ANSWERS: (A) 210 (B) SEM = 90 / sqrt(36) = 15 (C) Sample mean~N(210, 152). (D) .5000 (E) This sample mean would be one standard deviation above the expected mean. Thus, P(sample mean < 225) = .8413. (F) P(sample mean > 225) = 1 - .8413 = 0.1587. (G) No, since the probability of this occurrence is greater than .05. (H) P(sample mean > 240) = .0228. (I) Yes, since the probability of this occurrence is less than .05.
|2|0
|3|0
|4|00
|5|00
|6|00
|7|0
|8|0
Notice that this distribution is mound-shaped.
ANS:
Variable | standard deviation (s) (computed w/SPSS) |
n | sem = s / sqrt(n) |
AGE (years) | 2.95 | 654 | 2.95 / sqrt(654) = 0.115 |
FEV (l/sec) | .867059 | 654 | .867059 / sqrt(654) = 0.0339047 |
HEIGHT (in) | 5.704 | 654 | 5.704 / sqrt (654) = 0.2230 |
ANS:
Variable | Sample proportion | n | SEP = sqrt(pq/n) |
SEX (male) | .514 | 654 | sqrt[(.514)(.486)/(654)] = .0195 |
SMOKE (yes) | .099 | 654 | sqrt[(.099)(.901)/(654)] = .0117 |
ANS:
Solution A (from the point of view of the hypothesized population mean, �): We may think of possible sample means that
are likely to be derived from a population with mean 64.2. Most of these sample means will be within �2.9 margins of error
from the (hypothesized) population mean of 64.2. Therefore, most sample means will lie in the range 61.3 to 67.1. Since our
sample mean is 56.2, it is unlikely to have come from the hypothesized population.
ANS:
(6A) p^ = 7 / 525 = .0133
Solution A: (From the point of view of the assumed population proportion.) Most sample proportions will be within �0.01 of the assumed population proportion of 0.004, or between 0.000 (since you can't have a negative proportion) and 0.014. The sample proportion falls within this range of possibility.
Solution B: (From the point of view of the observed sample proportion.) We would expect that the likely incidence of this outcome in the population of nurse anesthetists is �d of the sample proportion. This range is equal to .0133 � (2)(.0050) = .0133 � .010 = .0033 to .0233. Since the nationwide norm of .004 is captured by this interval, it may very well have been part of this population.
ANS: See Vocabulary list at end of chapter
You do a study in which you estimate a mean cost of treatment per patient of $200 with a margin or error of $150. Explain to a health care manager what this means in plain terms. One of your goals is to convince her that additional money must be spent in order to collect more data.
(G) Alpha level
(H) (1 - alpha)100% confidence interval
(I) Margin of error
ANSWERS:
(A) Statistical inference - generalizing from the sample to the population with calculated degree of certainty.
(B) Induction - a logical process in which we attempt to draw inferences from the particular to the general
(C) Estimation - an inferential method that uses sample statistics to directly determine the probable value of a population parameter.
(D) Hypothesis [significance] testing - an inferential method that provides a way to assess the "statistical significance" of findings, allowing
categorical conclusions to specific questions.
(E) Parameter - a statistical characteristic of the population.
(F) Statistic - a mathematical summary of the sample.
(G) Alpha - the chance the researcher is willing to take of not capturing the parameter.
(H) (1 - alpha)100% confidence interval - a interval that has a (1 - alpha)100% chance of capturing the parameter
(I) Margin of error - the "plus or minus" wiggle room that defines half of the confidence interval.
ANSWERS: Differences: (1) "X bar" represents the sample mean, whereas µ represents the population mean. (2) "X bar" is a statistic and µ is a parameter. (3) "X bar" is a random variable, whereas µ is a constant. (4) After you collect data, "x bar" is known (calculated) and µ is unknown (and must be inferred). Similarity: Both "x bar" and µ represent arithmetic averages or the center / "expected value" of a distribution.
(B) What is the standard error of the sample mean?
(C) Calculate a 95% confidence interval for the population mean.
(D) What is the margin of error of the above confidence interval?
(E) Calculate a 90% confidence interval for the population mean.
(F) Why is the 90% confidence interval shorter than the 95% confidence interval?ANSWERS:
(A) "x bar" ~ N(µ, 16/25); note the sampling distribution is not centered on the sample mean, but is centered on the
(unknown) population mean.
(B) SE("x bar") = sqrt(16/25) = 0.80
(C) 95% confidence interval for µ = 55.5 ± (1.96)(0.80) = 55.5 ± 1.568 = (53.9, 57.1)
(D) 1.568
(E) 90% confidence interval for µ = 55.5 ± (1.645)(0.80) = 55.5 ± 1.316 = (54.2, 56.8)
(F) Because we are will to accept a greater chance of failing to capture µ.
(A) Calculate the estimated standard error of the sample mean.
(B) Calculate a 95% confidence interval for µ.
(C) Calculate a 90% confidence interval for µ.
ANS:
(A) se("x bar") = 5 / sqrt(34) = 0.8575.
(B) 95% confidence interval for µ = 25 ± (t33,.975)(0.8575) = 25 ± (2.035)(0.8575) = 25 ±
1.75 =
(C) 90% confidence interval for µ = 25 ± (t33,.95)(0.8575) = 25 ± (1.692)(0.8575) = 25 ±
1.45 =
(A) For the variable AGE (years), compute the sample mean and a 95% confidence interval for µ.
(B) For the variable FEV (l/sec), compute the sample mean and a 95% confidence interval for µ.
(C) For the variable HEIGHT (inches), compute the sample mean and a 95% confidence interval for µ.
(D) For the variables SEX (0 = female, 1 = male), compute the number of males (X), the sample size n, the
sample proportion ("p hat"), and a a 95% confidence intervals for population proportion p.
(E) For the variables SMOKE (0 = non-smoker, 1 = smoker), compute the number of smokers (X), n, the sample proportion
("p hat"), and a 95% confidence intervals for population proportion p.
ANSWERS:
(A) AGE (years): sample mean = 9.93 (95% confidence interval for µ: 9.70, 10.16)
(B) FEV (liters per second): sample mean = 2.6368 (95% confidence interval for µ: 2.5702, 2.7033)
(C) HEIGHT (inches): sample mean = 61.144 (95% confidence interval for µ: 60.706, 61.582)
(D) SEX (male): "p hat" = 318 / 654 = .4862 (95% confidence interval for p: .4473, .5253)
(E) SMOKE (current smoker): "p hat" = 65 / 654 = .099 (95% confidence interval for p: .077, .525).
(A) What specific pmf might be used to model the distribution of the number of smokers (X) in similar random samples of
size n = 100?
(B) Can a normal approximation be applied in this situation? (Justify your response.) If so, how will X be distributed?
(C) What is the point estimate of p, based on this sample?
(D) Assuming the normal approximation holds, describe the sampling distribution of sample proportion.
(E) What is the standard error estimate of the sampling distribution?
(F) Calculate a 95% confidence interval for p.
(G) What is the margin of error for the above confidence interval?
(H) How would you decrease your margin of error?
ANSWERS:
(A) A binomial random variable, so that X ~ b(100, p)
(B) Yes, since if we assume that p is about equal to p hat (14/100), then npq = (100)(.14)(.86) = 12.04, which is greater
than 5. We can therefore assume that X ~ N(14, 12.04)
(C) "p hat" = 14 / 100 = .14.
(D) "p hat" ~ N(p, pq/n); notice that the sampling distribution is centered on unknown value p.
(E) se("p hat") = sqrt ((.14)(.86) / (100)) = .0347
(F) 95% confidence interval for p = .14 ± (1.96)(.0347) = .14 ± .0680 = (.07, .21)
(G) margin of error = .0680 (or as they would say on the news, ±7%)
(H) Increasing the sample size would decrease your standard error and hence decrease the margin of error of your estimate.
ANS:
(A) p^ = 20 / 75 = .2667
(B) (p^)(q^)n = (.2667)(1 - .2667)(75) = 14.67. Therefore, the normal approximation can indeed be applied.
(C) 95% confidence interval for p = (.2667) ± (1.96)(sqrt[(.2667)(1 - .2667) / (75)]) = .2667 ±
(1.96)(.0511) ~= .27 ± .10 = (.17, .37)
(D) The Web calculator gives the 95% confidence interval (.17, .38), which is very close to the confidence interval
calculated by the normal approximation method.
Group | Mean | SD | n |
Normal | 1.35 | 0.5 | 40 |
COPD | 0.92 | 0.4 | 32 |
ANS:
(A) 95% confidence interval for µnormal = (sample mean) ± t(n-1, .975)(sem) = 1.35 ±
t(39,.975)(0.5/sqrt(40)) = 1.35 ± (2.02)(0.079) = 1.35 ± 0.16 = (1.19, 1.51)
(B) 95% confidence interval for µnormal = 0.92 ± t(31, .975)(0.4/sqrt(32)) = 0.92 ± (2.04)(0.071) =
0.92 ± 0.14 = (0.78, 1.06)
(C) Since the confidence intervals do not overlap, we may be confident that the population means differ.