QMB 3200 REVISION EXAM
QUESTIONS AND THEIR CORRECT
ANSWERS
In computing the standard error of the mean, the finite
population correction factor is not used when:
a. n ≥ 30
b. n/N ≤ 0.05
c. N/n ≤ 0.05
d. n/N > 0.05
b. n/N ≤ 0.05
A sample statistic, such as x bar , that estimates the value of
the corresponding population parameter is known as a:
a. population parameter
b. point estimator
c. Both a parameter and a population parameter are correct.
d. parameter
b. point estimator
The standard deviation of xbar is referred to as the
a. standard error of the mean
b. standard x
c. sample standard mean
d. sample mean deviation
a. standard error of the mean
,Which of the following is(are) point estimator(s)?
a. σ
b. μ
c. s
d. All of these answers are correct.
c. s
A probability distribution for all possible values of a sample
statistic is known as a
a. parameter
b. sample statistic
c. sampling distribution
d. simple random sample
c. sampling distribution
A population has a mean of 180 and a standard deviation of
24. A sample of 64 observations will be taken. The probability
that the mean from that sample will be between 183 and 186
is
a. 0.8185
b. 0.1359
c. 0.4772
d. 0.3413
b. 0.1359
1. st. error: 24/sqrt(64) = 3
2. z value (upper): (186-180)/3 = 2
,3. Norm.s.dist(1, true) = .9772
4. z value (lower): (183-180)/3= 1
5.Norm.s.dist(-1,true)= .8413
6. 0.9772-.8413=0.1359
The basis for using a normal probability distribution to
approximate the sampling distribution of xbar and pbar is:
a. the empirical rule
b. the central limit theorem
c. Chebyshev's theorem
d. Bayes' theorem
b. the central limit theorem
A finite population correction factor is needed in computing
the standard deviation of the sampling distribution of sample
means:
a. whenever the sample size is less than 5% of the population
size
b. whenever the population is infinite
c. whenever the sample size is more than 5% of the population
size
d. The correction factor is not necessary if the population has
a normal distribution
c. whenever the sample size is more than 5% of the population
size
A sample of 66 observations will be taken from a process (an
infinite population). The population proportion equals 0.12.
, The probability that the sample proportion will be less than
0.1768 is
a. 0.4222
b. 0.0778
c. 0.9222
d. 0.0568
c. 0.9222
1. Mean of the sample proportions = 0.12
2. St. dev. of pbar = sqrt [ p(1-p) / n] = sqrt [ (0.12)(1-.12) / 66]
= .04
μ = 0.12
σ = 0.04
3. standardize x to z = (x - μ) / σ = (0.1768-0.12) / 0.04 = 1.42
4. norm.s.dist(1.42,true) = 0.9222
A population has a mean of 53 and a standard deviation of 21.
A sample of 49 observations will be taken. The probability that
the sample mean will be greater than 57.95 is
a. 0
b. .0495
c. .4505
d. None of the alternative answers is correct.
b. .0495
1. 21/sqrt(49)=3
2. 57.93-53=4.95
QUESTIONS AND THEIR CORRECT
ANSWERS
In computing the standard error of the mean, the finite
population correction factor is not used when:
a. n ≥ 30
b. n/N ≤ 0.05
c. N/n ≤ 0.05
d. n/N > 0.05
b. n/N ≤ 0.05
A sample statistic, such as x bar , that estimates the value of
the corresponding population parameter is known as a:
a. population parameter
b. point estimator
c. Both a parameter and a population parameter are correct.
d. parameter
b. point estimator
The standard deviation of xbar is referred to as the
a. standard error of the mean
b. standard x
c. sample standard mean
d. sample mean deviation
a. standard error of the mean
,Which of the following is(are) point estimator(s)?
a. σ
b. μ
c. s
d. All of these answers are correct.
c. s
A probability distribution for all possible values of a sample
statistic is known as a
a. parameter
b. sample statistic
c. sampling distribution
d. simple random sample
c. sampling distribution
A population has a mean of 180 and a standard deviation of
24. A sample of 64 observations will be taken. The probability
that the mean from that sample will be between 183 and 186
is
a. 0.8185
b. 0.1359
c. 0.4772
d. 0.3413
b. 0.1359
1. st. error: 24/sqrt(64) = 3
2. z value (upper): (186-180)/3 = 2
,3. Norm.s.dist(1, true) = .9772
4. z value (lower): (183-180)/3= 1
5.Norm.s.dist(-1,true)= .8413
6. 0.9772-.8413=0.1359
The basis for using a normal probability distribution to
approximate the sampling distribution of xbar and pbar is:
a. the empirical rule
b. the central limit theorem
c. Chebyshev's theorem
d. Bayes' theorem
b. the central limit theorem
A finite population correction factor is needed in computing
the standard deviation of the sampling distribution of sample
means:
a. whenever the sample size is less than 5% of the population
size
b. whenever the population is infinite
c. whenever the sample size is more than 5% of the population
size
d. The correction factor is not necessary if the population has
a normal distribution
c. whenever the sample size is more than 5% of the population
size
A sample of 66 observations will be taken from a process (an
infinite population). The population proportion equals 0.12.
, The probability that the sample proportion will be less than
0.1768 is
a. 0.4222
b. 0.0778
c. 0.9222
d. 0.0568
c. 0.9222
1. Mean of the sample proportions = 0.12
2. St. dev. of pbar = sqrt [ p(1-p) / n] = sqrt [ (0.12)(1-.12) / 66]
= .04
μ = 0.12
σ = 0.04
3. standardize x to z = (x - μ) / σ = (0.1768-0.12) / 0.04 = 1.42
4. norm.s.dist(1.42,true) = 0.9222
A population has a mean of 53 and a standard deviation of 21.
A sample of 49 observations will be taken. The probability that
the sample mean will be greater than 57.95 is
a. 0
b. .0495
c. .4505
d. None of the alternative answers is correct.
b. .0495
1. 21/sqrt(49)=3
2. 57.93-53=4.95
