Proportions and Percentiles for Standard Normal ScoresStandard Proportion Standard Proportionscore, z below z Percentile score, z below z Percentile–6.00 0.000000001 0.0000001 0.03 0.51 51–5.20 0.0000001 0.00001 0.05 0.52 52–4.26 0.00001 0.001 0.08 0.53 53–3.00 0.0013 0.13 0.10 0.54 54–2.576 0.005 0.50 0.13 0.55 55–2.33 0.01 1 0.15 0.56 56–2.05 0.02 2 0.18 0.57 56–1.96 0.025 2.5 0.20 0.58 57–1.88 0.03 3 0.23 0.59 59–1.75 0.04 4 0.25 0.60 60–1.64 0.05 5 0.28 0.61 61–1.55 0.06 6 0.31 0.62 62–1.48 0.07 7 0.33 0.63 63–1.41 0.08 8 0.36 0.64 64–1.34 0.09 9 0.39 0.65 65–1.28 0.10 10 0.41 0.66 66–1.23 0.11 11 0.44 0.67 67–1.17 0.12 12 0.47 0.68 68–1.13 0.13 13 0.50 0.69 69–1.08 0.14 14 0.52 0.70 70–1.04 0.15 15 0.55 0.71 71–0.99 0.16 16 0.58 0.72 72–0.95 0.17 17 0.61 0.73 73–0.92 0.18 18 0.64 0.74 74–0.88 0.19 19 0.67 0.75 75–0.84 0.20 20 0.71 0.76 76–0.81 0.21 21 0.74 0.77 77–0.77 0.22 22 0.77 0.78 78–0.74 0.23 23 0.81 0.79 79–0.71 0.24 24 0.84 0.80 80–0.67 0.25 25 0.88 0.81 81–0.64 0.26 26 0.92 0.82 82–0.61 0.27 27 0.95 0.83 83–0.58 0.28 28 0.99 0.84 84–0.55 0.29 28 1.04 0.85 85–0.52 0.30 30 1.08 0.86 86–0.50 0.31 31 1.13 0.87 87–0.47 0.32 32 1.17 0.88 88–0.44 0.33 33 1.23 0.89 89–0.41 0.34 34 1.28 0.90 90–0.39 0.35 35 1.34 0.91 91–0.36 0.36 36 1.41 0.92 92–0.33 0.37 37 1.48 0.93 93–0.31 0.38 38 1.55 0.94 94–0.28 0.39 39 1.64 0.95 95–0.25 0.40 40 1.75 0.96 96–0.23 0.41 41 1.88 0.97 97–0.20 0.42 42 1.96 0.975 97.5–0.18 0.43 43 2.05 0.98 98–0.15 0.44 44 2.33 0.99 99–0.13 0.45 45 2.576 .995 99.5–0.10 0.46 46 3.01 0.9987 99.87–0.08 0.47 47 3.72 0.9999 99.99–0.05 0.48 48 4.26 0.99999 99.999–0.03 0.49 49 5.20 0.9999999 99.999990.00 0.50 50 6.00 0.999999999 99.9999999

Question 1. If Z is a standard normal randomvariable, what is P(−0.88 < Z < 0.47)?

(A) 0.19− 0.08

(B) 0.81− 0.32

(C) 0.88− 0.47

(D) 0.68− 0.19

(E) 0.81− 0.47

Question 2.

A study of caffeine levels of farmers and doctors reported thefollowing data:

Farmers DoctorsSample Size 25 25Sample Mean 13 10Sample S.D. 4 3

A 95% confidence interval for the difference in population meancaffeine levels (farmers’ mean minus doctors’ mean) is:

(A) (13− 10)± 2× (SE of the difference).

(B) (4− 3)± 1.64× (SE of the difference).

(C) (13− 10)± 1.64× (SE of the difference).

(D) (4− 3)± 2× (SE of the difference).

Question 3.

All other things remaining constant, if the sample size is multipliedby 16 then the standard error of the mean is multiplied by

(A) 1

(B) 1/4

(C) 1/16

(D) 2

(E) 16

Question 4.

In a study of caffeine levels of farmers and doctors, suppose that a95% confidence interval for the difference in the means did notcontain the value zero. We could infer that the population meansof farmers’ and doctors’ caffeine levels:

(A) Are close to each other; there is no significant differencebetween them.

(B) Have no relationship to each other; we have insufficientevidence to make an inference about their relative location.

(C) Are equal to each other; we are 95% confident that they areequal.

(D) Are significantly different from each other.

Question 5.

Suppose you have to cross a train track on your commute. Theprobability that you will have to wait for a train is 1/5, or .20. Ifyou don’t have to wait, the commute takes 15 minutes, but if youhave to wait, the commute takes 20 minutes. What is theexpected value of the time it takes you to commute?

(A) 20× 15 + 15, or 19 minutes

(B) 20× 15 + 15× 4

5 , or 16 minutes

(C) (20 + 15)× 15 , or 7 minutes

(D) 15+202 , or 17.5 minutes

(E) 20× 15 , or 4 minutes

Question 6.

A certain random variable X has possible values 0, 1, 2, and 3. IfP(X = 0), P(X = 1), and P(X = 2) are all equal to 0.2, what isP(X = 3)?

(A) 0.2

(B) 0.3

(C) 0.4

(D) 0.5

(E) 0.1

Question 7.

All other things remaining constant, if the sample size increases bya factor of nine then the confidence interval for the populationmean will:

(A) Become one-ninth as wide.

(B) Become one-third as wide.

(C) Become nine times as wide.

(D) Triple in width.

Question 8.

Phil decides to flip a fair coin repeatedly until he has seen 10heads. Let X be the total number of flips needed for this. Why isX not a binomial random variable?

(A) The total number of trials is not fixed.

(B) Actually, nothing is wrong; X really is a binomial randomvariable.

(C) X is not a discrete random variable.

(D) The probability of a success is not the same in each trial.

(E) The trials are not independent.

Question 9.

All other things remaining constant, if the sample size decreasesthen the standard error of the sample mean:

(A) Increases, levels off, and then increases again.

(B) Increases.

(C) Decreases.

(D) Will remain unchanged.

(E) Decreases and then increases.

Question 10.

The mean of a large number of sample proportions fromequally-sized random samples will be approximately:

(A) The true proportion of the population.

(B) The area below the normal curve and between -1.96 and+1.96.

(C) The square root of:(true proportion)× (1− true proportion)/(sample size).

(D) A proportion of the population which is never sampled.

(E) The square root of:(true proportion)× (sample size)/(1− true proportion).

Question 11.

If a result is statistically significant, this means that it is animportant result.

(A) True

(B) False

Question 12.

In a study of farmers’ caffeine levels, a random sample of 25farmers yielded a sample mean of 22 and a sample standarddeviation (S.D.) of 4. Therefore, the standard error of the mean(SEM) is:

(A) 4/√

25

(B) 22×√

25

(C) 4×√

25

(D) 22/√

25

Proportions and Percentiles for Standard Normal ScoresStandard Proportion Standard Proportionscore, z below z Percentile score, z below z Percentile–6.00 0.000000001 0.0000001 0.03 0.51 51–5.20 0.0000001 0.00001 0.05 0.52 52–4.26 0.00001 0.001 0.08 0.53 53–3.00 0.0013 0.13 0.10 0.54 54–2.576 0.005 0.50 0.13 0.55 55–2.33 0.01 1 0.15 0.56 56–2.05 0.02 2 0.18 0.57 56–1.96 0.025 2.5 0.20 0.58 57–1.88 0.03 3 0.23 0.59 59–1.75 0.04 4 0.25 0.60 60–1.64 0.05 5 0.28 0.61 61–1.55 0.06 6 0.31 0.62 62–1.48 0.07 7 0.33 0.63 63–1.41 0.08 8 0.36 0.64 64–1.34 0.09 9 0.39 0.65 65–1.28 0.10 10 0.41 0.66 66–1.23 0.11 11 0.44 0.67 67–1.17 0.12 12 0.47 0.68 68–1.13 0.13 13 0.50 0.69 69–1.08 0.14 14 0.52 0.70 70–1.04 0.15 15 0.55 0.71 71–0.99 0.16 16 0.58 0.72 72–0.95 0.17 17 0.61 0.73 73–0.92 0.18 18 0.64 0.74 74–0.88 0.19 19 0.67 0.75 75–0.84 0.20 20 0.71 0.76 76–0.81 0.21 21 0.74 0.77 77–0.77 0.22 22 0.77 0.78 78–0.74 0.23 23 0.81 0.79 79–0.71 0.24 24 0.84 0.80 80–0.67 0.25 25 0.88 0.81 81–0.64 0.26 26 0.92 0.82 82–0.61 0.27 27 0.95 0.83 83–0.58 0.28 28 0.99 0.84 84–0.55 0.29 28 1.04 0.85 85–0.52 0.30 30 1.08 0.86 86–0.50 0.31 31 1.13 0.87 87–0.47 0.32 32 1.17 0.88 88–0.44 0.33 33 1.23 0.89 89–0.41 0.34 34 1.28 0.90 90–0.39 0.35 35 1.34 0.91 91–0.36 0.36 36 1.41 0.92 92–0.33 0.37 37 1.48 0.93 93–0.31 0.38 38 1.55 0.94 94–0.28 0.39 39 1.64 0.95 95–0.25 0.40 40 1.75 0.96 96–0.23 0.41 41 1.88 0.97 97–0.20 0.42 42 1.96 0.975 97.5–0.18 0.43 43 2.05 0.98 98–0.15 0.44 44 2.33 0.99 99–0.13 0.45 45 2.576 .995 99.5–0.10 0.46 46 3.01 0.9987 99.87–0.08 0.47 47 3.72 0.9999 99.99–0.05 0.48 48 4.26 0.99999 99.999–0.03 0.49 49 5.20 0.9999999 99.999990.00 0.50 50 6.00 0.999999999 99.9999999

Question 13. Suppose that IQ scores arenormally distributed with a mean of 100 and astandard deviation of 16. What is theprobability that a randomly sampled individualhas an IQ score less than 108?

(A) 0

(B) 0.69

(C) 0.5

(D) 0.98

(E) 0.31

Question 14.

A statistical study considers the question of whether highlyeducated people are less likely to develop Alzheimer’s disease thanothers. In this study, the alternative hypothesis is:

(A) There is a relationship between level of education and thedevelopment of Alzheimer’s disease.

(B) Highly educated people are less likely than others to developAlzheimer’s disease.

(C) Highly educated people are certain of developing Alzheimer’sdisease.

(D) Insufficient information is given to allow us to determine thealternative hypothesis.

(E) There is no relationship between level of education and thedevelopment of Alzheimer’s disease.

Proportions and Percentiles for Standard Normal ScoresStandard Proportion Standard Proportionscore, z below z Percentile score, z below z Percentile–6.00 0.000000001 0.0000001 0.03 0.51 51–5.20 0.0000001 0.00001 0.05 0.52 52–4.26 0.00001 0.001 0.08 0.53 53–3.00 0.0013 0.13 0.10 0.54 54–2.576 0.005 0.50 0.13 0.55 55–2.33 0.01 1 0.15 0.56 56–2.05 0.02 2 0.18 0.57 56–1.96 0.025 2.5 0.20 0.58 57–1.88 0.03 3 0.23 0.59 59–1.75 0.04 4 0.25 0.60 60–1.64 0.05 5 0.28 0.61 61–1.55 0.06 6 0.31 0.62 62–1.48 0.07 7 0.33 0.63 63–1.41 0.08 8 0.36 0.64 64–1.34 0.09 9 0.39 0.65 65–1.28 0.10 10 0.41 0.66 66–1.23 0.11 11 0.44 0.67 67–1.17 0.12 12 0.47 0.68 68–1.13 0.13 13 0.50 0.69 69–1.08 0.14 14 0.52 0.70 70–1.04 0.15 15 0.55 0.71 71–0.99 0.16 16 0.58 0.72 72–0.95 0.17 17 0.61 0.73 73–0.92 0.18 18 0.64 0.74 74–0.88 0.19 19 0.67 0.75 75–0.84 0.20 20 0.71 0.76 76–0.81 0.21 21 0.74 0.77 77–0.77 0.22 22 0.77 0.78 78–0.74 0.23 23 0.81 0.79 79–0.71 0.24 24 0.84 0.80 80–0.67 0.25 25 0.88 0.81 81–0.64 0.26 26 0.92 0.82 82–0.61 0.27 27 0.95 0.83 83–0.58 0.28 28 0.99 0.84 84–0.55 0.29 28 1.04 0.85 85–0.52 0.30 30 1.08 0.86 86–0.50 0.31 31 1.13 0.87 87–0.47 0.32 32 1.17 0.88 88–0.44 0.33 33 1.23 0.89 89–0.41 0.34 34 1.28 0.90 90–0.39 0.35 35 1.34 0.91 91–0.36 0.36 36 1.41 0.92 92–0.33 0.37 37 1.48 0.93 93–0.31 0.38 38 1.55 0.94 94–0.28 0.39 39 1.64 0.95 95–0.25 0.40 40 1.75 0.96 96–0.23 0.41 41 1.88 0.97 97–0.20 0.42 42 1.96 0.975 97.5–0.18 0.43 43 2.05 0.98 98–0.15 0.44 44 2.33 0.99 99–0.13 0.45 45 2.576 .995 99.5–0.10 0.46 46 3.01 0.9987 99.87–0.08 0.47 47 3.72 0.9999 99.99–0.05 0.48 48 4.26 0.99999 99.999–0.03 0.49 49 5.20 0.9999999 99.999990.00 0.50 50 6.00 0.999999999 99.9999999

Question 15. To calculate a 99% confidenceinterval for a population proportion, we use:

(A) Sample proportion ± 2.576×S.E.

(B) Sample proportion ± 1.64×S.E.

(C) Sample proportion ± 2×S.E.

(D) Sample proportion ± 2.33×S.E.

Question 16.

The standard deviation of the histogram of a large number ofsample means is, approximately:

(A) (population standard deviation)/√

sample size.

(B) The area below the normal curve and between −2 and +2.

(C) The mean of a proportion of the population which is neversampled.

(D) The true mean of the population.

Question 17.

Suppose that X is a binomial random variable with mean 10 andstandard deviation 3. What is the correct normal approximation forP(X > 10) using the continuity correction? (Assume that Zdenotes a standard normal random variable.)

(A) P(Z > [3− 10]/10)

(B) P(Z > [10− 10]/3)

(C) P(Z > [10− 3]/10)

(D) P(Z > [9.5− 10]/3)

(E) P(Z > [10.5− 10]/3)

Question 18.

The mean of a large number of sample means from equally-sizedrandom samples will be approximately:

(A) (population standard deviation)/√

sample size.

(B) The mean of a proportion of the population which is neversampled.

(C) The true mean of the population.

(D) The area below the normal curve and between -1.96 and+1.96.

Question 19.

All other things remaining constant, if the population sizequadruples from 10 million to 40 million then the width of aconfidence interval will:

(A) Decrease by half.

(B) Remain unchanged.

(C) Increase by a factor of two.

(D) Increase and then decrease.

Question 20.

All other things remaining constant, if the sample size increases bya factor of 25 then the standard error of the mean:

(A) Becomes five times as large.

(B) Becomes one fifth as large.

(C) Becomes one twenty-fifth as large.

(D) Will remain unchanged.

(E) Becomes twenty-five times as large.

Question 21.

The histogram of sample means from a large number ofequally-sized random samples of size 50 will be shapedapproximately like a:

(A) Semi-circle.

(B) Skewed histogram, with a long right tail.

(C) Normal (bell-shaped) curve.

(D) Triangle.

Question 22.

In a random sample of 400 PSU graduates, 64% stated that theyprefer root beer (over birch beer). Therefore, a 90% confidenceinterval for the proportion of all PSU graduates who prefer rootbeer is:

(A) 0.64± 1.64×√.64/400

(B) 0.64± 2×√

.64/400

(C) 0.64± 2×√

.64× .36/400

(D) 0.64± 1.64×√.64× .36/400

Proportions and Percentiles for Standard Normal ScoresStandard Proportion Standard Proportionscore, z below z Percentile score, z below z Percentile–6.00 0.000000001 0.0000001 0.03 0.51 51–5.20 0.0000001 0.00001 0.05 0.52 52–4.26 0.00001 0.001 0.08 0.53 53–3.00 0.0013 0.13 0.10 0.54 54–2.576 0.005 0.50 0.13 0.55 55–2.33 0.01 1 0.15 0.56 56–2.05 0.02 2 0.18 0.57 56–1.96 0.025 2.5 0.20 0.58 57–1.88 0.03 3 0.23 0.59 59–1.75 0.04 4 0.25 0.60 60–1.64 0.05 5 0.28 0.61 61–1.55 0.06 6 0.31 0.62 62–1.48 0.07 7 0.33 0.63 63–1.41 0.08 8 0.36 0.64 64–1.34 0.09 9 0.39 0.65 65–1.28 0.10 10 0.41 0.66 66–1.23 0.11 11 0.44 0.67 67–1.17 0.12 12 0.47 0.68 68–1.13 0.13 13 0.50 0.69 69–1.08 0.14 14 0.52 0.70 70–1.04 0.15 15 0.55 0.71 71–0.99 0.16 16 0.58 0.72 72–0.95 0.17 17 0.61 0.73 73–0.92 0.18 18 0.64 0.74 74–0.88 0.19 19 0.67 0.75 75–0.84 0.20 20 0.71 0.76 76–0.81 0.21 21 0.74 0.77 77–0.77 0.22 22 0.77 0.78 78–0.74 0.23 23 0.81 0.79 79–0.71 0.24 24 0.84 0.80 80–0.67 0.25 25 0.88 0.81 81–0.64 0.26 26 0.92 0.82 82–0.61 0.27 27 0.95 0.83 83–0.58 0.28 28 0.99 0.84 84–0.55 0.29 28 1.04 0.85 85–0.52 0.30 30 1.08 0.86 86–0.50 0.31 31 1.13 0.87 87–0.47 0.32 32 1.17 0.88 88–0.44 0.33 33 1.23 0.89 89–0.41 0.34 34 1.28 0.90 90–0.39 0.35 35 1.34 0.91 91–0.36 0.36 36 1.41 0.92 92–0.33 0.37 37 1.48 0.93 93–0.31 0.38 38 1.55 0.94 94–0.28 0.39 39 1.64 0.95 95–0.25 0.40 40 1.75 0.96 96–0.23 0.41 41 1.88 0.97 97–0.20 0.42 42 1.96 0.975 97.5–0.18 0.43 43 2.05 0.98 98–0.15 0.44 44 2.33 0.99 99–0.13 0.45 45 2.576 .995 99.5–0.10 0.46 46 3.01 0.9987 99.87–0.08 0.47 47 3.72 0.9999 99.99–0.05 0.48 48 4.26 0.99999 99.999–0.03 0.49 49 5.20 0.9999999 99.999990.00 0.50 50 6.00 0.999999999 99.9999999

Note: “population mean” should be changedhere to “population proportion” because meansuse t∗ multipliers while proportions use z∗

multipliers.

Question 23. A confidence interval for apopulation mean is given as “Sample mean ±2.33×SEM.” The corresponding level ofconfidence is:

(A) 98%

(B) 99%

(C) 95%

(D) 64%

Question 24.

All other things remaining constant, which of the following sampleproportions will result in the widest confidence interval?

(A) .1

(B) .4

(C) .2

(D) .5

(E) .3

Question 25.

The standard deviation of the histogram of a large number ofsample proportions is, approximately:

(A) The square root of:(true proportion)× (sample size)/(1− true proportion).

(B) The area below the normal curve and between -1.96 and+1.96.

(C) The square root of:(true proportion)× (1− true proportion)/(sample size).

(D) A proportion of the population which is never sampled.

(E) The true proportion of the population.

Question 26.

A confidence interval for a population proportion is a range ofnumbers which:

(A) Is certain to contain the population proportion.

(B) Increases in width as the sample size increases.

(C) Is always the same for different samples and has a 95%probability of containing the true population proportion.

(D) Is a plausible range of values for the population proportion.

Question 27.

In a randomized experiment, if the p-value is very small whencomparing the treatment and control groups, we can infer that thetreatment caused the difference:

(A) True

(B) False

Question 28.

A researcher interviewed 507 randomly chosen PSU students andfound that 59% of the students in his sample like to play chess.Consider the research question of whether or not a majority of PSUstudents like to play chess. The test for this research question is:

(A) A one-sided test.

(B) Neither a one-sided nor two-sided test.

(C) Both a one-sided and two-sided test.

(D) A two-sided test.

Question 29.

In a different question, it is stated that the interval 21.5 to 23.0 isa 95% confidence interval for the population mean. We mayconclude that:

(A) 23.3 is a plausible value for the population mean.

(B) 19.7 is a plausible value for the population mean.

(C) 21.8 is not a plausible value for the population mean.

(D) 22.3 is a plausible value for the population mean.

Question 30.

A researcher repeatedly collects random samples of size 1,600 andcomputes a 95% confidence interval for the population mean usingeach sample. Over the long run, the proportion of confidenceintervals which will fail to capture the population mean is:

(A) None of the above.

(B) 1/√

1, 600, or 1/40, 2.5%

(C) 95%

(D) 5%

Question 31.

All other things remaining constant, increasing the confidencecoefficient causes the width of a confidence interval to:

(A) Decrease.

(B) Remain unchanged.

(C) Increase.

(D) Increase for a while and then decrease.

Question 32.

A student claims that, for any data set of size two or more, thestandard error of the mean (SEM) is smaller than the samplestandard deviation. This claim is:

(A) Not always false; it depends on the sample size.

(B) Always false.

(C) Not always true; it depends on the actual numbers in thesample.

(D) Always true.

Question 33.

A random sample of 25 farmers was examined in a study ofcaffeine levels. From the data collected, a 95% confidence intervalfor the population mean caffeine level was calculated to be 21.5 to23.0. We can conclude that:

(A) If we repeatedly sample the entire population then, about95% of the time, the population mean will fall between 21.5 and23.0.

(B) If we repeatedly collect random samples of size 25 andcalculate the corresponding confidence intervals then, over the longrun, 95% of these intervals will capture the population mean and5% will fail to capture the population mean.

(C) For any random sample of 25 farmers, the resulting samplemean will always fall between 21.5 and 23.0.

(D) None of the above.

(E) 95% of all farmers have caffeine levels between 21.5 and 23.0.

Question 34.

A study of caffeine levels of farmers and doctors reported thefollowing data:

Farmers DoctorsSample Size 25 25Sample Mean 13 10Sample S.D. 4 3

The standard error of the difference between the two samplemeans is:(A) (Sample Mean1 − Sample Mean2)/

√SD1 − SD2 = (13− 10)/

√4− 3 = 3.0

(B)√

(SEM1)2 + (SEM2)2 =√

( 4√25

)2 + ( 3√25

)2 = 1.0

(C) The standard deviation of the data in the combined samples.

(D)√

(Sample Mean1)2/SD1 + (Sample Mean2)

2/SD2 =√

( 13√25

)2 + ( 10√25

)2 = 3.28

Question 35.

The histogram of sample proportions from a large number ofequally-sized random samples will be shaped approximately like a:

(A) Triangle.

(B) Normal (bell-shaped) curve.

(C) Semi-circle.

(D) Skewed curve, with a long right tail.

Question 36.

If we roll a fair 6-sided die 12 times and Y is the number of timeswe roll the number 5, what is the mean of Y ?

(A) 4

(B) 6

(C) 2

(D) 3

(E) 5

Question 37.

Fiona receives a beautiful four-sided die for her eighteenthbirthday. After playing with it for two hours, she starts to suspectthat her die is more favorable to rolling “1” than to any othernumber. Fiona’s alternative hypothesis is

(A) One-sided

(B) Two-sided

Question 38.

Fiona’s null hypothesis is that

(A) The die rolls “1” with probability greater than 0.25.

(B) The die rolls “1” with probability less than 0.25.

(C) The die rolls “1” with probability equal to 0.25.

(D) The die rolls “1” with probability not equal to 0.25.

Question 39.

Fiona chooses for her test statistic the standardized scorecorresponding to the sample proportion of 1’s obtained in 300 rollsof her die. The standardized score is

Sample proportion - Population proportion

S.D. of the sample proportion

Fiona rolled her die 300 times and obtained a “1” on 93, or 31%,of her rolls.The value of Fiona’s test statistic is:

(A) 300−93√(0.25)(1−0.25)

300

= 8,280

(B) 0.31−0.25√(0.25)(1−0.25)

300

= 2.4

(C) 0.25−0.31√(0.25)(1−0.25)

300

= -2.4

(D) 0.31−0.25√(0.31)(1−0.31)

300

= 2.25

Question 40.

It is known that in the presidential elections of 1992, 56% of alleligible adults actually voted. If we collect a large number of simplerandom samples each of size 1,600 adults then, about 95% of thetime, the sample proportion of adults who voted will fall between:

(A) .56± 3×√

.56(1− .56)/1600, or 0.524 and 0.596.

(B) .56± 4×√

.56(1− .56)/1600, or 0.512 and 0.608.

(C) .56± 2×√

.56(1− .56)/1600, or 0.536 and 0.584.

(D) .56± 1×√

.56(1− .56)/1600, or 0.548 and 0.572.