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Solutions for Chapter 9

Updated 30 June 2015 (What’s New?)
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← Exercises for Ch 9 

1 You make probability statements about things that can change if you repeat the experiment. There’s a 1/6 chance of rolling doubles, because you’ll get doubles about 1/6 of the times that you roll two dice. But the mean of the population is one definite number. It doesn’t change from one experiment to the next. Your estimate changes, because it’s based on your sample and no sample is perfect. But the thing you’re trying to estimate, mean or proportion, is what it is even though you don’t know it exactly.

(Statisticians would say, “the population mean or proportion is not a random variable.” By that, they mean just what I said in less technical language.)

2

Answer: A confidence interval for numeric data is an estimate of the average, and tells you nothing about individuals. Correct his conclusion to I’m 90% confident that the average food expense for all TC3 students is between $45.20 and $60.14 per week..

Remark: Use all or a similar word to show that you’re estimating the mean for the population, not just the sample of 40 students. There’s no need to estimate the mean of the sample, because you know the exact sample mean for your sample.

Remark: Be clear in your mind that you’re estimating the average spending per student at $45–60 a week. Some individual students will quite likely spend outside that range, so your interpretation shouldn’t say anything about individual student spending.

3

Answer: It’s the use of the word average. When you collect data points that are all yes/no or success/failure, you have a sample proportion , equal to the number of successes divided by sample size, and you can estimate a population proportion. There is no “average” with non-numeric data.

Your 90% confidence estimate is simply that 27% to 40% usually or always prepare their own food.

4

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements: random sample, OK. 10n = 10×40 = 400 is less than total number of batteries made; OK. n = 40 >30, OK.

TInterval 1756, 142, 40, .95

(1710.6, 1801.4)

Neveready is 95% confident that the average Neveready A cell, operating a wireless mouse, lasts 1711 to 1801 minutes (28½ to 30 hours).

Common mistake: Don’t make any statement about 95% of the batteries! Your CI is about your estimate of one number, the average life of all batteries. Your CI has a margin of error of ±15 minutes; the 95% range for all batteries would be about 4 to 5 hours.

5 (a) = 5067/10000 = 0.5067

Don’t make the term “point estimate” harder than it is! The point estimate for the population mean (or proportion, standard deviation, etc.) is just the sample mean (or proportion, standard deviation, etc.).

(b) The sample is his actual data, the 10,000 flips. Therefore the sample size is n = 10,000. The population is what he wants to know about, all possible flips. The population size is infinite or “indefinitely large”.

6

This is sample size for a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases. Since you have no prior estimate, use 0.5 for .

With the MATH200A program (recommended): If you’re not using the program:
MATH200a/sample size/binomial,  = .5, E = .035, C-Level = .95, sample size is at least 784 invNorm(1-.025, 0, 1) yields 1.959963986.  Divided by .035 is 55.99897103.  Squared is 3135.884756.  Times .5, times 1 minus .5, is 783.9711891. The formula is n=Find z-sub-alpha-over-2, divide by E, and square. Multiply by p-hat and then by 1 minus p-hat..
1−α = .95 ⇒ α/2 = 0.025.
z0.025 = invNorm(1−.025, 0, 1)
Divide by .035, square the result, and multiply by .5*(1−.5).
Answer: at least 784. Remember — you’re not rounding, you’re going up to a whole number.
7

This is a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases.

Requirements:

Common mistake: Don’t say “n > 30” or “n ≥ 30”. That’s true, but it doesn’t help you with binomial data. For computing a confidence interval about a proportion from binomial data, the “sample size large enough” condition is at least 10 successes and at least 10 failures, not sample size at least 30.

1-PropZInt 40, 100, .9 → (.31942, .48058),  = .4

31.9% to 48.1% of all claims at that office have been open for more than a year (90% confidence).

8

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

TInterval 17.7, 1.8, 40, .95 → (17.124, 18.276)
She’s 95% confident that the average of all her commutes is 17.1 to 18.3 minutes.

9

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

TInterval L6, 1, .95 → (62.918, 65.016), =63.96666667, s=1.894226818, n=15

The average height of women aged 20–29 is 62.9 to 65.0 inches (95% confidence).

Remark: Since adult women’s heights are known to be normally distributed, you could get away without checking for normality and outliers in this sample. But it does no harm to check every time.

10

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

TInterval L5, 1, .9 → (97.757, 98.343),  = 98.05, s =.7155828558 → 0.72, n = 18.
(a) Fred is 90% confident that the average body temperature of healthy male students is 97.8 to 98.3 °F.

(b) He’s 90% confident that the average body temperature is not more than 98.3°, so 98.6° as normal (average) temperature is inconsistent with his data.

(c) E = 98.343−98.05 = 0.3°, or E = 98.05−97.757 = 0.3°, or (98.343−97.757)/2 = 0.3°.

With the MATH200A program (recommended): If you’re not using the program:
(d) MATH200A/Sample size/Num unknown σ: s=.7155828558, E=.1, C-Level=.95, n≥202. He will need at least 202 in his sample. (d) Confidence level = 1−α = 0.95 ⇒ α = 0.05 ⇒ α/2 = 0.025.
computations as described in text z0.025 = invNorm(1−.025)

Multiply by s, divide by E, and square the result. This gives 197. But the t distribution is more spread out than the normal (z) distribution, so you probably want to bump that number up a bit, say to 200 or so.

11

This problem is about a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases.

(a) Requirements check:

1-PropZInt, 219, 500, .9 → (.4015, .4745),  = .438

You’re 90% confident that 40.2% to 47.5% of Metropolis adults aged 50–75 have had a colonoscopy in the past ten years.

(b) MATH200A/sample size/binomial,  = .438, E = .02, C-Level = .9 → at least 1665

12

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

TInterval L4, 1, .95 → (179.86, 198.93),  = 189.40, s = 20.37, n = 20

You’re 95% confident that the average of all cash deposits is between $179.86 and $198.93.

Common mistake: Don’t say that 95% of deposits are between those values — if you look at the sample you’ll see that’s pretty unlikely. You’re estimating the average, not the individual deposits in the population.

13

This is a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases.

Requirements check:

1-PropZInt 520, 1000, .95 → (.48904, .55096),  = .52

With 95% confidence, 48.9% to 55.1% of voters voted Snake. At the 95% confidence level, we can’t tell whether more or less than 50% of voters voted for Abe Snake.

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