Practice Test 3 –Bus 2023
Directions: For each question find the answer that is the best solution provided. There is only one correct answer.
1. A random variable that can assume only a finite number of values is referred to as a(n)
a. infinite sequence
b. finite sequence
c. discrete random variable
d. discrete probability function
ANSWER: c
2. The number of customers that enter a store during one day is an example of
a. a continuous random variable
b. a discrete random variable
c. either a continuous or a discrete random variable, depending on the number of the customers
d. either a continuous or a discrete random variable, depending on the gender of the customers
ANSWER: b
3. An experiment consists of measuring the speed of automobiles on a highway by the use of radar equipment. The random variable in this experiment is speed, measured in miles per hour. This random variable is a
a. discrete random variable
b. continuous random variable
c. complex random variable
d. None of these alternatives is correct.
ANSWER: b
4. A measure of the average value of a random variable is called a(n)
a. variance
b. standard deviation
c. expected value
d. None of the alternative answers is correct.
ANSWER: c Note: So we can write E(x) = μ
5. The expected value for a binomial probability distribution is
a. E(x) = pn(1 - n)
b. E(x) = p(1 - p)
c. E(x) = np
d. E(x) = np(1 - p)
ANSWER: c
6. The variance for the binomial probability distribution is
a. Var(x) = p(1 - p)
b. Var(x) = np
c. Var(x) = n(1 - p)
d. Var(x) = np(1 - p)
ANSWER: d
7. Assume that you have a binomial experiment with p = 0.4 and a sample size of 50. The variance of this distribution is
a. 20
b. 12
c. 3.46
d. Not enough information is given to answer this question.
ANSWER: b
Recall that variance of the binomial is Var(x) = np(1 - p) = 50(0.4)(0.6) = 12
8. Twenty percent of the students in a class of 100 are planning to go to graduate school. The standard deviation of this binomial distribution is
a. 20
b. 16
c. 4
d. 2
ANSWER: c
Recall that variance of the binomial is Var(x) = np(1 - p) = 100(0.2)(0.8) = 16
Std(x) = = 4
Exhibit 5-8
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
9. Refer to Exhibit 5-8. What is the random variable in this experiment?
a. the 60% of female students
b. the random sample of 8 students
c. the number of female students out of 8
d. the student body size
ANSWER: c
10. Refer to Exhibit 5-8. What is the probability that among the students in the sample exactly two are female?
a. 0.0896
b. 0.2936
c. 0.0413
d. 0.0007
ANSWER: c
Recall that P( X = x) = ( ) px (1-p)(n-x) à P( X = 2) = ( ) )0.62 (0.4)(8-2) = 0.0413
11. Refer to Exhibit 5-8. What is the probability that among the students in the sample at least 7 are female?
a. 0.1064
b. 0.0896
c. 0.0168
d. 0.8936
ANSWER: a
Note: This is the same set up as above but now you need P (X ≥7) = P (7) + P(8)
12. Refer to Exhibit 5-8. What is the probability that among the students in the sample at least 6 are male?
a. 0.0413
b. 0.0079
c. 0.0007
d. 0.0499
ANSWER: d
Note: Same set up, but now the percentage of inters is not females, but males. P (X ≥6) = P(6) + P (7) + P(8). Make sure to use p=0.40 since the probability of interest is males.
13. A numerical measure from a sample, such as a sample mean, is known as
a. a statistic
b. a parameter
c. the mean deviation
d. the central limit theorem
ANSWER: a
14. A subset of a population selected to represent the population is a
a. subset
b. sample
c. small population
d. None of the alternative answers is correct.
ANSWER: b
15. A simple random sample of 5 observations from a population containing 400 elements was taken, and the following values were obtained.
12 18 19 20 21
A point estimate of the population mean is
a. 400
b. 18
c. 20
d. 10
ANSWER: b
Note: Since this is a sample that is supposed to be representative of the population its mean is an approximate of the population mean. So we get that E () = μ.
16. The sampling error is the
a. same as the standard error
b. absolute value of the difference between an unbiased point estimate and the corresponding population parameter
c. error caused by selecting a bad sample
d. standard deviation multiplied by the sample size
ANSWER: b
17. The probability distribution of all possible values of the sample mean is called the
a. central probability distribution
b. sampling distribution of the sample mean
c. random variation
d. standard error
ANSWER: b
18. From a population of 200 elements, the standard deviation is known to be 14. A sample of 49 elements is selected. It is determined that the sample mean is 56. The standard error of the mean is
a. 3
b. 2
c. greater than 2
d. less than 2
ANSWER: d
Recall that σ = σ / = 14 / = 14 / 7 = 2
19. From a population of 500 elements, a sample of 225 elements is selected. It is known that the variance of the population is 900. The standard error of the mean is approximately
a. 1.1022
b. 2
c. 30
d. 1.4847
ANSWER: d
Note: Same process as above but now a finite population
σ = * (σ /) = * (30 /) = 1.4847
20. As the sample size increases, the
a. standard deviation of the population decreases
b. population mean increases
c. standard error of the mean decreases
d. standard error of the mean increases
ANSWER: c
21. Random samples of size 36 are taken from an infinite population whose mean and standard deviation are 20 and 15, respectively. The distribution of the population is unknown. The mean and the standard error of the distribution of sample mean are
a. 36 and 15
b. 20 and 15
c. 20 and 0.417
d. 20 and 2.5
ANSWER: d
Recall that the mean of the sampling distribution and population are the same. We also know σ = σ / = 15 / = 15 / 6 = 2.5
22. A theorem that allows us to use the normal probability distribution to approximate the sampling distribution of sample means and sample proportions whenever the sample size is large is known as the
a. approximation theorem
b. normal probability theorem
c. central limit theorem
d. central normality theorem
ANSWER: c
23. 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. 0.1359
b. 0.8185
c. 0.3413
d. 0.4772
ANSWER: a
To make a probability statement we need to simply calculate the z-values for each.
Z183 = (X – μ ) / σ where σ = σ / = 24 / = 24/8 = 3
= (183 – 180) / 3 = 1
Z186 = (X – μ ) / σ = (186 – 180) / 3 = 2
P ( 1 < Z < 2 ) = 0.1359
24. A population has a mean of 84 and a standard deviation of 12. A sample of 36 observations will be taken. The probability that the sample mean will be between 80.54 and 88.9 is
a. 0.0347
b. 0.7200
c. 0.9511
d. None of the alternative ANSWERS is correct.
ANSWER: c
-same format at above.
25. An estimate of a population parameter that provides an interval believed to contain the value of the parameter is known as the
a. confidence level
b. interval estimate
c. parameter value
d. population estimate
ANSWER: b
26. The confidence associated with an interval estimate is called the
a. level of significance
b. degree of association
c. confidence level
d. precision
ANSWER: c
27. The z value for a 97.8% confidence interval estimation is
a. 2.02
b. 1.96
c. 2.00
d. 2.29
ANSWER: d
-simply use the z-table and find the z value that gives the middle 97.8%
28. It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a .95 probability that the sample mean will provide a margin of error of
a. 7.84 or less
b. 31.36 or less
c. 344.96 or less
d. 1,936 or less
ANSWER: a
Recall that the margin of error is the difference between the sample value and population parameter and it is bounded by Zα/2* σ = (1.96*() / ( ) < 7.84
Exhibit 8-1
In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.2 hours.
29. Refer to Exhibit 8-1. The standard error of the mean is
a. 7.5
b. 0.014
c. 0.160
d. 0.133
ANSWER: d
Recall that σ = σ / = 1.2/9 = 0.133
30. Refer to Exhibit 8-1. With a 0.95 probability, the margin of error is approximately
a. 0.26
b. 1.96
c. 0.21
d. 1.64
ANSWER: a
Margin of Error = Zα/2* σ = 1.96*0.133≈0.26
31. Refer to Exhibit 8-1. If the sample mean is 9 hours, then the 95% confidence interval is
a. 7.04 to 110.96 hours
b. 7.36 to 10.64 hours
c. 7.80 to 10.20 hours
d. 8.74 to 9.26 hours
ANSWER: d
Recall that ± zα/2 σ. So just apply the formula to get 8.74 to 9.26. This means we are 95% sure the true population mean falls within these values.
32. If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect the
a. width of the confidence interval to increase
b. width of the confidence interval to decrease
c. width of the confidence interval to remain the same
d. sample size to increase
33. In hypothesis testing, the hypothesis tentatively assumed to be true is
a. the alternative hypothesis
b. the null hypothesis
c. either the null or the alternative
d. None of the other answers are correct.
ANSWER: b
34. Your investment executive claims that the average yearly rate of return on the stocks she recommends is at least 10.0%. You plan on taking a sample to test her claim. The correct set of hypotheses is
a. H0: m < 10.0% Ha: m ³ 10.0%
b. H0: m £ 10.0% Ha: m > 10.0%
c. H0: m > 10.0% Ha: m £ 10.0%
d. H0: m ³ 10.0% Ha: m < 10.0%
ANSWER: d
35. A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. Any overfilling or underfilling results in the shutdown and readjustment of the machine. To determine whether or not the machine is properly adjusted, the correct set of hypotheses is
a. H0: m < 12 Ha: m £ 12
b. H0: m £ 12 Ha: m > 12
c. H0: m ¹ 12 Ha: m = 12
d. H0: m = 12 Ha: m ¹ 12
ANSWER: d
36. If a hypothesis test leads to the rejection of the null hypothesis, a
a. Type II error must have been committed
b. Type II error may have been committed
c. Type I error must have been committed
d. Type I error may have been committed
ANSWER: d
37. The probability of making a Type I error is denoted by
a. a
b. b
c. 1 - a
d. 1 - b
ANSWER: a
38. A one-tailed test is a
a. hypothesis test in which rejection region is in both tails of the sampling distribution
b. hypothesis test in which rejection region is in one tail of the sampling distribution
c. hypothesis test in which rejection region is only in the left tail of the sampling distribution
d. hypothesis test in which rejection region is only in the right tail of the sampling distribution
ANSWER: b
Exhibit 9-1
n = 36 = 24.6 s = 12 H0: m £ 20
Ha: m > 20
39. Refer to Exhibit 9-1. The test statistic equals
a. 2.3
b. 0.38
c. -2.3
d. -0.38
ANSWER: a
Since we have that the standard deviations is unknown but we have n< 30 it is a
z-stat = ( – μ) / σ = (20-24.6) / (12/) = 4.6*6 / 12 = 2.3
40. Refer to Exhibit 9-1. The p-value is
a. 0.5107
b. 0.0214
c. 0.0107
d. 2.1
ANSWER: c
41. Refer to Exhibit 9-1. If the test is done at a .05 level of significance, the null hypothesis should
a. not be rejected
b. be rejected
c. Not enough information is given to answer this question.
d. None of the other answers are correct.
ANSWER: b
Short Answer:
Question 1: Binomial Distribution
a. Thirty-two percent of the students in a management class are graduate students. A random sample of 5 students is selected. Using the binomial probability function, determine the probability that the sample contains exactly 2 graduate students?
b. A production process produces 2% defective parts. A sample of 5 parts from the production is selected. What is the probability that the sample contains exactly two defective parts? Use the binomial probability function and show your computations to answer this question.
c. When a particular machine is functioning properly, 80% of the items produced are non-defective. If three items are examined, what is the probability that one is defective? Use the binomial probability function to answer this question.