Rule, 99.7% of data values are within
3 standard deviations from the mean: between 79 − 3(3.89) ≈ 67.3 inches and
79 + 3(3.89) ≈ 90.7 inches, or between 5 ft 7 in and 7 ft 7 in.
3. The percent of fat calories that a person in America consumes each day is normally
distributed with a mean of about 36 and a standard deviation of 10. Suppose that one
individual is randomly chosen. Let X = percent of fat calories. Chapter 6 Homework 16 (edited).*
(a) X ∼
Solution: X ∼ N (36, 10)
(b) Find the probability that the percent of fat calories a person consumes is more than
40. Graph the situation. Shade in the appropriate area. Interpret your answer.
Solution: P (x > 40) = normalcdf(40, 109 , 36, 10) = 0.345
The diagram should be a bell curve centered a the mean, 36. A vertical line
should be drawn at 40, with the area under the curve to the right of 40 shaded.
Interpretation: 34.5% of Americans consume more than 40 fat calories per day.
(c) Find the maximum number for the bottom quartile of fat calories. Round to the
nearest calorie. Sketch the graph and write the probability statement. Interpret
what you have found.
Solution: Q1 = P25 = invnorm(0.25, 36, 10) ≈ 29 fat calories.
The graph should be a bell curve centered at 36. Draw a vertical line to the
left of the mean. Label the area to the left of the vertical line: 0.25. Label the
point where the vertical line touches the x-axis: x = P25 .
Interpretation: 25% of Americans consume less than 29 fat calories per day.
(d) Ten percent of Americans consume more than
fat calories each day.
Solution: We need the boundary value that has 10% of data values above it,
which means 90% of data values are below it. In other words, we need P90 =
invnorm(0.9, 36, 10) ≈ 49.
Chapter 7
4. Random variable X has mean µ = 34 and standard deviation σ = 4. The distribution
of X is unknown. Random variable X = the mean of a sample of size eight from X.
Can we apply the Central Limit Theorem to determine the distribution of X? If not,
explain why not. If so, provide the distribution of X.
Elementary Statistics
Practice Test 3 - Page 4
Chapters 5, 6, 7, 8
Solution: No. Because the distribution of X is unknown and the sample size (eight)
is small (less than 30), we cannot determine the distribution of X. (According to
the Central Limit Theorem, X is normally distributed ONLY IF X is normally
distrbuted OR if the sample size is “large enough” (30 or more).)
5. According to the Internal Revenue Service, the average length of time for an individual
to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040
is 10.53 hours. Let us assume that the completion times are normally distributed with
a standard deviation of two hours. Suppose we randomly sample 36 taxpayers.
(a) In words, X =
Solution: X = the length of time for a single individual to complete IRS form
1040
(b) In words, X =
Solution: X = the mean length of time for a sample of 36 individuals to
complete the form
(c) X ∼
√
Solution: X ∼ N (10.53, 2/ 36) = N (10.53, 1/3)
(d) Would you be surprised if one taxpayer finished his or her Form 1040 in more than
12 hours? In a complete sentence, explain why. Justify your answer with numerical
evidence.
Solution: The probability that one taxpayer finished his or her form in more
than 12 hours is P (x > 12) = normalcdf(12, 109 , 10.53, 2) ≈ 0.231. There is
about a 23% chance that a single taxpayer takes longer than 12 hours, which
isn’t very unlikely at all, so NO, I would not be surprised.
(e) Would you be surprised if the 36 taxpayers finished their Form 1040s in an average
of more than 12 hours? Explain why or why not in complete sentences. Justify
your answer with numerical evidence.
Solution: The probability that 36 taxpayers finished their form in an average of
more than 12 hours is P (x > 12) = normalcdf(12, 109 , 10.53, 1/3) ≈ 5 × 10−6 ≈
0.000. Because the probability of this happening is so low (close to 0% chance),
YES I would be surprised if 36 taxpayers took longer than 12 hours on average.
(f) If a random sample of 36 taxpayers took 13 hours on average to complete the form,
what would that imply about the quoted mean of 10.53 hours? Explain.
Elementary Statistics
Practice Test 3 - Page 5
Chapters 5, 6, 7, 8
Solution: If the mean is really 10.53 hours, it is extremely unlikely that 36
taxpayers would take so long to complete the form on average based on the
probability found in part (e). There is statistical evidence that the mean time
is actually longer than 10.53 hours.
Chapter 8
6. For each scenario, are the methods of inferential statistics that we have learned in chapters 8 and 9 appropriate? If not, explain. If so, identify the (i) margin of error formula,
(ii) confidence interval calculator function and only if you have already studied chapter
9, (iii) hypothesis test calculator function to be used in each case.
(a) We are interested in the average length of time elementary school students are on
the school bus each day. Our sample of 250 ride lengths appears to be normally
distributed.
Solution: (i) EBM = t-score √sn (ii) TInterval (iii) TTest
(b) We are interested in the percentage of elementary school students who ride the bus
each day. In a preliminary study, 70 out of 97 students were bus riders.
q
0 0
Solution: (i) EBP = z-score pnq (ii) 1-PropZInt (iii) 1-PropZTest
(c) We are interested in the average number of penalties committed in a Champions
League soccer match. Our sample of 9 penalty counts appears to be skewed right.
Solution: Our methods of inferential statistics are inappropriate. The requirements for the Central Limit Theorem are not satisfied: the sample is small and
the sample does not appear to come from a normally distributed population.
(d) We are interested in the mean IQ of nursing students. We know that IQs are
normally distributed with a standard deviation of 15. We have a sample of 20
nursing student IQs.
Solution: (i) EBM = z-score √σn (ii) ZInterval (iii) ZTest
(e) We are interested in the proportion of aspiring actors who land a major role. Our
sample of 500 actors includes 3 who have landed a major role.
Solution: Our methods of inferential statistics are not appropriate. The sample
3
proportion p0 = 500
= 0.006 is very small, so this binomial distribution cannot
be approximated by a normal distribution.
Elementary Statistics
Practice Test 3 - Page 6
Chapters 5, 6, 7, 8
7. The standard deviation of the weights of elephants is known to be approximately 15
pounds. We wish to construct a 95% confidence interval for the mean weight of newborn
elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds.
The sample standard deviation is 11 pounds. Chapter 8 Practice 1-5 (edited).*
(a) Identify the following: x̄, σ, s, n, CL, α.
Solution: x̄ = 244 lb, σ = 15 lb, s = 11 lb, n = 50, CL = .95, α = 0.05
(b) What is the best point estimate for the mean weight of a newborn elephant?
Solution: µ ≈ x̄ = 244 lb
(c) Find the z-score or t-score that you would need for the margin of error formula for
a 98% confidence interval. Label as a z-score or t-score. Use 3 decimal places.
Solution: z-score = invnorm(.95+.025) = 1.960 (We should use a z-score because we are estimating a population mean µ and population standard deviation
σ is known.)
(d) Find the margin of error for a 95% confidence interval estimate of the mean weight
of a newborn elephant.
Solution: EBM = (z-score) √σn = 1.960
√15
50
≈ 4.158.
(e) Express the 95% confidence interval estimate for the mean weight of a newborn
elephant using ± notation. Round to one decimal place as needed.
Solution: 244 ± 4.2 pounds
(f) Express the 95% confidence interval estimate for the mean weight of a newborn
elephant using interval notation. Round to one decimal place as needed.
Solution: (239.8, 248.2) pounds
(g) Use the appropriate calculator function to check your answer to the previous part.
Solution: Use ZInterval with the “STATS” option: σ = 15, x̄ = 244, n =
50, c − level = 0.95.
(h) Write a sentence interpreting the confidence interval that you have found.
Solution: With 95% confidence, the mean weight of a newborn elephant is
between 239.8 and 248.2 pounds.
Elementary Statistics
Practice Test 3 - Page 7
Chapters 5, 6, 7, 8
(i) What will happen to the confidence interval obtained if 500 newborn elephants are
weighed instead of 50? Why?
Solution: The confidence interval will be narrower. With a larger sample, the
estimate will be more precise because there is more data to draw from.
8. A hospital is trying to cut down on emergency room wait times. It is interested in the
the mean amount of time patients must wait before being called back to be examined.
An investigation committee randomly surveyed 70 patients. The sample mean was 1.5
hours with a sample standard deviation of 0.5 hours Chapter 8 Practice 38-42 (edited).*
(a) Are we estimating a population mean µ or population proportion p?
Solution: population mean µ
(b) Find the z-score or t-score that you would need for the margin of error formula for
a 98% confidence interval. Label as a z-score or t-score. Use 3 decimal places.
Solution: t-score = 2.382. Method 1: invT(0.98+0.02/2, 69) = invT(0.99, 69)
= 2.382. Method 2: If your calculator doees not have an invT function find the
t-score on the T-distribution chart corresponding to a cumulative area of 0.99
and 69 degrees of freedom.
(c) Use the formula to find the margin of error. Construct a 98% confidence interval
for the population mean time spent waiting. Round your ANSWER to the nearest
tenth of an hour (one decimal place).
≈ 0.14235. The confidence
Solution: EBM = (t-score) √sn = 2.382 √0.5
70
interval is 1.5 ± 0.1 hours or (1.5 − 0.1, 1.5 + 0.1) = (1.4, 1.6) hours.
(d) Use the appropriate calculator function to check your answer to the previous part.
Solution: Use TInterval with the “STATS” option.
(e) Interpret the confidence interval you have found.
Solution: With 98% confidence, the mean emergency room wait time is between 1.4 hours and 1.6 hours.
9. The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.2
15.0
15.5
15.7
15.9
15.0
15.5
15.7
(a) Find the best point estimate for the mean amount of juice in all such bottles.
Elementary Statistics
Practice Test 3 - Page 8
Chapters 5, 6, 7, 8
Solution: µ ≈ x̄ = 15.44 ounces. (The best point estimate for population
mean is sample mean. We use 2 decimal places since the data has one decimal
place.)
(b) Construct a 95% confidence interval for the mean amount of juice in all such bottles. Express your answer using interval notation and “±” notation. Assume the
population is normally distributed.
Solution: (15.16, 15.72) ounces or 15.44 ± 0.28 ounces.
Method 1: Calculate EBM = t-score √sn then construct the confidence interval:
(x̄ − EBM, x̄ + EBM ) and x̄ ± EBM .
Method 2: Use TInterval with the “DATA” option to get the answer in interval
notation. Subtract the endpoints of the interval and divide by 2 to get the
EBM.
10. A poll of 1,200 voters asked what the most significant issue was in the upcoming election.
Sixty-five percent answered the economy. We are interested in the population proportion
of voters who feel the economy is the most important. Chapter 8 Practice 74-78 (edited).*
(a) Find the best point estimate for the percentage of voters who feel the economy is
the most important issue.
Solution: p ≈ p0 = 0.65 or 65%. (The best point estimate for population
proportion is sample proportion.)
(b) Find the number of voters out of the 1,200 polled who feel the economy is the most
important issue.
Solution: x ≈ np0 = 1200(0.65) = 780 out of 1,200 voters feel the economy is
the most important issue.
(c) Find a 90% confidence interval for the percentage of all voters who feel the economy
is the most important issue. Use one decimal point for percentages.
Solution: (62.7%, 67.3%) or 65 ± 2.3%.
q
q
(0.65)(0.35)
p0 q 0
Method 1: Calculate EBP = (z-score) n = (1.645)
.
1200
Method 2: Use 1-PropZInt with x = 780, n = 1200, and CL = 0.90.
(d) What would happen to the confidence interval if we chose to use 95% confidence
instead of 90% confidence? Explain.
Elementary Statistics
Practice Test 3 - Page 9
Chapters 5, 6, 7, 8
Solution: The confidence interval would be wider. In order to be more confident we are capturing the true population proportion, we must include a wider
range of values around the point estimate. This increase in the margin of error
comes from increasing the z-score in the EBP formula.
11. How many students must be randomly selected to estimate the mean weekly earnings
of U.S. college students? We want 99% confidence that the sample mean is within $5 of
the population mean, and the population standard deviation is known to be $63. Show
your work.
Solution: 1054 students. Use the margin of error formula EBM = (z-score) √σn . Set
EBM = 5, σ = 63 and z-score = invnorm(0.995) = 2.576 and solve for n. Remember
to round sample size UP.
12. A political polling organization wishes to estimate the percentage of voters who favor
a particular candidate in the presidential race. How many voters would need to be
sampled to have 95% confidence that the estimate is within 6 percentage points of the
actual percentage of all voters? Show your work.
q
0 0
Solution: 267 voters. Use the margin of error formula EBP = (z-score) pnq . Set
EBP = 6% = 0.06, z-score = invnorm(0.975) = 1.960, p0 = q 0 = 0.5, and solve for n.
Remember to round sample size UP.
*Problems are from Barbara Illowsky & Susan Dean.
https://itun.es/us/kFeL1.l
“Introductory Statistics.” OpenStax College, 2013.
iBooks.