Math 115 Statistics: Concepts and Applications

Math 115 Statistics: Concepts and Applications

Test 3

(84 points) Read each problem carefully. Write clearly. Good luck!

SOLUTIONS

I. (12 points) Suppose you choose a Normfield household at random and ask how many cars that household owns. Here are the probabilities:

Number of cars 0 1 2 3 4

Probability0.060.250.450.180.06

(a) What’s the probability that a Normfield household has at most two cars?

.06 + .25 + .45 = 0.76.

(b) What is the expected number of cars owned by a Normfield household?

0*.06 + 1*.25 + 2*.45 + 3*.18 + 4*.06 = 1.93

(c) Normfield has about 2,500 households. If all the cars in Normfield are owned by households, about how many cars are there in Normfield?

2500 x 1.93 = 4,825 cars.

(d) There are five households on Penny Lane in Normfield. What’s the probability that every household on Penny Lane owns at least one car?

The probability that one household owns at least one car is 1-0.06 = .94. The probability

That five households own at least one car (assuming independence) is .94^5 = 0.73.

II. (17points) It is claimed that a majority of all college students oppose the death penalty. A poll of 650 students chosen at random finds that 349 students say that they are opposed to the death penalty.

(a) Identify the parameter and the statistic.

The parameter is the proportion of all college students who oppose the death penalty. In symbols, it is p. The statistic is the sample proportion, p-hat, the proportion of the students in the sample who oppose the death penalty. For our study, p-hat = 349/650 = 0.537.

(b) Compute a 95% confidence interval for the proportion of college students who oppose the death penalty.

The standard deviation here is = 0.02. The 95% confidence interval is or (.50, .58).

(c) Interpret your confidence interval in words that your roommate who has never taken statistics can understand.

We are 95% confident that the true proportion of college students who oppose the death penalty lies between 50% and 58%. The real meaning of this statement is that we have constructed the confidence interval using a procedure which, in repeated samples, will capture the true parameter value 95% of the time.

(d) Your confidence interval gives STRONG evidence to support the claim that a majority of students oppose the death penalty.

(e)Suppose another poll is taken of 650 different students (chosen randomly from the same population). This time a 90% confidence interval is computed. Compared to the first interval you would expect that this new interval would be NARROWER.

III.(14 points) A nationwide random survey of 256 adults asked about attitudes toward “alternative medicine” such as acupuncture, massage therapy, and herbal therapy. Among the respondents, 92 said they would use alternative medicine if traditional medicine was not producing the results they wanted. Does this survey provide good evidence that more than 1/3 of all adults would use alternative medicine if traditional medicine did not produce the results they wanted?

(a) State the hypotheses to be tested.

Ho: p = 1/3 versus Ha: p > 1/3.

(b) If your null hypothesis is true, what is the sampling distribution of the sample proportion p-hat? Sketch this distribution.

The sampling distribution is a normal model centered at the mean of 1/3 and with standard deviation equal to =0.03.

(c) Compute the P-value for this test as best you can without using a table.

The sample proportion is 92/256 = 0.36. Thus the test statistic value is (.36 - .33)/.03 = 1.

The P-value is the probability of obtaining a result that’s one standard deviation to the right of the mean. By the 68-95-99.7 rule that probability is equal to (1-.68)/2 = 16% (draw the picture!).

(d) Interpret the result. State your conclusion in the context of the problem and in a way that someone who does not know statistics can understand.

This is not a “small” P-value. There is not enough evidence to reject the null hypothesis. The survey does not provide good evidence that more than 1/3 of all adults would use alternative medicine.

IV. (15 points) A nursery claims that 65% of its dogwood seedlings survive when planted under normal conditions. You plant three such seedlings.

(a) Explain how to simulate the growth of three seedlings if the nursery’s claim is correct.

First simulate one seedling. Let 00-64 represent survival and let 65-99 represent death. Take three pairs of numbers to represent planting three seedlings

(b) You would like to know the probability that at least two of the three seedlings survive. Use the following entries from a random digits table

9629907196986422063923185562826992914125

4574041807655613330207051936231813209547

and 10 repetitions to estimate this probability.

The simulation is

DSD / DDD / SSS / SSS / SSD / DDD / SSS / DSS / SDS / SSS with numbers surviving

1 / 0 / 3 / 3 / 2 / 0 / 3 / 2 / 2 / 3.

Of the ten simulations, 7 had at least two survivals. So the estimated probability is 7/10 = 70%.

(c) Use your simulation in (b) to estimate the expected number of seedlings that survive (when three are planted).

The estimated expected value is (1 + 0 + 3 + 3 + 2 + 0 + 3 + 2 + 2 + 3) / 10 = 1.9.

(d) If you were able to do 10,000 repetitions, instead of 10, your estimate in (c) would be much closer to the true value. This is a consequence of what famous law?

The law of large numbers.

(e) (Extra Credit) Find the (exact) expected number of seedlings that survive.

We first need to find the probabilities corresponding to 0, 1, 2 and 3 seedlings surviving. These are

0 / 1 / 2 / 3
.35^3 = .04 / 3 (.35^2 )( .65) = .24 / 3(.35)(.65^2) = .44 / .65^3 = .27

The exact expected value is 0*.04 + 1*.24 + 2*.44 + 3 * .27 = 1.93.

V. (15 points) SHORTS.

A. A poker player is dealt poor hands for several hours. He decides to bet heavily on the last hand of the evening on the grounds that after many bad hands he is due for a winner.

(a) He’s right, because the winnings have to average out.

(b) He’s wrong, because successive deals are independent of each other.

(d) He’s wrong, because his expected winnings are $0 and he’s below that now.

B. Which is more likely: to get exactly 50,000 heads in 100,000 fair coin tosses or to get exactly 50 heads in 100 fair coin tosses?

50 heads in 100 fair coin tosses.

C. George wants to know the proportion of Carleton students who come from Texas. He calls up the college office which keeps that information and learns that the proportion is 10%. George then computes the standard deviation of 0.007 (based on 1,800 students) and reports a 95% confidence interval for the proportion as 0.10 ± .014. Has George done anything wrong?

George already has the parameter! There is no need to do any inference!

D. In statistical inference, what’s the advantage of a confidence interval over a hypothesis test?

A confidence interval is more informative than a test because it is actually estimating a population parameter. It is easier to interpret. It gives a measure of the magnitude of the parameter rather than just a yes-no decision.

E. The prosecutor’s fallacy arises from

(i) Confounding of variables

(ii) Misinterpreting conditional probabilities

(iii) Confusing statistical significance with practical significance

(iv) Assuming events are independent when they are not.

(v) Simpson’s paradox

VI. (10 points) In a recent survey, 933 people were asked two questions: Do you favor or oppose the death penalty for persons convicted of murder? Do you think the use of marijuana should be made legal or not? The results are shown in this table. We wonder if there is evidence of an association between attitudes on the death penalty and attitudes toward legalizing marijuana.

Marijuana should
be legal / Marijuana should
not be legal
Favor death penalty / 152 / 561
Oppose death penalty / 61 / 159

(a) If attitudes toward legalizing marijuana had no relationship to attitudes on the death penalty, of the 933 people, how many would you expect to both favor the death penalty and favor legalizing marijuana?

First take the marginal row and columns sums and then the overall total to get

(152+561) x (152+61) / 933 = 162.7749. About 163 people.

(b) Here is the SPSS output from the chi-square test.

Draw a picture which shows the sampling distribution of the chi-square statistic, the observed value of the chi-square statistic, and the P-value.

The picture would show the right-skewed chi-square distribution curve. The observed value is 3.92. The area under the curve to the right of that point is the P-value, which is 0.048.

The P-value is small, and less than 5%. We reject the null hypothesis. There is strong evidence that there is a relationship between attitudes toward the death penalty and attitudes toward legalizing marijuana.