(4) Binomial distributions have two parameters. Name these
ANS: n (number of independent trials); p (probability of success for each trail)
(5) The normal distribution has two two parameters . Name these and identify the symbols used to denote them.
ANS: � (mean or expected value); s2 (variance)
(6) Other than their source (population vs. sample), identify a way in which the sample mean differs from the population mean.
ANS: The sample mean is a random variable. The population mean is a constant.
(1) Suppose that the probability of selecting a diabetic at random from a population is 1 in 10 (0.10). A random sample of n = 4 is selected.
(A) What is the probability of selecting 3 diabetics in a given sample?
Pr(X = 3) = .0036
(B) What is the probability of selecting 4 diabetics?
Pr(X = 4) = .0001
(C) What is the probability of selecting at least 3?
Pr(X >=3 ) = Pr(X = 3) + Pr(X = 4) = .0036 + .0001 = .0037
ALTERNATIVE SOLUTION: Pr(X >= 3) = 1 - Pr(X < 2) = 1 - .9963 = .0037
(2) Regarding standard normal random variable Z:
(A) What is the probability that a Z picked at random will less than 1.96? _________0.975
(B) What is the probability that a Z picked at random will be greater than +1.96? _________0.025
(C) What is the probability that a Z picked at random will lie between -1.96 and +1.96? ________0.95
(D) What is the probability that a Z picked at random will be either less than -1.96 or greater than +1.96? _________0.05
(3) Determine the following z percentiles:
(Answers provided)
(A) z.25 = -0.67
(B) z.85 = 1.04
(C) z.995 = 2.58
(4) What is the probability of observing a mean of at least 220 in a sample on n = 25 from a normal population with � = 200 and s = 49?
ANS:
SEM = 49 / sqrt(25) = 9.8
z = (220 - 200) / 9.8 = 2.04
Pr(X > 220) = Pr(Z >2.04) = 1 - .9793 = .0207
(Comment: You should draw the sampling distribution; not shown on Web because of technical limitations!)
(5) The data below represent a random sample of lengths of stay at a long term care facility. Test whether the mean length of stays is significantly different from 45 days. Include all hypothesis testing steps. Let alpha = .05. Show all calculations. Interpret your results.
Data: 17, 45, 31, 28, 27
Descriptive statistics
n = 5
mean = 29.60
s = 10.0896
sem = 10.0896 / sqrt(5) = 4.5122
H0: � = 45 vs. H1: � not = 45
a = .05
tstat = (29.6 - 45) / (4.5122) = -3.41
df = 5 - 1 = 4
.02 < p < .05 [Show t distribution and areas under the curve here.]
Reject H0 (Data suggest a length of stay that is significantly less than 45 days.)
(6) A sample of n = 12 shows a mean of 85 with s = 12.67.
(A) Calculate a 95% CI for �.
ANS:
sem = 12.67 / sqrt(12) = 3.66
85 � (t11,1-.05/2)(12.67/3.66) = 85 � (2.20)(3.66) = 85 � 8.1 = (76.9, 93.1)
(B) Calculate a 99% CI for �.
ANS:
85 � (t11,1-.01/2)(12.67/3.66) = 85 � (3.11)(3.66) = 85 � 11.4 = (73.6, 96.4)