TC3 → Stan Brown → Statistics → Practice Problems—Solutions
revised 1 Aug 2012

# Solutions to Practice Problems for Statistics

Summary: These are answers to the Practice Problems for Statistics, with a few comments.

## Section A: Concept Questions

Write your answer to each question. There’s no work to be shown. Don’t bother with a complete sentence if you can answer with a word, number, or phrase.

1Two events A and B are disjoint. Is it possible for those same events to be independent as well? Give an example, or explain why it’s impossible.

Answer: Disjoint events cannot be independent. Why? Disjoint events, by definition, can’t happen on the same trial. That means if A happens, P(B) = 0. But if A and B are independent, whether A happens has no effect on the probability of B. With disjoint events, whether A happens does affect the probability of B. Therefore disjoint events can’t be independent.

2Gasoline pumped from a supplier’s pipeline is supposed to have an octane rating of 87.5. To test this, a random sample was taken on 13 consecutive days and the octane measured in a lab.

(a) The data would best be analyzed as an example of
A. one population proportion
B. two populations, difference in proportions
C. one population mean
D. two populations, difference in means, paired data
E. two populations, difference in means, unpaired data
F. goodness of fit
G. contingency table

(b) Which two tests must you perform on your sample data before doing the analysis mentioned above? (In other words, how would you make sure that the sample meets the requirements?)

(b) For numeric data with sample size under 30, you check for outliers by making a box-whisker plot and check for normality by making a normal probability plot.

3The two main types of data are qualitative and quantitative. Give the shorter name for each, and give an example of each.

Answer: qualitative = attribute. Examples: political party affiliation, gender
quantitative = numeric. Examples: height, number of children

Common mistake: Binomial, categorical, discrete, and continuous are subtypes of the above data types, but are not shorter names for them, just as Fresno is not a shorter name for California.

4The probability of rolling a 6 on an honest die is 1/6. If you roll an honest die ten times and none of the rolls comes up 6, is the probability of rolling a 6 on the next roll less than 1/6, equal to 1/6, or greater than 1/6? Explain why.

Answer: equal to 1/6. The die has no memory: each trial is independent of all the others.

The Gambler’s Fallacy is believing that the die is somehow “due for a 6”. The Law of Large Numbers says that in the long run the proportion of 6’s will tend toward 1/6, but it doesn’t tell us anything at all about any particular roll.

5In a large elementary school, you select two age-matched groups of students. Group 1 follows the normal schedule. Group 2 (with parents’ permission) spends 30 minutes a day learning to play a musical instrument. You want to show that learning a musical instrument makes a student less likely to get into trouble. You consider a student in trouble if s/he was sent to the principal’s office at any time during the year.
(a) Write your hypotheses, in symbols.
(b) Identify either the case number or the specific TI-83 test you would use.

Answer: (a) pop. 1 = control, pop 2 = music
H0: p2 = p1 and H1: p2 < p1
or: H0: p2 – p1 = 0 and H1: p2 – p1 < 0
(or any equivalent statements)
(b) Case 5, Difference between Two Pop. Proportions; or 2-PropZTest

Common mistake: You must specify which is population 1 and which is population 2.

Common mistake: The data type is binomial: a student is in trouble, or not. There are no means, so μ is incorrect in the hypotheses.

6Imagine rolling five standard dice. You compute the probability of rolling no 3s, one 3, and so on up to five 3s. Is this a binomial probability distribution? With reference to the definition of a binomial PD, why or why not?

Answer: A binomial PD is one where (a) there are a fixed number of trials; (b) there are only two outcomes, success and failure; and (c) the probability of success is the same from trial to trial (or the trials are independent).

(a) You roll five dice: five trials (n=5).  (b) Success is “rolling a 3” and failure is “rolling anything other than 3”.  (c) The dice are independent: p = 1/6 for each die regardless of what turns up on any other die. Therefore this is a binomial PD.

7Over the course of many statistical experiments, which one of these values for the significance level would enable you to prove the most results?
A. 5%        B. 1%        C. 0.1%        D. Significance level has no effect on how likely you are to prove a hypothesis.

Remark: The significance level α is the level of risk of a Type I error that you can live with. If you can live with more risk, you can reach more conclusions.

8A key step in hypothesis testing is computing a p-value and comparing it to your preselected α. After you do that, which of the following conclusions would be possible, depending on the specific values of p and α? (Write the letter of each correct answer; there may be more than one.)
A. Accept H0, reject H1
B. Reject H0, accept H1
C. Fail to accept H0, no conclusion
D. Fail to reject H0, no conclusion

Answer: B,D — B if p<α, D if p>α

9 Distinguish disjoint events, mutually exclusive events, and complementary events. Give an example of each.

Answer: “Disjoint” means the same as “mutually exclusive”: two events that can’t happen at the same time. Example: rolling a die and getting a 3 or a 6.

Complementary events can’t happen at the same time and one or the other must happen. Example: rolling a die and getting an odd or an even. Complementary events are a subtype of disjoint events.

10 When is a histogram an appropriate graphical method of presentation?

Answer: For any numeric data set of moderate to large size. If the variable is discrete with only a few different answers, you would use an ungrouped histogram. For continuous data, or discrete data with many different answers, you would use a grouped histogram.

For a small set of numeric data, you might prefer a stemplot.

11For what type of events does P(A or B) = P(A) + P(B)? Give an example.

Answer: For mutually exclusive (disjoint) events. Example: if you draw one card from a standard deck, the probability that it is red is ½. The probability that it is a club is ¼. The events are disjoint; therefore the probability that it is red or a club is ½+¼ = ¾.

12In a χ² goodness-of-fit test, which of the following is/are true?
(A question with this many technical alternatives will not be on the exam. Just use it to test your own understanding of χ².)
A. The hypotheses are stated in words rather than relating some population parameter to a number.
B. The null hypothesis is always some variation on “the observed sample matches the model reasonably closely.”
C. The alternative hypothesis is always some variation on “our model is good.”
D. Instead of a p-value, we compare the value of χ² to α to draw a conclusion.
E. Degrees of freedom equals the number of cells in our model.
F. If the difference between our observed results and our expected results could likely have occurred by random chance, we reject the null hypothesis.

Remark: C is wrong because “model good” is H0. D is also wrong: every hypothesis test, without exception, compares a p-value to α. For E, df is number of cells minus 1. F is backward: in every hypothesis test you reject H0 when your sample is very unlikely to have occurred by random chance.

13What are the two types of numeric data called? Explain the difference, and give an example of each.

Answer: Continuous data are measurements and answer “how much” questions. Examples: height, salary
Discrete data usually count things and answer “how many” questions. Example: number of credit hours carried

14Suppose the null hypothesis is that a machine is producing the allowed 1% proportion of defectives (H0: p = 0.01). Your experiment could end in one of several conclusions, depending on your sample data. List the letters of all possible conclusions from those below. (The actual conclusion would depend on α, the choice of H1, and the calculated p-value. Not all possible conclusions are listed below.)
A. The machine is producing exactly the acceptable proportion of defectives.
B. The machine is producing no more defectives than acceptable.
C. The machine is producing too many defectives.
D. Unable to prove anything either way.

Remark: As stated, what you can prove depends partly on your H1. There are three things it could be:

• If H1: p > 0.01 (machine producing too many defectives) and you calculate p-value<α, your conclusion is C.
• If H1: p ≠ 0.01 (machine not operating as specified) and you calculate p-value<α, your conclusion depends on the sample data. Your conclusion is either C above or the unlisted conclusion “the machine is producing fewer defectives than allowed”, in other words performing better than specified. See p < α in Two-Tailed Test: What Does It Tell You? for reminders about interpreting a two-tailed test when p-value<α.
• There is little reason to choose H1: p < 0.01 (the machine is performing better than specified), but if that is H1 and p-value<α then your conclusion is that the machine is performing better than specified; that conclusion is not listed above.

Regardless of H1, if p-value>α your conclusion will be D or similar to it.

Common mistake: Conclusion A is impossible because it’s the null hypothesis and you never accept the null hypothesis.

Conclusion B is also impossible. Why? because “no more than” translates to ≤. But you can’t have ≤ in H1, and H1 is the only hypothesis that can be accepted (“proved”) in a hypothesis test.

15How can you avoid making a Type I error in a hypothesis test?

Answer: You can’t. You can reduce the likelihood of a Type I error by setting the significance level α to a lower number, but the possibility of a Type I error is inherent in the sampling process.

Remark: A Type I error is a wrong result, but it is not necessarily the result of a mistake by the experimenter or statistician.

16Which one or more of the following describe the p-value in an experiment?
A. the probability of a correct decision
B. the probability that we are right to reject the null hypothesis
C. the probability that rejecting the null hypothesis is an error
D. the probability that our sample results could have been obtained by random selection if the null hypothesis is true
E. the probability of a Type I error
F. the probability of a Type II error

17Data are gathered and a computation is done to answer the question “As near as we can tell, how much does the average high-school student spend on lunch?” This computation would be part of
A. hypothesis test
B. sample size
C. confidence interval
D. none of the above

Remark: There is no specific claim, so this is not a hypothesis test.

18Linear correlation coefficients must lie between what two values? What value indicates “no linear correlation”? Does this mean no correlation at all?

Answer: r must be between −1 and +1 inclusive. (Symbolically, −1 ≤ r ≤ +1.) A value of r = 0 indicates no linear correlation. But this doesn’t necessarily mean no correlation, because another type of correlation might still be present. Example: the noontime height of the sun in the sky plotted against day of the year will show near zero linear correlation but very strong sine-wave correlation.

19“Four out of five dentists surveyed recommend Trident sugarless gum for their patients who chew gum.” Which of these is the correct symbol for “four out of five dentists surveyed”?
μ      π      σ      p           po      x           s

Answer: , proportion of a sample (In this case,  = 4/5 = 0.8 or 80%.)

20A poll concludes that 26.9% of TC3 students are satisfied with the food service. What is the type of the original data gathered?

Answer: attribute or qualitative, specifically binomial (“Are you satisfied with the food service?”)

21For what sort of data would you typically prefer a pie chart? Why?

Answer: Attribute (qualitative) data, either binomial or categorical. This compact form makes it easy to compare the relative sizes of all the categories. (Note that word “typically”. While “attribute data? pie chart!” is a good rule of thumb, in particular cases a different presentation might be better, or grouped numeric data might be presented in a pie chart.)

Caution: The percentages must add to 100%. Therefore you must have complete data on all categories to display a pie chart. Also, if multiple responses from one subject are allowed, then a pie chart isn’t suitable, and you should use some other presentation, such as a bar chart.

22 The mean is usually the best measure of center of numerical data. But under certain circumstances the mean is not representative and you prefer a different measure of center. Which circumstances, and which measure of center?

Answer: When the data are skewed, prefer the median.

23Usually you make what you want to prove the alternative hypothesis, not the null hypothesis. Why?

Answer: Because you can never accept the null hypothesis; only the alternative hypothesis can be accepted.

24A company wishes to claim, “People who eat our shredded wheat for breakfast every day for a month lose more than ten points on their cholesterol.” One or more of the following state the null and alternative hypotheses correctly. Which one(s)?

 A. H0 > 10       H1 ≤ 10 B. H0: x̅ > 10     H1: x̅ ≤ 10 C. H0: μ > 10     H1: μ ≤ 10 D. H0: x > 10     H1: x ≤ 10 E. H0 = 10       H1 > 10 F. H0: x̅ = 10     H1: x̅ > 10 G. H0: μ = 10     H1: μ > 10 H. H0: x = 10     H1: x > 10 I. H0 ≤ 10       H1 > 10 J. H0: x̅ ≤ 10     H1: x̅ > 10 K. H0: μ ≤ 10     H1: μ > 10 L. H0: x ≤ 10     H1: x > 10

Answer: G at TC3, but K at some other colleges; it depends on the textbook.

Remark: This problem tests for several very common mistakes by students. Always make sure that

• Your hypotheses include a population parameter (rules out A, E, and I because they have no symbol; rules out B, D, F, H, J, L because their symbols aren’t a population parameter)
• The “=” sign is in H0 (rules out A–D)

This leaves you with G and K as possibilities. Either can be correct, depending on your textbook. For example, Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011) always puts a plain = sign in H0 regardless of H1, so for TC3 students the correct answer is G. Students at some other institutions might have K as the correct answer.

25Which of the following is a Type I error?
A. failing to reject the null hypothesis when it is true
B. failing to reject the null hypothesis when it is false
C. rejecting the null hypothesis when it is true
D. rejecting the null hypothesis when it is false

26Compare an experiment and an observational study.

Answer: In an experiment, you assign subjects to two or more treatment groups, and through techniques like randomization or matched pairs you control for variables other than the one you’re interested in. By contrast, in an observational study you gather current or past data, with no element of control; the possibility of lurking variables severely limits the type of conclusions you can draw. In particular, you can’t conclude anything about causation from an observational study.

27Our symbol for level of confidence in a confidence interval is
α        α/2        1–α        z(α/2)        E
(If none of these, supply the correct symbol.)

Answer: 1–α, or (1−α)100% is also acceptable.

28You gather a random sample of selling prices of 2006 Honda Civics. Which selection on your TI-83 would be used to test the claim “In the U.S., 2006 Honda Civics sell, on average, for more than \$2,000”?
A. Z-Test     B. T-Test     C. 1-PropZTest     D. 1-PropTTest     E. χ²-Test     F. none of these

Remark: This is Case 1, not Case 0, since you do not know the standard deviation for the selling price of all 2006 Honda Civics in the U.S.

29Compare descriptive and inferential statistics, and give an example of each.

Answer: descriptive: presentation of actual sample measurements
inferential: estimate or statement about population made on the basis of sample measurements

Example: “812 of 1000 Americans surveyed said they believe in ghosts” is an example of descriptive statistics: the numbers of yeses and noes in the sample were counted. “78.8% to 83.6% of Americans believe in ghosts (95% confidence)” is an example of inferential statistics: sample data were used to make an estimate about the population. “More than 60% of Americans believe in ghosts” is another example of inferential statistics: sample data were used to test a claim and make a statement about a population.

30You find that your maximum error of estimate (margin of error) is ±3.3 at a confidence level of 95%. At 90% confidence, what would be the maximum error of estimate?
A. more than 3.3         B. 3.3         C. less than 3.3         D. can’t say without more information.

Remark: Remember that the confidence interval derives from the central 95% or 90% of the normal distribution. The central 90% is obviously less wide than the central 95%, so the interval will be less wide.

31Compare “sample” and “population”; give an example.

Answer: A sample is a subgroup of the population, specifically the subgroup from which you take measurements. The population is the entire group of interest.

Example: You want to know the average amount of money a full-time TC3 student spends on books in a semester. The population is all full-time TC3 students. You randomly select a group of students and ask each one how much s/he spent on books this semester. That group is your sample.

32You take a random sample of Lamborghini owners and a random sample of Subaru owners. Which selection on your TI-83 would be used to answer the question “How much more do Lamborghini owners spend per year on maintenance than Subaru owners?”
A. ZInterval     B. TInterval     C. 2-SampZInt     D. 2-SampTInt     E. 2-PropZInt     F. none of these

Remark: This is unpaired numeric data, Case 4.

33You believe that more than 25% of high-school students experienced strong peer pressure to have sex. To test this belief, you survey 500 randomly selected graduating seniors nationwide and find that 150 of them say that they did feel such pressure.

(a) The data would best be analyzed as an example of
A. one population proportion
B. two populations, difference in proportions
C. one population mean
D. two populations, difference in means, paired data
E. two populations, difference in means, unpaired data
F. goodness of fit
G. contingency table

(b) Which two tests must you perform on your sample data before doing the analysis mentioned above? (In other words, how would you make sure that the sample meets the requirements?)

Answer: (a) A This is binomial because each respondent was asked “Did you feel strong peer pressure to have sex?” There is one population, high-school seniors, so this is Case 2.

(b) For binomial data, requirements are slightly different between CI and HT. Here you are doing a hypothesis test. First check that npo(1−po) ≥ 10. Here n=500 and po=.25, and therefore 500×.25×(1−.25) ≅ 94 > 10.

You also check that the sample is not too large: 20n≤N. 20×500 = 10,000, and far more than 10,000 students graduate from US high schools each year.

Common mistake: Some students answer this question with “n > 30”. That’s true, but not relevant here. Sample size 30 is important for numeric data, not binomial data.

## Section B. Problems

Red
die
White dieRed die
total
123456
15475875004626216903407
26096554975356516843631
35145404684385876293176
44625074144135096112916
55515624995066586723448
65635985194876096463422
White die
total
32463449289728413635393220000
34 Skip this problem: it uses techniques we did not study.
In 1850, the Swiss astronomer Wolf rolled two dice 20,000 times to determine whether they were biased. His data are shown at right. (For example, there were 2841 rolls when the white die came up 4. There were 611 rolls when the white die came up 6 and the red die came up 4.)

(a) What is P(2 on red | 4 on white)?

Answer: P(2 on red | 4 on white) = 535/2841 or about 0.1883

Remark: Remember: “probability of one equals proportion of all.” The question is equivalent to “What proportion of rolls with a white 4 also showed a red 2?”.

Common mistake: 535/3631 is what you get if you read the condition backward. Remember that the “given that” condition follows the vertical bar.

Alternative solution: You could do this by probability formulas, if you prefer. P(2 on red | 4 on white) = P(2 on red and 4 on white) ÷ P(4 on white) = (535/20000) ÷ (2841/20000) = (535/20000) × (20000/2841) = 535/2841

(b) What is P(5 on white and 1 on red)?

Answer: There’s really no need for a formula here: you just read off the answer from the relevant cell.

P(5 on white and 1 on red) = 621/20000 or about 0.0311

Common mistake: Some students misapply the formula P(A and B) = P(A)×P(B) here and write (3635/20000)×(3407/20000). That formula applies only when A and B are independent. You might expect that “5 on white” and “1 on red” are independent events, but you can’t just assume it. In this case they are very nearly independent but not quite:

P(5 on white) = 3635/20000 = 0.1818, but P(5 on white|1 on red) = 621/3407 = .1823.

Alternative solution: I don’t recommend formulas at all for this problem, but if you insist on a formula here it is:

P(A and B) = P(A) × P(B | A)

P(5 on white and 1 on red) = P(5 on white) × P(1 on red | 5 on white)

P(5 on white and 1 on red) = (3635/20000) × (621/3635) = 621/20000

(c) What is P(5 on white or red)?

Answer: P(5 on white or 5 on red) = P(5 on white) + P(5 on red) − P(5 on both) = (3635/20000) + (3448/20000) − (658/20000) = (3635+3448−658)/20000 = 6425/20000 or about 0.3213

Common mistake: Some students misapply the formula P(A or B) = P(A)+P(B) here to get (3635+3448)/20000. That formula applies only when A and B are disjoint (mutually exclusive). But they are not mutually exclusive, because it’s very possible to have a 5 come up on both dice at the same time.

Common mistake: Some students forget how to add and subtract ractions and give the incorrect answer (3635/20000) + (3448/20000) − (658/20000) = (3635+3448−658)/60000 = 6425/60000. When you add and subtract fractions, the denominators must be the same to begin with, and the answer has the same denominator as the components.

(d) At the 0.05 significance level, is the white die biased? (Hint: what would you expect if the white die is not biased?)

Solution: If the white die is unbiased, then you would expect 20000/6 of each number. This is a model of 1:1:1:1:1:1, and you use Case 6.

(1) H0: The six faces are equally likely: the die is fair. H1: Some faces are more likely than others: the die is biased. α = 0.05 Model in L1 is 1,1,1,1,1,1; white-die total row in L2. MATH200A/GOF. Results: χ² = 270.96, df = 5, pval = 1.7E-56 L3 shows 3333.3 in each slot, so all E’s (expected values) are ≥5. And any dice throws are intrinsically random, assuming you shake them well. p < α. Reject H0 and accept H1. At the 0.05 level of significance, the white die is biased, meaning that the different faces are not equally likely to come up.
35You are testing the assertion, “Judge Judy is more friendly to plaintiffs than Judge Wapner was.” Since it would be tedious to tabulate the hundreds or thousands of decisions each judge has handed down, you randomly select 32 of each judge’s decisions. Judge Judy’s average award to plaintiffs was \$650 (standard deviation = \$250) and Judge Wapner’s was \$580 (standard deviation = \$260). Assume that the amounts are normally distributed without outliers. Using a significance level of 0.05, can you conclude that Judge Judy does indeed give higher awards on average?

Solution: Numeric data, two populations, independent samples with σ unknown: Case 4 (2-SampTTest).

Common mistake: You cannot do a 2-SampZTest because you do not know the standard deviations of the two populations.

(1) Population 1 = Judge Judy’s decisions; Population 2 = Judge Wapner’s decisions H0: μ1 = μ2, no difference in awards H1: μ1 > μ2, Judge Judy gives higher awards α = 0.05 random sample Sample sizes are both above 30, so there’s no worry about whether the population data are normal. 2-SampTTest: x̅1=650, s1=250, n1=32, x̅2=580, s2=260, n2=32, μ1>μ2, Pooled: No Results: t=1.10, pval = .1383 p > α. Fail to reject H0. At the 0,05 level of significance, we can’t tell whether Judge Judy was more friendly to plaintiffs (average award higher than Judge Wapner’s) or not.
36Weights of frozen turkeys at one large market were normally distributed with a mean of 14.8 pounds and a standard deviation of 2.1 pounds. If there were 10,000 turkeys in the market, how many choices would a shopper have who wanted a bird 20.5 pounds or larger? (Hint: begin by figuring the percentage or proportion of turkeys in that weight range.)

Solution: normalcdf(20.5, 10^99, 14.8, 2.1) = .00332. Then multiply by population size 10,000 to obtain 33.2, or about 33 turkeys.

37(from Johnson & Kuby’s Just the Essentials of Elementary Statistics 2/e problem 9.26) “The addition of a new accelerator is claimed to decrease the drying time of latex paint by more than 4%. Several test samples were conducted with the following percentage decrease in drying time:
“5.2    6.4    3.8    6.3    4.1    2.8    3.2    4.7
“If we assume that the percentage decrease in drying time is normally distributed”
(a) Test the claim, at the .05 level.

Solution: This is one-population numeric data, and you don’t know the standard deviation of the population: Case 1. Put the data in L1, and 1-VarStats L1 tells that  = 4.56, s = 1.34, n = 8.

(1) H0: μ = 4, 4% or less improvement in drying time H1: μ > 4, better than 4% decrease in drying time Remark: Why is a decrease in drying time tested with > and not μo Results: t = 1.19, p = 0.1370 p > α. Fail to reject H0. At the 0.05 significance level, we can’t tell whether the average drying time improved by more than 4% or not.

(b) “Find the 95% confidence interval for the true mean decrease in the drying time based on this sample.”

For part (b), there’s no need to repeat the requirements check or to write down all the sample statistics again.
TInterval: C-Level=.95
Results: (3.4418, 5.6832)

With 95% confidence, the true mean decrease in drying time is between 3.4% and 5.7%.

3828% of a certain breed of rabbits are born with long hair. Assume that the distribution is random, and consider a litter of five rabbits.

(a) What is the probability that none of the rabbits in the litter have long hair?

Solution: This is a binomial probability distribution: each rabbit has long hair or not, and the probability for any given rabbit doesn’t change if the previous rabbit had long hair. Use MATH200A part 3.

n = 5, p = 0.28, from = 0, to = 0. Answer: 0.1935

Alternative solution: If you don’t have the program, you can compute the probability that one rabbit has short hair (1−.28 = 0.72), then that all the rabbits have short hair (0.72^5 = 0.1935), which is the same as the probability that none of the rabbits have long hair.

(b) What is the probability that one or more in a litter have long hair?

Solution: The complement of “one or more” is none, so you can use the previous answer.

P(one or more) = 1−P(none) = 1−0.1935 = 0.8065

Alternative solution: MATH200A part 3 with n=5, p=.28, from=1, to=5; probability = 0.8065

(c) What is the probability that four or five of them have long hair?

Solution: Again, use MATH200A part 3 to compute binomial probability: n = 5, p = 0.28, from = 4, to = 5. Answer: 0.0238

Alternative solution: If you don’t have the program, do binompdf(5, .28) and store into L3, then sum(L3,5,6) or L3(5)+L3(6) = 0.0238.  Avoid the dreaded off-by-one error! For x=4 and x=5 you want L3(5) and L3(6), not L3(4) and L3(5).

For n=5, P(x≥4) = 1−P(x≤3). So you can also compute the probability as 1−binomcdf(5, .28, 3) = 0.0238.

(d) What is the average number (mean) of long-haired rabbits you expect in a litter of five?

Solution: For this problem you must know the formula:

μ = np = 5×0.28 = 1.4 per litter of 5, on average

39An aptitude test is known to have a mean score of 37.5 with standard deviation of 3.5. A company administers this test to applicants, and requires a standard score of z = 1.5 or better. For Jane to be considered, she needs at least what test score on the aptitude test?

Answer: z = (x−μ)/σ and you know z = 1.5, μ = 37.5, σ = 3.5. Solve for x = 42.75

40A survey asked a number of professionals, “Which of the following is your most common choice for breakfast?” Using the following data from a random survey, determine whether doctors choose breakfasts in different proportions from other self-employed professionals, to a .05 significance level.

```        Cereal  Pastry   Eggs   Other   No bfst  Total
Doctors     85      22     47      60        17    231
Others     185      90    160     135        35    605
Total      270     112    207     195        52    836```

Solution: This is Case 7, a 2×5 table. (The total row and total column aren’t part of the data.)

Remark: It might be tempting to do this problem as a goodness-of-fit, Case 6, taking the Others row as the model and the doctors’ choices as the observed values. But that would be wrong. Both the Doctors row and the Others row are experimental data, and both have some sampling error around the true proportions. If you take the Others row as the model, you’re saying that the true proportions for all non-doctors are precisely the same as the proportions in this sample. That’s rather unlikely.

(1) H0: Doctors eat different breakfasts in the same proportions as others. H1: Doctors eat different breakfasts in different proportions from others. α = 0.05 χ²-Test gives χ² = 9.71, df = 4, p=0.0455 random sample Matrix B shows that all the expected counts are ≥5. (As an alternative, you could use MATH200A part 7.) p < α. Reject H0 and accept H1. Yes, doctors do choose breakfast differently from other self-employed professionals, at the 0.05 significance level.
41 Suppose that the mean adult male height is 5′10″ (70″) and the standard deviation is 1.4″.

(a) If a particular man’s z-score is −1.2, what is his actual height to the nearest 0.1″?

Answer: z = (x−μ)/σ ⇒ −1.2 = (x−70)/1.4 ⇒ x = 68.3″
or: x = zσ + μ ⇒ x = −1.2×1.4 + 70 = 68.3″

(b) Using the Empirical Rule, what percentile is a height of 68.6″?

Answer: z = −1. By the empirical rule, 68% of data lie between z = ±1 and therefore 32% lie outside those bounds. 32%/2 = 16% lie below z = −1. Therefore 68.6″ is the 16th percentile.

(c) By the empirical rule, what proportion of adult men are shorter than 72.8″?

Answer: z = +2. By the empirical rule, 95% of men fall between z = −2 and z = +2, so 5% fall below z = −2 or above z = +2. Half of those, 2.5%, fall above z = +2, so 97.5% fall below z = +2. 97.5% of men are shorter than 72.8″.

life, hrcount
500–6506
650–80018
800–95060
950–110089
1100–125029
1250–140017

42The length of life of a random sample of incandescent light bulbs was obtained, and the results are in the table at right.

(a) Plot a histogram of the data.

Solution: The histogram is shown at left. You must show the scale for both axes and label both axes. The scale for the horizontal axis is predetermined: you label the edges of the histogram bars and not their centers. You have some latitude for the scale of the vertical axis, as long as you include zero, show consistent divisions, and have your highest mark greater than 89. For example, 0 to 100 in increments of 20 would also work.

(b) What is the size of the sample, with its proper symbol?

Solution: Compute the class marks or midpoints: 575, 725, and so on. Put them in L1 and the frequencies in L2. Use `1-VarStats L1,L2` and get n = 219.
See Sample Statistics on TI-83/84.

(c) What are the mean, standard deviation, and variance? (Use the proper symbols and round to one decimal place.)

Solution: Further data from `1-VarStats L1,L2`:  = 990.1 and s = 167.3
s² = 27982.90813 ⇒ s² = 27982.9

Common mistake: If you answered = 950 you probably did `1-VarStats L1` instead of `1-VarStats L1,L2`. Your calculator depends on you to supply one list when you have a simple list of numbers and two lists when you have a frequency distribution.

Common mistake: If you answered 27989.3 for the variance, you squared the rounded number. Never use rounded numbers for further calculation. Either enter the unrounded s and square it, or use [`VARS`] [`5`] [`3`] to paste the value of s automatically.

(d) What is the relative frequency of the 1100–1250 class?

Solution: f/n = 29/219 ≅ 0.13 or 13%

43One way to set speed limits is to observe a random sample of drivers. The speed limit is set at the 85th percentile, which is the speed such that 85% of drivers are going slower and 15% are going faster. What speed corresponds to that 85th percentile, assuming drivers’ speeds are normally distributed with μ = 57.6 and σ = 5.2 mph?

Solution: invNorm(0.85, 57.6, 5.2) = 62.98945357 → 63.0 mph

44You’re planning a survey to see what fraction of people who live in Virgil would take the bus if the county added a route between Greek Peak and downtown Cortland via routes 392 and 215.
(a) You think the answer is only about 20% of them. If you need 90% confidence in an answer to within ±4%, how many people will you need to survey?

Solution: This is binomial data (each person either would or would not take the bus), hence Case 2, One population proportion. Use MATH200A part 5.

(a)  = .2, E = 0.04, C-Level = 0.90.

Common mistake: The margin of error is E = 4% = 0.04, not 0.4.

Alternative solution: See How Big a Sample Do I Need? or your textbook and use the formula at right. With the estimated population proportion  = 0.2 in the formula, you get zα/2 = z0.05 = invNorm(1−0.05) = 1.6449, and n = 270.5543 → 271

(b) What if you have no idea of the answer? How many would you need to survey then?

Solution: If you have no prior estimate, use  = 0.5. The other inputs are the same, and the answer is 423

45A 1992 study showed the mean cost for all homes in Sassafras County to be \$70,000 with standard deviation \$5,500. This year, you survey 35 randomly selected homes that were sold, and you find a mean of \$72,050.
(a) Compute the value of the test statistic for the mean of this sample.
(b) Compute the value of P( ≥ 72,050), the probability of getting a sample mean this large or larger, if the mean price for all Sassafras County houses is still \$70,000.
(c) At the .05 level, has the mean price of a Sassafras County house increased since the study was done? (Use your answer from (b); you don’t need to do the full hypothesis test.)

Solution: The sampling distribution for n=35 is a normal distribution with μ = \$70,000, σ = \$5500/√35 (about \$930).
(a) z = (72050−70000)÷(5500/√35) ⇒ z = 2.2051
(b) P( ≥ 72,050) = normalcdf(72050, 10^99, 70000, 5500/√35) = 0.0137
(c) p < α: Yes, the mean price has increased.

Remark: You could also solve (a) and (b) by doing a Z test with μo=70000, σ=5500, =72050, n=35, μ>μo and read off z and p as above.

46 Some popular fast-food items were compared for calories and fat, and the results are shown below:

 Calories (x) Fat (y) 270 420 210 450 130 310 290 450 446 640 233 9 20 10 22 6 25 7 20 20 38 11

(a) Make a scatter plot on your TI-83. Do you expect a positive, negative, or zero correlation? Why?

Solution: For the procedure, see Step 1 of Scatter Plot, Correlation, and Regression on TI-83/84. Your plot should look like the one at right.

You expect positive correlation because points trend upward to the right (or, because y tends to increase as x increases). Even before plotting, you could probably predict a positive correlation because you assume higher calories come from fat; but you can’t just assume that without running the numbers.

(b) Find the correlation coefficient and the equation of the line of best fit and write them down. Round to four decimal places and use proper symbols.

Solution: See Step 2 of Scatter Plot, Correlation, and Regression on TI-83/84.
r = .8863314629 → r = 0.8862
a = .0586751909 → a = 0.0587
b = −3.440073602 → b = −3.4401

ŷ = 0.0587x − 3.4401

Common mistake: The symbol is ŷ, not y.

(c) Give the value of the y intercept and interpret its meaning.

Answer: The y intercept is −3.4401. It is the number of grams of fat you expect in the average zero-calorie serving of fast food. Clearly this is not a meaningful concept.

Remark: Remember that you can’t trust the regression outside the neighborhood of the data points. Here x varies from 130 to 640. The y intercept occurs at x = 0. That is pretty far outside the neighborhood of the data points, so it’s not surprising that its value is absurd.

(d) Using the regression equation or your TI-83 graph, how many grams of fat would you predict for an item of 310 calories? Explain why this is different from the actual data point (310 calories, 25 grams).

Solution: See Finding ŷ from a Regression on TI-83/84. Trace at x = 310 and read off ŷ = 14.749... ≅ 14.7 grams fat. This is different from the actual data point (x=310, y=25) because ŷ is based on a trend reflecting all the data. It predicts the average fat content for all 310-calorie fast-food items.

Alternative solution: ŷ = .0586751909(310) − 3.440073602 = 14.749 ≅ 14.7.

(e) What is the value of the residual for the data point (310,25)?

Solution: The residual at any (x,y) is y−ŷ. At x = 310, y = 25 and ŷ = 14.7 from the previous part. The residual is y−ŷ = 10.3

Remark: If there were multiple data points at x = 310, you would calculate one residual for each point.

(f) What is the value of the coefficient of determination in this regression? What does it mean?

Answer: From the `LinReg(ax+b)` output, R² = 0.7855834621 → R² = 0.7856 About 79% of the variation in fat content is associated with variation in calorie content. The other 21% comes from lurking variables such as protein and carbohydrate count and from sampling error.

(g) The decision point for n = 11 is 0.602. What if anything can you say about the correlation for all fast foods?

Solution: See Decision Points for Correlation Coefficient. Since 0.8862 is positive and 0.8862 > 0.602, you can say that there is some positive correlation in the population, and higher-calorie fast foods do tend to be higher in fat.

47Aluminum plates produced by a company are normally distributed with a mean thickness of 2.0 mm and a standard deviation of 0.1 mm. If 6% of the plates are too thick, what is the cutoff point between “too thick” and “acceptable?”

Solution: invNorm(1-.06, 2.0, 0.1) = 2.1555, about 2.16 mm

48Many people took a physical fitness course. Seven of them were randomly selected and were tested for how many sit-ups they could do. The same seven were re-tested after the course. From the data below, can you conclude that improvement took place among the general run of people who took the course? Use α = 0.01.

```         Anne    Bill   Chance   Deb      Ed     Frank   Grace
Before    29      22      25      29      26      24      31
After     30      26      25      35      33      36      32```

Solution: This is paired data, Case 3. (Each individual gives you two numbers, Before and After.)

(1) d = After − Before H0: μd = 0, no improvement H1: μd > 0, improvement in number of sit-ups Remark: Why After−Before instead of the other way round? Since we expect After to be greater than Before, doing it this way you can expect the d’s to be mostly positive (if H1 is true). Also, it feels more natural to set things up so that an improvement is a positive number. But if you do d=Before−After and H1:μd<0, you get the same p-value. α = 0.01 random sample Enter the seven differences — 1, 4, 0, 6, 7, 12, 1 — into a statistics list. A normal probability plot (MATH200A part 4) shows a straight line (r=.957), so the data are normal. The modified box-whisker plot (MATH200A part 2) shows no outliers. The plots are shown here for comparison to yours, but you don’t need to copy these plots to an exam paper. T-Test: μo=0, List:L4, Freq:1, μ>μo Results: t = 2.74, p = 0.0169, x̅ = 4.4, s = 4.3, n = 7 p > α. Fail to reject H0. At the 0.01 significance level, we can’t say whether the physical fitness course improves people’s ability to do sit-ups or not.
49 You’re auditing a bank. The bank’s internal accountants tell you that the average deposit is \$189.56 and the standard deviation is \$45.00. You take a random sample of 400 deposits and find an average of \$200.00. How likely is it, if the bank’s accountants are correct, that you would get a random sample of that size with a mean of \$200.00 or more?

Solution: This is a straightforward question about sampling distributions.

μ = \$189.56, σ = \$45.00, n = 400. The standard error of the mean is σ = \$45.00/√400 = \$2.25.

P( ≥ 200.00) = normalcdf(200, 10^99, 189.56, 45/√(400)) = 1.7439×10-6 or about 0.000 002, two chances in a million.

Common mistake: This problem is about the distribution of sample means, with σ = 45/√400. If you just use σ = 45 you’ve missed the whole point of the problem.

Alternative solution: One definition of p-value is the probability of getting a sample like the one you got, or more extreme, if H0 is true. You could also take the given μ and σ as null hypothesis and do a Z-Test with μo = 189.56, σ = 45,  = 200, n = 400, μ > μo. The p-value is the same: 1.7439E-6.

Studio/efficiency18.2%75
1 bedroom18.2%60
2 bedrooms40.4%105
3 bedrooms18.2%45
Over 3 bedrooms5.0%15
Total100.0%300
50(adapted from Johnson & Kuby’s Just the Essentials of Elementary Statistics 2/e problem 11.15)  A survey was taken nationally to see what size vacation home people preferred. A separate survey was taken in Nebraska. Both were random samples. Do the Nebraska results differ significantly (0.05 level) from the national results?

Solution: Here you have a model (the US population) and you’re testing an observed sample (Nebraska) for consistency with that model. One tipoff is that you are given the size of the Nebraska sample but for the US you have only percentages, not actual numbers of people. This is Case 6, goodness of fit to a model.

(1) H0: Nebraska proportions are the same as national proportions. H1: Nebraska proportions are different from national proportions. α = 0.05 US percentages in L1, Nebraska observed counts in L2. MATH200A part 6. The result is χ² = 12.0093 → 12.01, df = 4, p-value = 0.0173 Common mistake: Some students convert the Nebraska numbers to percentages and perform a χ² test that way. The χ² test model can equally well be percentages or whole numbers, but the observed numbers must be actual counts. random sample L3 shows the expected values, and they are all above 5. p < α. Reject H0 and accept H1. Yes, at the 0.05 significance level Nebraska housing patterns are different from those for the U.S. as a whole.
51An experiment was designed to test the effectiveness of a short course that teaches diabetic self care. Fifty diabetic patients were enrolled in the course, and fifty others served as a control group. (Patients were randomly assigned between the two groups.) Six months after the course, blood sugar levels were tested and results obtained as follows:
Diabetic course group: mean = 6.5, standard deviation = 0.7
Control group: mean = 7.1, standard deviation = 0.9
At a significance level of 0.01, does the diabetic course succeed in lowering patients’ blood sugar?

Solution: This is unpaired numeric data, Case 4.

(1) Population 1 = Course, Population 2 = No course H0: μ1 = μ2, no benefit from diabetic course H1: μ1 < μ2, reduced blood sugar from diabetic course α = 0.01 Independent random samples, both n’s >30 2-SampTTest: x̅1=6.5, s1=.7, n1=50, x̅2=7.1, s2=.9, n2=50, μ1<μ2, Pooled:No Results: t=−3.72, p=1.7E-4 or 0.0002 (Though we do not, some classes use the preliminary 2-SampFTest. That test gives p=0.0816>0.05. Those classes would use Pooled:Yes in 2-SampTTest and get p=0.00016551 and the same conclusion.) p < α. Reject H0 and accept H1. At the 0.01 level of significance, the course in diabetic self care does lower patients’ blood sugar, on average.
52(Johnson & Kuby’s Just the Essentials of Elementary Statistics 2/e problem 9.36)  “A study in the journal PAIN, October 1994, reported on six patients with chronic myofascial pain syndrome. The mean duration of pain had been 3.0 years for the 6 patients and the standard deviation had been 0.5 year. Test the hypothesis that the mean pain duration of all patients who might have been selected for this study [meaning, of all persons who suffer from this condition] was greater than 2.5 years. Use α = 0.05. Assume that the sample is a random sample, normally distributed with no outliers.

Solution: This is a test on the mean of one population, with population standard deviation unknown: Case 1.

Common mistake: The standard deviation 0.5 was the standard deviation of the sample, so this is Case 1 not Case 0, and you must use a T-Test and can’t use a Z-Test.

(1) H0: μ = 2.5 years H1: μ > 2.5 years α = 0.05 random sample, normal with no outliers (given) T-Test: μo=2.5, x̅=3, s=.5, n=6, μ>μo Results: t = 2.45, p = 0.0290 p < α. Reject H0 and accept H1. Yes, at the 0.05 significance level, the mean duration of pain for all persons with the condition is greater than 2.5 years.
53In a survey of working parents, 200 men and 200 women were randomly selected and asked, “Have you refused a promotion because it would mean less time with your family?” Of the men, 60 said yes; 48 of the women said yes.

(a) Obviously more men in the sample refused promotions. But can you conclude at the 0.05 significance level that a higher percentage of all working men have refused promotions, versus the percentage of all working women?

Solution: Each man or woman was asked a yes/no question, so you have binomial data for two populations: Case 5.

(1) Population 1 = men, Population 2 = women H0: p1 = p2 H1: p1 > p2, more men refuse promotions α = 0.05 2-PropZTest: x1=60, n1=200, c2=48, n2=200, p1>p2 Results: z = 1.35, p = .0883, p̂1=.3, p̂2=.24, p̂=.27 independent random samples For each sample, 20n = 20×200 = 4000 is far less than the total number of men or women. (n1+n2)p̂(1−p̂) = (200+200)×.27×(1−.27) ≅ 79 ≥ 10. p > α. Fail to reject H0. At the 0.05 level of significance, we can’t determine whether the percentage of men who have refused promotions to spend time with their family is more than, the same as, or less than the percentage of women.

(b) In an English sentence, state a 95% confidence interval for the difference in percentages of men and women who refuse promotions.

Solution: 2-PropZInt with the above inputs and C-Level=.95 gives (−.0268, .14482). The English sentence needs to state both magnitude and direction, something like this: Regarding men and women who refused promotion for family reasons, we’re 95% confident that the the difference in percentages is between 2.7% more men and 14.5% more women.

Common mistake: With two-population confidence intervals, you must state the direction of the difference, not just the size of the difference.

54Ten thousand students take a test, and their scores are normally distributed. If the middle 95% of them score between 70 and 130, what are the mean and standard deviation?

Solution: This problem depends on the Empirical Rule and knowing that the normal distribution is symmetric.

If the middle 95% runs from 70 to 130, then the mean must be μ = (70+130)÷2 ⇒ μ = 100

95% of any population are within 2 standard deviations of the mean. The range 70 to 100 (or 100 to 130) is therefore two s.d.
2σ = 100−70 = 30 ⇒ σ = 15

55An insurance company advertises that 75% of its claims are settled within two months of being filed. The state insurance commission thinks the percentage is less than 75, and sets out to prove it. First a small study is done. For this preliminary study, the commissioner can live with a 5% chance of making a Type I error. The commission staff randomly selects 65 claims, and finds out that 40 were settled within two months. Based on this study, can you say that less than 75% of claims are settled within two months?

Solution: This is binomial data, Case 2. (The members of the sample are claims, and each claim either is settled or is not.)

(1) H0: p = .75 H1: p < .75 α = 0.05 random sample 20n = 20×65 = 1300, obviously less than the total number of claims filed in the state. npo(1−po) = 65×.75×(1−.75) ≅ 12 ≥ 10. 1-PropZTest: po=.75, x=40, n=65, prop
56 Skip this problem: it uses techniques we did not study.
A shoe store gets its shoes from just two companies, 40% from A and 60% from B. 2.5% of pairs from Brand A are mislabeled, and 1.5% of pairs from Brand B are mislabeled. Find the probability that a randomly selected pair of shoes in the store is mislabeled.

Solution: P(mislabeled) = P(Brand A and mislabeled) + P(Brand B and mislabeled) because those are disjoint events. But whether a pair is mislabeled is dependent on the brand, so

P(Brand A and mislabeled) = P(Brand A) × P(mislabeled | Brand A)

and similarly for brand B.

P(mislabeled) = 0.40 × 0.025 + 0.60 × 0.015 = 0.019 or just under 2%

Alternative solution: The formulas can be confusing, and often there’s a way to do without them. You could also do this as a matter of proportions:

Out of 1000 shoes, 400 are Brand A and 600 are Brand B.

Out of 400 Brand A shoes, 2.5% are mislabeled. 0.025×400 = 10 brand A shoes mislabeled.

Out of 600 Brand B shoes, 1.5% are mislabeled. 0.015×600 = 9 brand B shoes mislabeled.

Out of 1000 shoes, 10 + 9 = 19 are mislabeled. 19/1000 is 1.9% or 0.019.

This is even easier to do if you set up a two-way table, as shown below. The values in bold face are given in the problem, and those in light face are derived from them.

Brand A Brand B Total 40% × 2.5% = 1% 60% × 1.5% = 0.9% 1% + 0.9% = 1.9% 40% − 1% = 39% 60% − 0.9% = 59.1% 39% + 59.1% = 98.1% 40% 60% 100%
57Ten randomly selected men compared two brands of razors. Each man shaved one side of his face with brand A and the other side with brand B. (They flipped coins to decide which razor to use on which side.) Each tester assigned a “smoothness score” of 1 to 10 to each side after shaving. The scores are as shown below. Determine whether there is a difference in smoothness performance between the two razors, using α = 0.10.
```            Man   1   2   3   4   5   6   7   8   9  10
A score   7   8   3   5   4   4   9   8   7   4
B score   5   6   3   4   6   5   6   7   3   4
d = A-B   2   2   0   1  -2  -1   3   1   4   0```

Solution: This is paired numeric data, Case 3.

Common mistake: You must do this as paired data. Doing it as unpaired data will not give the correct p-value.

(1) d = A−B H0: μd = 0, no difference in smoothness H1: μd ≠ 0, a difference in smoothness Remark: You must define d as part of your hypotheses. α = 0.10 random sample Compute the ten differences (positive or negative, as shown above) and put them in a statistics list. Use MATH200A part 4 for the normal probability plot to show data are normal. MATH200A part 2 gives a modified boxplot showing no outliers. TTest: μo=0, List:L1, Freq: 1, μ≠μo Results: t = 1.73, p = 0.1173, x̅ = 1, s = 1.83, n = 10 p > α. Fail to reject H0. At the 0.10 level of significance, it’s impossible to say whether the two brands of razors give equally smooth shaves or not.
58In August 2009, the National Geographic News Web site reported that 90% of US currency was tainted with cocaine.
(a) If you drew a random sample of two bills, what is the chance that exactly one of them is tainted with cocaine?
(b) You have ten bills, and you’ve been told that 90% of these ten bills are tainted with cocaine. If you draw two of the ten bills at random, what is the chance that exactly one of your two is tainted with cocaine?

Remark: The key to this is recognizing the difference between with and without replacement. While (a) and (b) are both technically without replacement, recall that when the sample is less than 5% of a large population, as it is in (a), you treat the sample as drawn with replacement. But in (b), the sample of two is drawn from a population of only ten bills, so you must use computations for without replacement.

Solution: (a) Use MATH200A part 3 with n=2, p=.9, from=1, to=1. Answer: 0.18

Alternative solution: The probability that exactly one is tainted is sum of two probabilities: (i) that the first is tainted and the second is not, and (ii) that the first is not tainted and the second is. Symbolically,

P(exactly one) = P(first and secondC) + P(firstC and second)

P(exactly one) = 0.9×0.1 + 0.1×0.9

P(exactly one) = 0.09 + 0.09 = 0.18

Solution: (b) When sampling without repacement, the probabilities change. You have the same two scenarios — first but not second, and not first but second — but the numbers are different.

P(exactly one) = P(first and secondC) + P(firstC and second)

P(exactly one) = (9/10)×(1/9) + (1/10)×(9/9)

P(exactly one) = 1/10 + 1/10 = 2/10 = 0.2

Common mistake: Many, many students forget that both possible orders have to be considered: first but not second, and second but not first.

Common mistake: You can’t use binomial distribution in part (b), because when sampling without replacement the probability changes from one trial to the next.

59Fifteen farms were randomly selected from a large agricultural region. Each farm’s yield of wheat per acre was measured. For the 15 farms, the mean yield per acre was 85.5 bushels and the standard deviation was 10.0 bushels. Find a 90% confidence interval for the mean yield per acre for all farms in this region, assuming yield per acre is normally distributed and there were no outliers in the sample.

Solution: This is numeric data for one population with σ unknown: Case 1. Requirements are met because the original population (yields per acre) is normal. The T-Interval yields (80.952, 90.048). 81.0 < μ < 90.0 (90% confidence) or 85.5±4.5 (90% confidence)

60You draw five cards from a deck, without replacement, and record the number of aces you drew. Then you replace the five cards and shuffle the deck thoroughly. If you repeat this experiment many times, is the number of aces in five cards drawn a binomial distribution? Why or why not?

Answer: No, because the probabilities on the five trials are not independent.

For example, if the first card is an ace then the probability the second card is also an ace is 3/51, but if the first card is not an ace then the probability that the second card is an ace is 4/51. Symbolically, P(A2|A1) = 3/51 but P(A2| not A1) = 4/51.

61 In a survey of 300 people from Tompkins County, 128 of them preferred to rent or stream a movie on Saturday night rather than watch broadcast or cable TV. In Cortland County, 135 of 400 people surveyed preferred a movie. You’re interested in the difference of proportion in movie renters for Tompkins County over Cortland County.
(a) What is the point estimate for that difference?
(b) Find the 98% confidence interval for the difference in the two proportions for all residents of the counties.
(c) What is the maximum error of estimate, at the 98% confidence level?

Solution: This is two-population binomial data, Case 5.

(a) T = 128/300 = 0.4267. C = 135/400 = 0.3375. TC = 0.0892 or about 8.9%

Remark: The point estimate is descriptive statistics, and requirements don’t enter into it. But the confidence interval is inferential statistics, so you must verify that n(1−) is ≥10 for each sample, and that n ≤ 0.05N for each sample.

(b) 2-PropZInt: The 98% confidence interval is 0.0029 to 0.1754 (about 0.3% to 17.5%), meaning that with 98% confidence Tompkins viewers are more likely than Cortland viewers, by 0.3 to 17.5 percentage points, to prefer a movie over TV.

(c) E = 0.1754−0.0892 = 0.0862 or about 8.6%

You could also compute it as 0.0892−0.0029 = 0.0863 or (0.1754−0.0029)/2 = 0.0853. All three methods get the same answer except for a rounding difference.

Germinated Didn’t 80 20 135 15

62Two batches of seeds were randomly drawn from the same lot, and one batch was given a special treatment. Consider the data for germination shown at right. At significance level 0.05, does the treatment make any difference in how likely seeds are to germinate?

Solution: This is binomial data for two populations, Case 5. (The members of the samples are seeds, and a given seed either germinated or didn’t.)

(1) Population 1 = no treatment, Population 2 = special treatment H0 p1 = p2, no difference in germination rates H1 p1 ≠ p2, there’s a difference in germination rates α = 0.05 2-PropZTest: x1=80, n1=80+20, x2=135, n2=135+15, p1≠p2 Results: z = −2.23, pval = 0.0256, p̂1 = .8, p̂2 = .9, p̂ = .86 independent random samples 20n1=20×100=2000; 20n2=20×150=3000; obviously there are far more than 3000 seeds of this type. (n1+n2)p̂(1−p̂) = (100+150)×.86×(1−.86) ≅ 30 ≥ 10 p < α. Reject H0 and accept H1. Yes, at the 0.05 significance level, the special treatment made a difference in germination rate. Specifically, seeds with the special treatment were more likely to germinate than seeds that were not treated. Remark: p < α in Two-Tailed Test: What Does It Tell You? explains how you can reach a one-tailed result from a two-tailed test.

Alternative solution: You could also do this as a test of homogeneity, Case 7. The χ²-Test gives χ² = 4.98, df = 1, p=0.0256

63A booster rocket has six gaskets, each with a 97% reliability rating. If any gasket fails, the launch fails and the rocket will explode. (This actually happened to the space shuttle Challenger.) Assuming that the gaskets hold or fail independently, what is the chance of an explosion?

Solution: An explosion occurs if one or more gaskets fail. Rather than compute the probability of one failing, the probability of two failing, and so on, recognize that the complement is your friend in “at-least” and “at-most” problems and compute the probability of al siz gaskets holding, which is the probability of no explosion, then subtract from 1.

Probability of a given gasket holding = 0.97 (given)

Probability of all gaskets holding = 0.97^6

Probability of explosion (one or more gaskets fail) = 1−0.97^6 = 0.1670

Common mistake: It is wrong to compute the probability of failure as 6×0.03 = 0.18. 6×0.03 is 0.03+0.03+0.03+0.03+0.03+0.03, but you can add probabilities like that only when the events are disjoint. Gasket failures are not disjoint, since it is possible for more than one to fail.

Alternative solution: This can also be solved as a binomial probability problem, since you have six trials, each with success or failure, and they are independent. Use MATH200A part 3 to compute binomial probability. n = 6, p = 0.97, and an explosion will occur if the number of successes is from 0 to 5 (anything less than all six gaskets holding). This gives the same answer, 0.1670.

## What’s New

• 1 Aug 2012: Add MATH200A as alternative in the breakfast problem.
• 6 Sep 2011: Mark problems to be skipped because they require probability rules not covered in class.
• (intervening changes suppressed)
• 11 Nov 2007: new document, mostly an amalgamation of the old separate documents for descriptive and inferential statistics

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