final exam winter
(a) Point estimate = 0.236; Margin of error = 0.282
(b) Point estimate = 0.236; Margin or error = 0.046
(c) Point estimate unknown; Margin of error = 0.023
(d) Point estimate = 0.259; Margin or error = 0.046
(e) Point estimate = 0.259; Margin of error = 0.023
(a) About 95% of the shoppers have between a 23.6% and a 28.2% chance of purchasing cookies.
(b) There is a 95% probability that the sample proportion lies between 0.236 and 0.282.
(c) If a second sample was taken, there is a 95% chance that its confidence interval would contain 0.25.
(d) This confidence interval indicates that more than 25% of shoppers buy cookies.
(e) We are reasonably certain the true proportion of shoppers who purchase cookies is between 24% and 28%.
1.64 (100/√ n) <= 10 (100/√ n)<=6.098 (100/6.098)^2=271
(e) The standard error cannot be calculated unless we know the standard deviation.
√((.56 x .44)/50) = .0702
(b)Lack of randomness in sampling
(c)Skew in the population distribution
(d)the presence of extreme outliers
(e)None of the above.
(a) The interval would be wider, because the standard error would be larger.
(b) The interval would be narrower, because the standard error would be smaller.
(c) The interval would be wider, because the critical z* would be larger.
(d) The interval would be narrower, because the critical z* would smaller.
(e) The interval would be about the same width, because the standard error would be smaller, but the critical z* would be larger. c
Keywords: wider, critical z*, larger
this year follows a normal distribution with mean ju and standard deviation a = 100. You read a
report that says, “on the basis of a simple random sample of 100 high school seniors that took the
SAT-M test this year, a confidence interval for ja is 512.00 ± 25.76.” The confidence level for this
Forty percent of the trees show some signs of damage. Which of the following statements is correct?
(a) 40% is a parameter
(b) 40% is a statistic
(c). 40% of all trees in the forest show some signs of damage
(d). More than 40% of the trees in the forest show some signs of damage
(e). Less than 40% of the trees in the forest show some signs of damage
keywords: sample, statistic
(a) the probability that we obtain the statistic in repeated random samples.
(b) the mechanism that determines whether randomization was effective.
(c) the distribution of values taken by a statistic in all possible samples of the same sample size from the same population.
(d) the extent to which the sample results differ systematically from the truth.
(e) the distribution of values in a sample of size n from the population
keywords: all possible samples, same sample size
(a) the survey used to obtain the statistic was designed so as to avoid even the hint of racial or sexual prejudice.
(b) the mean of its sampling distribution is equal to the true value of the parameter being estimated.
(c)both the person who calculated the statistic and the subjects whose responses make up the statistic were truthful.
(d)the value from any sample is equal to the parameter being estimated.
(e)it is used for honest purposes only.
keywords: true mean equal
I. The population distribution
II. The distribution of sample data
III. The sampling distribution
(a) I only
(b) II only
(c) III only
(d) II and III
(e) all three distributions
The distribution of sample data
statements is true?
(a)σX + Y = σX+σY
(b)Var (X – Y) = Var (X) + Var (Y)
(c)Var (a + bX) = b Var (X)
(d)σX – Y = σX – σY
(e)Var (X + Y) = Var (X) + Var (Y)
A) 0.2119 B) 0.2881 C) 0.7881 D) 0.845
(a) The expected value of a geometric random variable is determined by the formula p(1-p)^n-1
(b) The distribution of every binomial random variable is skewed right
(c) If X is a geometric random variable and the
probability of success is .85, then the probability distribution of X will be skewed left, snince 85 is closer to 1 than to 0.
(d) An important difference between binomial and geometric random variables is that there is a fixed number of trials in a binomial setting, and the number of trials varies in a geometric setting.
number of computers in a randomly-selected household, and we omit the rare cases of more than 5 computers, then X has the following distribution:
X 0 1 2 3 4 5
P(X) 0.24 0.37 0.20 0.11 0.05 0.03
What is the probability that a randomly chosen household has at least two personal computers?
P(x=0 or x=1) =1
X 0 1 2
P(X) 2k 3k 13k 2k
Where k is a positive constant. The probability P(X < 2.0) is equal to A. 0.90. B. 0.25. C. 0.65. D. 0.15. E. 1.00.
p(X<2)=P(x=0 or x=1)=5k=.25
(b)(10) (1)^3 (7)^5
(3) (8) (8)
Binomial probability formula: P(k successes in n trials when p = success in one trial) is
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