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Practice Final Exam
From a May 2000 L.A. Times: " Pregnant women traveling to developing countries should take special precautions to ward off mosquitoes because they are twice as likely to be bitten by the sometimes- disease-carrying insects, according to British researchers. "Dr. Steve Lindsay and his colleagues at the University of Durham in northern England studied 36 pregnant women and 36 non-pregnant women in rural Gambia. Each night, three women from each group slept alone under a bed net. Researchers counted the number of mosquitoes attracted to each woman. The team reported in Saturday's Lance (http://www.thelancet.com) that the pregnant women attracted twice as many mosquitoes."
a) Is this a controlled study or an observational study, and why?
Observational: the "treatment" is pregnancy. The observed response variable
is the number of mosquitoes attracted to a woman. Because researchers did
not assign some woman to be pregnant and others not, this cannot be
classified as a controlled stud.
b) Suppose we let X be a random variable representing the number of mosquitoes attracted to a pregnant woman. Is X binomial? Why or why not?
No. Let's go through the steps.
1) Outcomes of trials are yes/no.
A "trial" here would be whether or not a mosquito is attracted to a
woman. So you could say that the mosquito either lands on the net or does
not.
2) There are a fixed number of trials. No. There could be any number of
mosquitoes on the net.
So it fails to be binomial.
- The table below shows the number of transfer applicants to UCLA in Fall '99, and the number of these that ended up enrolling in Fall '99. The data come from the Student Affairs Information & Research Office. Transfer applicants are students coming in from other colleges, including other UC's, Cal States, and community colleges.
Did not Enroll
Enrolled Total Applicants Native American 59 16 75 African American
Latino 289 106 395 Chicano 665 214 879 Asian 1916 595 2511 White 2780 870 3650 Other/unknown 788 231 1019 Total 6749 2098 8847
a) What is the probability that a randomly selected transfer student will enroll at UCLA?
Let A be the event "transfer student enrolls". There are 8847 transfer
students, and 2098 enrolled. Therefore P(A) = 2098/8847 =.
b) Let A be the event that a randomly selected transfer applicant is Chicano, and let B be the event that he or she enrolls. Find P(A), P(B).
P(A) = 879/8847 =.
P(B) = 214/879 =.
(Note: B is a somewhat ambiguous event. When I gave this final, it required
some elaboration. What it is asking is, what's the probability a randomly
selected Chicano applicant will enroll.
c) What is the probability a randomly selected transfer applicant is Chicano and will enroll?
P(Chicano and Enrolls) = 214/8847 =.
d) According to these data, are the events that the selected transfer applicant is Chicano mutually exclusive from the event that he or she enrolls? Or these events independent? Explain each of your answers.
Mutually exclusive: No. It is possible that a transfer applicant be both
Chicano and enroll. Otherwise, the answer to part (c) would be 0. It is not 0,
so these are not mutually exclusive.
Independent. If so, then P(Chicano and Enroll) = P(chicano) * P(enroll).
P(Chicano) = .0993. P(enroll) = .23714. So P(chicano)*P(enroll) = .0022.
But from part c, we found P(chicano and enroll) = .024. So these are not
equal, and therefore the events are NOT independent.
- Does a child's height at age two predict his height at age 18? The data below are based on 66 boys and come from "Physical Growth of California Boys and Girls from Birth to
a) (12 points) What is the regression line?
slope = SD(y) * r / SD(x) = 6.5175*.5708/3.3203 = 1.
intercept = ybar - slopexbar = 178.98 - 1.12088.371 = 80.
yhat = 80 + 1.120 x
or
height at 18 = 80 + 1.120 * height at 2 (measured in cm.)
b) Consider the children who are 85 cm tall at age two. What is the estimated average height of these children at age 18?
estimated height = 80 + 1.120 *85= 175.2 cm
c) Interpret the slope of the regression line in the context of this study.
Boys who are one cm taller than others at age two are, on average, 1.12 cm
taller than that "other" group at age 18.
- Equal time for girls: Here are summary statistics for the heights at age 18 (cm) for 70 girls as reported by the same study as in question (3). Variable Average Std. Dev Minimum Median Maximum Ht18 166.54 6.0749 153.6 166.75 183. First Quartile: 163 cm Third Quartile: 170.
150 160 170 180 190 Ht
a) Consider the height of the 18-year-old girls. Which, if either, is a better summary of the center of this distribution: the mean or the median. Why?
The distribution of this sample is fairly symmetric, so the sample median and
the average will be nearly equal. So it doesn't matter which you use.
b) Sketch the box plot for height at 18 years. Label important features clearly. (Are there any outliers? If so, you don't have enough information to say precisely where they are, so indicate whether or not there might be, if this is appropriate for this boxplot.)
Can't do this on the web. But your box would have a line through the center
at 166.75. The top of the box would be at 170.2, the bottom of the box at
163. The IQR is 170.2 - 163 = 7.2.
The top whisker will go at 170.2 + 1.5*IQR or at the maximum value,
whichever gives the shortest whisker. 170.2 + 1.5*IQR = 181.3, and this is
shorter than the maximum value (at 183.2). So the whisker goes at 181.3,
and there is at least one outlier.
The bottom whisker will go at 163 - 1.5*7.2 = 152.2 or the minimum, which
is 153.6. This time, the minimum value gives the shorter distance, so the
whisker goes at 153.6 and there are no outliers at this end of the box.
c) According to this data, how tall is a girl whose height in standard units is -1.4?
-1.4 means she is 1.4 SDs below average. 1 SD is 6.0749, so 1.4 is 8.
inches below average. So she is 166.54 - 8.59486 = 158.04 cm tall.
d) One inch is 0.394 centimeters (approximately). What is the average height in inches? What is the SD? (Two questions)
166.54 cm is 166.54 cm * (.394 cm/inch) = 65.6 inches
The SD is .394 * 6.0749 = 2.39 inches.
e) This study also recorded the girls' weight at age 2 and at age 18. Which do you think has the greatest SD: girls' weight at age 2 or girls' weight at age 18? Why?
Age 18 had the greatest SD. As people grow older, there is more time for them to diversify.
- From an April 2000 L.A. Times Poll: "The Rampart police corruption scandal is contributing to a malaise in Los Angeles, helping to raise questions about the city's health and image, devastating public impressions
a) What is the probability that a randomly selected two-year old will weigh more than 14 kilograms?
Let X represent the weight of a randomly selected two-year-old.
P(X > 14) = P(Z > (14-13)/1.6) = P(Z > .625) = 1 - P(Z <= .625) = 1 -.
Or, depending on how you round-off .625: 1-.7357 = .2643.
b) Take a random sample (with replacement) of 10 two-year olds. What is the expected value of their average weight?
Let Xbar represent the average of the 10 observations.
E(Xbar) = 13kg.
c) Take a random sample (with replacement) of 10 two-year olds. What is the SD of their average weight?
SD(Xbar) = sigma/sqrt(n) = 1.6/sqrt(10) = 0.5060 kg
Thus, we expect our average to be 13 kg, give-or-take .5060 kg.
MORE
d) Take a random sample (with replacement) of 10 two-year olds. What is the probability that the average weight will be more than 14 kg?
P(Xbar >= 14) = P(Z >= (14 - 13)/.5060) = P(Z >= 1.98) = 1 - P(Z < 1.98)
Note that the calculations are essentially the same as for part (a). Only the
SD of Xbar is smaller (.506 compared to 1.6), and so Xbar is less likely to be
far away from the mean value of 13.
e) Suppose we want the probability that the average weight of a random sample of two-year olds will be greater than 14 kg to be 1%. How many two-year olds should we sample (with replacement.)? In other words, we want P(avg > 14) = 0.01.
P(Xbar > 14) = 0.
means P( Z > (14 -13) / (1.6/sqrt(n)) ) = 0.
means P( Z < 1/(1.6/sqrt(n)) ) =.
From the table, we find that if z = 2.33, then P(Z < 2.33) = .99.
So it must be that 2.33 = 1/(1.6/sqrt(n))
Doing some algebra:
(2.33 * 1.6) = sqrt(n)
sqrt(n) = 3.
So n = (3.728)^2 = 13.898 so you need at least 14 observations.
The algebra might be less confusing if you write it out by hand, rather than
rely on reading what's on the screen.
- pH level is a scale that determines the acidity of a liquid. The scale ranges from 0 (very acidic) to 14 (no acidity). A score of 7 is considered "neutral." The mean pH level of patients who are having morning surgery is 3.94, with an SD of 2.51. This is important, because patients with a high levels of stomach acid (low pH) run the risk of getting something called Mendelson's Syndrome, an often fatal condition that results when anesthetized patients inhale the contents of their stomach. Do afternoon surgery patients have lower pH levels? (If so, they could be at greater risk than morning patients). A random sample of 49 afternoon surgery patients had an average pH of 2.93. Assume the population SD was still 2.51. Let represent the mean pH level of the population of afternoon patients.
a) Write the null and alternative hypotheses for this study.
H0: mean afternoon patients = 3.
Ha: mean afternoon patients < 3.
b) What is the observed value of the test statistic?
This is somewhat different from the problems we've done in class. In this
case, you are told that you know the population SD is 2.51. So there is no
need to estimate it with the sample standard deviation. (And you're not given
enough data to estimate it anyways.) So rather than use the T statistic, we
can use the Z-statistic:
Z = (Xbar - mu)/sigma/sqrt(n)
Z(observed) = (2.93 - 3.94)/(2.51/sqrt(49) = -2.
In "english", this means that the observed value of Xbar was 2.82 standard
deviations lower than what was expected.
c) What is the p-value for the test statistic?
p-value = P(Xbar < 2.93) = P(Z < -2.82) = 0.
d) Using a significance level of 5%, what would your conclusion be?
Reject the null hypothesis: there is evidence that afternoon patients have a
significantly lower mean ph level than morning patients. We reject because
the p-value is less than 0.05.
