BrownMath.com → Stats w/o Tears → Review Prob. Solutions

# Stats without TearsSolutions to Review Problems

Updated 1 Jan 2016

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## Problem Set 1: Short Answers

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.

1 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.
2 (a) C
(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.
3 qualitative = attribute, non-numeric, categorical. Examples: political party affiliation, gender.
quantitative = numeric. Examples: height, number of children.

Common mistake: Binomial is a subtype of qualitative data so it’s not really a synonym. Discrete and continuous are subtypes of numeric data.

4 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.

5 (a) pop. 1 = control, pop 2 = music
H0: p2 = p1 and H1: p2 < p1
Or: H0: p2 – p1 = 0 and H1: p2 – p1 < 0
(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.

6 Check this against the definition:
• Are there a fixed number of trials? Yes, you are rolling five dice, n = 5.
• Are there only two outcomes, success and failure? Yes, each die is either a 3 or not.
• Is the probability of success the same from trial to trial; are the trials independent? Yes, p = 1/6 for each die, and the dice are independent.

This is a binomial PD.

7 A

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.

8 B,D — B if p<α, D if p>α
9 “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 For any set of continuous data, or discrete data with many different values. If the variable is discrete with only a few different answers, you could use a bar graph or an ungrouped histogram.

For a small- to moderate-sized set of numeric data, you might prefer a stemplot.

11 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.
12 A, B

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.

13 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
14 C, D

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.

15 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.

16 (a) The population, the group you want to know something about, is all churchgoers. Common mistake: Not churchgoers who think evolution should be taught, but all churchgoers. “Churchgoers who think evolution should be taught” is a subgroup of that population, and you want to know what proportion of the whole population is in that subgroup.

(b) The size is unknown, but certainly in the millions. You also could call it infinite, or uncountable. Common mistake: Don’t confuse size of population with size of sample. The population size is not the 487 from whom you got surveys, and it’s not the 321 churchgoers in your sample.

(c) The sample size n is the 321 churchgoers from whom you collected surveys. Yes, you collected 487 surveys in all, but you have to disregard the 166 that didn’t come from churchgoers, because they are not your target group. Common mistake: 227 isn’t the sample size either. It’s x, the number of successes within the sample.

(d) No. You want to know the attitudes of churchgoers, so it is correct sampling technique to include only churchgoers in your sample.

If you wanted to know about Americans in general, then it would be selection bias to include only churchgoers, since they are more likely than non-churchgoers to oppose teaching evolution in public schools.

17 In your experiment, there was some difference between the average performance of Drug A and Drug B. The p-value is the chance of getting a difference that large or larger if Drug A and Drug B are actually equally effective. Using the number, you can say that if there is no difference between Drug A and Drug B, then there’s a 6.78% probability of getting this big a difference between samples, or even a bigger difference.

Common mistake: Your answer will probably be worded differently from that, but be careful that it is a conditional probability: If H0 is true, then there’s a p-value chance of getting a sample this extreme or more so. The p-value isn’t the chance that H0 is true.

Remark: If you are at all shaky about this, review What Does the p-Value Mean?

18 (a) There are a fixed n = 100 trials, and the probability of success is p = 0.08 on every trial. This is a binomial distribution.
MATH200A part 3, or `binomcdf(100,.08,5)`

(b) This is a binomial distribution, for exactly the same reasons.
MATH200A part 3, or `binompdf(100,.08,5)`

(c) The probability of success is p = 0.08 on every trial, but you don’t have a fixed number of trials. This is a geometric distribution.
`geometpdf(.08,5)`

19 C

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

20 r must be between −1 and +1 inclusive. (Symbolically, −1 ≤ r ≤ +1, or | 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.
21 , proportion of a sample (In this case,  = 4/5 = 0.8 or 80%.)
22 attribute or qualitative, specifically binomial (“Are you satisfied with the food service?”)
23 Attribute (qualitative or categorical) data. This compact form of graph makes it easy to compare the relative sizes of all the categories. (A bar graph is also a common choice for qualitative data.)

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 graph.

24 When the data are skewed, prefer the median.
25 Because you can never accept the null hypothesis; only the alternative hypothesis can be accepted.
26 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, the textbooks used at TC3 always put 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.

27 C
28 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.
29 1–α, or (1−α)100% is also acceptable.
30 B

Remark: The Z-Test is wrong because you don’t know the SD of the selling price of all 2006 Honda Civics in the US. The 1-PropZTest and χ²-test are for non-numeric data. There is no such thing as a 1-PropTTest.

31 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.

32 C

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.

33 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.

34 D

Remark: This is unpaired numeric data, Case 4.

35 (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.

• Random sample? ✔
• At least 10 successes and 10 failures expected? npo = 500×0.25 = 125, and 500−125 = 375, both >10. ✔

Common mistake: For hypothesis test, you need expected successes and failures. It’s incorrect to use actual successes (150) and failures (350).

• Check that the sample is not too large: 10n = 10×500 = 5000, and far more than 5000 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.

## Problem Set 2: Calculations

36 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, p-value = .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.

Some instructors do a preliminary F-test. It gives p=0.9089>0.05, so after that test you would use Pooled:Yes in the 2-SampTTest and get p=0.1553.

37 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.
38

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 CRIT(.9054). Therefore data are ND. Box-whisker (MATH200A part 2) shows no outliers.  (You don’t have to show these graphs on your exam paper; just show the numeric test for normality and mention that the modified boxplot shows no outliers.) T-Test: μo=4, x̅=4.5625, s=1.34..., n=8, μ>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) TInterval: C-Level=.95
Results: (3.4418, 5.6832)

(There’s no need to repeat the requirements check or to write down all the sample statistics again.)

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

39 (a) 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) 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) 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) For this problem you must know the formula:

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

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

Common mistake: 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 (a) z = (x−μ)/σ ⇒ −1.2 = (x−70)/2.4 ⇒ x = 67.1″
or: x = zσ + μ ⇒ x = −1.2×2.4 + 70 = 67.1″

(b) 70−67.6 = 2.4″, and therefore z = −1. By the Empirical Rule, 68% of data lie between z = ±1. Therefore 100−68 = 32% lie outside z = ±1 and 32%/2 = 16% lie below z = −1. Therefore 67.6″ is the 16th percentile.

Alternative solution: Use the big chart to add up the proportion of men below 67.6″ or below z = −1. That is 0.15+2.35+13.5 = 16%.

(c) z = (74.8−70)/2.4 = +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 100−2.5 = 97.5% fall below z = +2. 97.5% of men are shorter than 74.8″.

Alternative solution: You could also use the big chart to find that P(z > 2) = 2.35+0.15 = 2.5%, and then P(z < 2) = 100−2.5 = 97.5%.

42 (a) 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) 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 Summary Numbers on the TI-83.

(c) Further data from `1-VarStats L1,L2`:  = 990.1 and s = 167.3

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.

(d) f/n = 29/219 ≈ 0.13 or 13%

43 The 85th percentile is the speed such that 85% of drivers are going slower and 15% are going faster.

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

44 (a) This is binomial data (each person either would or would not take the bus), hence Case 2, One population proportion.

MATH200A/sample size/binomial:  = .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 Sample Size by Formula 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) If you have no prior estimate, use  = 0.5. The other inputs are the same, and the answer is 423

45 (a) For the procedure, see Step 1 of Scatterplot, 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) See Step 2 of Scatterplot, 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) 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) 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) 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) 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) 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.

46 invNorm(1-.06, 2.0, 0.1) = 2.1555, about 2.16 mm
47 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 with r(.957) > CRIT(.8978), 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.
48 (a) normalcdf(-10^99, 24, 27, 4) = .2266272794 → 0.2266 or about a 23% chance

(b) normalcdf(-10^99, 24, 27, 4/√5) = .0467662315 → 0.0468 or about a 5% chance

49 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 preferences are the same as national proportions. H1: Nebraska preferences 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 preferences in vacation homes are different from those for the US as a whole.
50 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.

(b) For two-population numeric data, paired data do a good job of controlling for lurking variables. You would test each person’s blood sugar, then enroll all thirty patients in the course and test their blood sugar six months after the end of the course. Your variable d is blood sugar after the course minus blood sugar before, and your H1 is μd < 0.

One potential problem is that all 30 patients receive a heightened level of attention, so you have to worry about the placebo effect. (With the original experiment, the control group did not receive the extra attention of being in the course, so any difference from the attention is accounted for in the different results between control group and treatment group.)

It seems unlikely that the placebo effect would linger for six months after the end of a short course, but you can’t rule out the possibility. There are two answers to that. You could re-test the patients after a year, or two years. Or, you could ask whether it really matters why patients do better. If they do better because of the course itself, or because of the attention, either way they’re doing better. A short course is relatively inexpensive. If it works, why look a gift horse in the mouth? In fact, medicine is beginning to take advantage of the placebo effect in some treatments.

51 This is a test on the mean of one population, with population standard deviation unknown: Case 1.
(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.
52 (a) 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, men and women equally likely to refuse promotions H1: p1 > p2, men more likely to refuse promotions α = 0.05 independent random samples For each sample, 10n = 10×200 = 2000 is far less than the total number of men or women. Men: 60 yes, 200−60 = 140 no; women: 48 yes, 200−48 = 152 no; all are ≥ 10. (The formal requirement uses the blended proportion p̂ = (60+48)/(200+200) = .27, so men have .27×200 = 54 expected yes and 200−54 = 146 expected no, and women have the same; again, all are ≥ 10.) 2-PropZTest: x1=60, n1=200, x2=48, n2=200, p1>p2 Results: z=1.351474757 → z = 1.35, p=.0882717604 → p-value = .0883, p̂1=.3, p̂2=.24, p̂=.27 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) 2-PropZInt with the above inputs and C-Level=.95 gives (−.0268, .14682). 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 men were between 2.7 percentage points less likely than women, and 14.7 percentage points more likely.

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

53 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 SD. 2σ = 100−70 = 30 → σ = 15

54 This is binomial data, Case 2. (The members of the sample are insurance claims, and each claim either is settled or is not.)
(1) H0: p = .75 H1: p < .75 α = 0.05 random sample 10n = 10×65 = 650, obviously less than the total number of claims filed in the state. 65×0.75 = 48.75 expected successes and 65−48.75 = 16.25 expected failures, both ≥ 10. Common mistake: Don’t use the actual successes and failures, 40 and 65−40 = 25. That would be right for a confidence interval, but for a hypothesis test you assume H0 is true and so you must use the proportion 0.75 from your null hypothesis. 1-PropZTest: po=.75, x=40, n=65, prop
55 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%
56

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. T-Test: μo=0, List:L3, 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.
57 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

You could also use binompdf(2, .9, 1) = 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 replacement, 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.

58 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)
59 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.

60 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 each sample is random, each sample has at least 10 successes and 10 failures, and each sample is less than 10% of the population it came from.

(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.

61 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.) Note: sample sizes are 80+20 = 100 and 135+15 = 150.
(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 independent random samples 10n1=10×100=1000; 10n2=10×150=2000; obviously there are far more than 3000 seeds of this type. In sample 1, 80 successes and 20 failures; in sample 2, 135 successes and 15 failures; all are at least 10. (The formal requirement uses the blended proportion p̂ = (80+135)/(100+150) = 0.86 to find expected successes and failures. For sample 1, 0.86×100 = 86 and 100−86 = 14; for sample 2, 0.86×150 = 129 and 150−129 = 21. All are at least 10.) 2-PropZTest: x1=80, n1=80+20, x2=135, n2=135+15, p1≠p2 Results: z = −2.23, p-value = 0.0256, p̂1 = .8, p̂2 = .9, p̂ = .86 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

## What’s New

• 1 Jan 2016: retake screen shots here and here for the new version of MATH200A Program part 4.
• 5–9 Dec 2015: Correct a requirements check.
• (intervening changes suppressed)
• 11 Nov 2007: New document, mostly an amalgamation of the old separate review problems for descriptive and inferential statistics.
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