Note 1: Exam has been moved to Thursday. (Students who want to take the exam on Tuesday need to show up at the regular class time.)

Note 2: These are just some exam questions for you to practice on. The exact difficulty and make-up of your actual exam may vary.

Practice Exam Answers

1) Assume that you are applying for three jobs, one with the firm of Smith, Carlyle, and Snookums, one with the firm of Dewy, Cheatham, and Howe, and the last with the firm House of Pancakes. Assume the probability that you get a job offer from Smith et al. is 0.3, the probability of getting an offer from Dewey et al. is .6, and the probability of getting an offer from House et al is 0.8.

a) What is the probability that you will get a job offer from Smith et al or Dewey et al?

Event A: Job offer from Smith, Event B: Job offer from Dewey Event C: Job offer from HOP

P(A) = .3, P(B) = .6, P(C) = .8

P(A or B) = P(A)+ P(B)- P(A&B) = .3 +.6 - (.3 x .6) = .9 - .18 = .72

b) What is the probability that at least one job offer if you apply at all three firms.
P(A or B or C) = 1 -P(not A & not B & not C) = 1 -(.7 x .4 x .2) = 1 - .056 = .944

c) For this part of the question forget about the House et al job. Consider only the Smith and Dewey jobs. Assume that you prefer to work at Smith, and will take that job if offered. If you aren't offered a job at Smith, but are offered one at Dewey you will take the Dewey job. What is the probability that you will end up working at Dewey et al?

Since events are independent:

P(not A & B) = P(not A) x P(B) = .7 x .6 = .42

2) (Challenging.) A brand of flashlight battery has normally distributed lifetimes with a mean of 30 hours and a standard deviation of 5 hours. A supermarket purchases 500 of these batteries from the manufacturer. What is the probability that at least 80% of them will last longer than 25 hours?

First must get the probability of a random battery lasting longer than 25 hours:

z = (30 -25) / 5 = -1 Prob(z > -1) = .8413

Then we need the probability of 400 (80% of 500) or more "successes" in 500 trials, with prob. of success for each trial equal to .8413. Use normal approximation to the binomial:

Mean = n x p = 500 x .8413 = 420.65

Standard deviation = sqrt (n x p x (1 - p)) = sqrt (500 x .8413 x .1587) = 8.17

z = (400 - 420.65) / 8.17 = - 2.53

Prob. of z > -2.53 = .9943

3) Consider a random variable X whose population distribution is given by the following probability histogram:

The mean and standard deviation for this population (assume the population is infinite) are given as follows:

m_x= 7.22 ; s _x = 5.11

(a) For a random sample of size n = 147 from this population, the Central Limit Theorem can be used to determine the (approximate) sampling distribution of the sample mean. State the shape of this distribution, and provide its mean and standard error.

Normal with mean = 7.22 and standard error = 5.11/sqrt(147) = .421

(b) Now, pretend that the mean of the above mentioned population is unknown. For the random sample of size n = 147 above, you find that the sample mean is 6.33. How far off would you expect this estimate to be from the true population mean?

.421

(c) Knowing what the real population mean is, do you think the value of the sample mean you observed in (b) is unusual? Briefly support your answer.

(7.22 -6.33)/.421 = 2.11, so yes, obtaining a sample mean more than 2 standard errors away from the population mean is unusual. It happens .0348 of the time according to Table A.

4) In a family of 8 children, what is the chance that there will be exactly two boys? Assume that the gender of each child is equally likely to be male or female. (This is not a trick or a paradox! It is a straightforward probability problem.)

Binomial with p = .5, n = 8, and k = 2.

(8! / 2!6!) x .5² x .5⁶ = 28 x .25 x .0156 = .109

5) In a certain city, 80% of all drivers have auto insurance. Those who have auto insurance have a 2% chance of being involved in an accident, while those drivers without insurance have an accident rate of only 7%. Suppose a driver hits you. What is the probability that this driver has auto insurance?

Event A: driver has insurance Event B: driver has accident

P(A) = .80 P(B|A) = .02 P(B|not A) = .07

We want P(A|B).

Bayes rule: P(A|B) = P(B|A)P(A) / {P(B|A)P(A) + P(B| not A)P(not A) = (.02)(.80) / {(.02)(.80) + (.07)(.20) = .16 / {.16 + .14} = .533

6) Which hypothesis- the null or the alternative- is the researcher usually most interested in supporting? Which hypothesis does he end up testing? Why?

Researcher is usually interested in supporting the alternative hypothesis which corresponds to a real effect, difference, or relationship. The researcher tests the null hypothesis because it is a point hypothesis that is readily tested, unlike the alternative, which is actually a composite of a whole range of different possibilities

7) If you change alpha from .05 to .10, what effect will that have on the power of your study?

It will increase your power since it will make it easier to reject the null.

8) Mary believes that brown-eyed women like her tend to be smarter than the average person. She randomly selects 5 brown-eyed women and gives them an IQ test. Their IQ. s are: 105, 112, 98, 122, and 103. Is this evidence enough to show intelligence superiority for brown-eyed women, that is, that their mean IQ is different than the national mean of 100? (Use a = .05.)

mean = 108, s = 9.30, standard error = 9.30 /sqrt(5) = 4.16

t = (108 - 100) /4.16 = 1.92

critical value of t (df = 4) for a = .05 is 2.776.

Since |1.92| < 2.776 we fail to reject null, and cannot conclude that brown-eyed women have superior intelligence.

9) Compute the 99% confidence interval for the mean IQ of brown-eyed women based on the data from question 8.

95% CI = 108 +/- 4.604 (4.16) = 108 +/- 19.15 = 88.85 to 127.15

10) What are sampling distributions and what use are they?

Sampling distributions are theoretically derived probability distributions of sample statistics that would obtain if the null were true. They provide a benchmark against which we can compare our sample statistic to see if it is consistent with the null.

11) What is the difference between a standard deviation and a standard error?

A standard deviation is a numerical measure of the variability in scores for individual observations. The standard error is a measure of the variability expected in a sample statistic (like a mean or sample proportion) if samples of a certain size are repeatedly drawn.

12) You read a newspaper article that states that researchers have found that daily meditation causes a decrease in blood pressure for people with high blood pressure. Assume that the researchers did an excellent job of designing a flawless double-blind experimental study with a very large, representative sample, and their results were significant for a = .01. Could you legitimately conclude that meditation is an effective intervention for high blood pressure? If so, why? If not, what additional information would you want?

You would want to know the size of the effect- exactly how much was blood pressure lowered on average by meditation. Just because a finding is statistically significant does not mean that it is a large enough effect to be of practical importance.

13) If the random variable X has a m = 10 and a s =5, and the random variable Y has a m=20 and a s = 5, what are the values of m and s for the random variable Z = X -Y? What about for the random variable V = X + Y?

Mean(X-Y) = 10-20 = -10, s_X-Y = sqrt(5² +5²) = 7.07

Mean(X+Y) = 10+20 = 30, s_X+Y = sqrt(5² +5²) = 7.07