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.

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.

Answer: (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.

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.

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.

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.

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.

Answer: 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.

Answer: B,D — B if p<α, D if p>α

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.

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.

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.

Answer: 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.

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

Answer: 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.

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.

Answer: C, D, E

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

Answer: C

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

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.

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

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

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.

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

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

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.

Answer: C

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.

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

Answer: B

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.

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.

Answer: 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.

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.

Answer: D

Remark: This is unpaired numeric data, Case 4.

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 np_o(1−p_o) ≥ 10. Here n=500 and p_o=.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

(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.

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.

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.

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%.

(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

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

        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.

(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″.

(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: x̅ = 990.1 and s = 167.3
s² = 27982.90813 ⇒ s² = 27982.9

Common mistake: If you answered x̅ = 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%

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

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

Solution: The sampling distribution for n=35 is a normal distribution with μ = $70,000, σ_x̅ = $5500/√35 (about $930).
(a) z = (72050−70000)÷(5500/√35) ⇒ z = 2.2051
(b) P(x̅ ≥ 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 μ_o=70000, σ=5500, x̅=72050, n=35, μ>μ_o and read off z and p as above.

(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.

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

         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.)

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_x̅ = $45.00/√400 = $2.25.

P(x̅ ≥ 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_x̅ = 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 μ_o = 189.56, σ = 45, x̅ = 200, n = 400, μ > μ_o. The p-value is the same: 1.7439E-6.

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.

Solution: This is unpaired numeric data, Case 4.

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.

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

(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.

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

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

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.

            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.

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 second^C) + P(first^C 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 second^C) + P(first^C 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.

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)

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.

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

(a) p̂_T = 128/300 = 0.4267. p̂_C = 135/400 = 0.3375. p̂_T−p̂_C = 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 np̂(1−p̂) 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.

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.)

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

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