Macro Study guide EX2

As the number of degrees of freedom for a t distribution increases, the difference between the t distribution and the standard normal distribution Select one: a. becomes larger b. becomes smaller c. stays the same d. becomes negative

b

In interval estimation, the t distribution is applicable only when Select one: a. the population has a mean of less than 30 b. the sample standard deviation is used to estimate the population standard deviation c. the variance of the population is known d. the standard deviation of the population is known

b

The following data was collected from a simple random sample from a process (an infinite population). 13 15 14 16 12 Refer to Exhibit 7-1. The point estimate of the population mean (Hint: page 16 of ppt file) Select one: a. is 5 b. is 14 c. is 4 d. cannot be determined because the population is infinite

b

Using an α = 0.04 a confidence interval for a population proportion is determined to be 0.65 to 0.75. If the level of significance is decreased, the interval for the population proportion (Hint: check section 8.1 on the textbook for the definition of level of significance) Select one: a. becomes narrower b. becomes wider c. does not change d. remains the same

b

A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is Select one: a. 20.5 to 26.5 b. 24.4 to 25.6 c. 23.0 to 27.0 d. 20.0 to 30.0

b =25+2.387*2/SQRT(64) t value The correct answer is: 24.4 to 25.6

As the sample size increases, the Select one: 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

c

Excel's NORM.DIST function can be used to compute Select one: a. cumulative probabilities for a standard normal z value b. the standard normal z value given a cumulative probability c. cumulative probabilities for a normally distributed x value d. the normally distributed x value given a cumulative probability

c

If an interval estimate is said to be constructed at the 90% confidence level, the confidence coefficient would be Select one: a. 0.1 b. 0.95 c. 0.9 d. 0.05

c

Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What is the probability that a randomly selected item will weigh more than 10 ounces? Select one: a. 0.3413 b. 0.8413 c. 0.1587 d. 0.5000

c It's type B question. z-score=1. Using the tale, at z-score =1, the are to the left is .8413. So the area to the right, the probability bigger than the z-score is, 1-.8413. The correct answer is: 0.1587

A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. Refer to Exhibit 8-6. If we want to determine a 95% confidence interval for the average hourly income, the value of "t" statistics is Select one: a. 1.96 b. 1.64 c. 1.28 d. 1.993

d

Excel's NORM.INV function can be used to compute Select one: a. cumulative probabilities for a standard normal z value b. the standard normal z value given a cumulative probability c. cumulative probabilities for a normally distributed x value d. the normally distributed x value given a cumulative probability

d

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. Refer to Exhibit 8-1. If the sample mean is 9 hours, then the 95% confidence interval is Select one: a. 7.04 to 110.96 hours b. 7.36 to 10.64 hours c. 7.80 to 10.20 hours d. 8.61 to 9.39 hours

d

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. Refer to Exhibit 8-1. The standard error of the mean is Select one: a. 7.50 b. 0.39 c. 2.00 d. 0.20

d

A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. Refer to Exhibit 8-6. The value of the margin of error at 95% confidence is Select one: a. 80.83 b. 7 c. 0.8083 d. 1.611

d =1.993*7/SQRT(75)

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 Select one: a. 0 b. .0495 c. .4505 d. None of the alternative answers is correct.

the z score of 57.95=(57.95-53)/(21/square root of 49)=1.65, On z table Pr(z<1.65)=0.9505, so the probability bigger than 57.95=1-0.9505. The correct answer is: .0495

"DRUGS R US" is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that the bottles of vitamins marked 6 ounces vary in content with a standard deviation of 0.3 ounces. Assume the contents of the bottles are normally distributed. Ninety-five percent of the bottles will contain at least how many ounces? (Use the Standard Normal Cumulative Probability Table. Check page 32 on ppt file for an example. Just enter the number without any units. Keep four decimal places.)

1, the z value is -1.645 where 5% area to the left. 2. ==6-1.645*0.3 The correct answer is: 5.5065

Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 20 gallons and a standard deviation of 8 gallons. If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than .10, what should the reorder point be? (Use the Standard Normal Cumulative Probability Table. If you can not find the exactly probability on the table, check page 32 of the ppt file. Just enter the number without any units. Keep two decimal places.)

1, the z value is 1.285 where 10% area to the right. 2. =20+1.285*8 The correct answer is: 30.28

Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 20 gallons and a standard deviation of 8 gallons. If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than .025, what should the reorder point be? (Use the Standard Normal Cumulative Probability Table, just enter the number without any units. Keep two decimal places.)

1. z value is 1.96 where 2.5% area to the right. 2. =20+1.96*8 The correct answer is: 35.68

In a sample of 400 voters, 360 indicated they favor the incumbent governor. The 95% confidence interval of voters not favoring the incumbent is Select one: a. 0.871 to 0.929 b. 0.120 to 0.280 c. 0.765 to 0.835 d. 0.071 to 0.129

=0.1+1.96*SQRT((0.1)*(0.9)/400) z value The correct answer is: 0.071 to 0.129

A sample of 225 elements from a population with a standard deviation of 75 is selected. The sample mean is 180. The 95% confidence interval for μ is Select one: a. 105.0 to 225.0 b. 175.0 to 185.0 c. 100.0 to 200.0 d. 170.2 to 189.8

=180+1.96*75/SQRT(225) z value The correct answer is: 170.2 to 189.8

A continuous random variable may assume Select one: a. all values in an interval or collection of intervals b. only integer values in an interval or collection of intervals c. only fractional values in an interval or collection of intervals d. all the positive integer values in an interval

A

A normal probability distribution Select one: a. is a continuous probability distribution b. is a discrete probability distribution c. can be either continuous or discrete d. always has a standard deviation of 1

A

Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What is the random variable in this experiment? Select one: a. the weight of items produced by a machine b. 8 ounces c. 2 ounces d. the normal distribution

A

Given that z is a standard normal random variable, what is the value of z if the area to the right of z is 0.1401? Use the Standard Normal Cumulative Probability Table. Select one: a. 1.08 b. 0.1401 c. 2.16 d. -1.08

A

The fact that the sampling distribution of the sample mean can be approximated by a normal probability distribution whenever the sample size is large is based on the Select one: a. central limit theorem b. fact that there are tables of areas for the normal distribution c. assumption that the population has a normal distribution d. All of these answers are correct.

A

The sampling distribution of the sample mean Select one: a. is the probability distribution showing all possible values of the sample mean b. is used as a point estimator of the population mean μ c. is an unbiased estimator d. shows the distribution of all possible values of μ

A

X is a normally distributed random variable with a mean of 12 and a standard deviation of 3. The probability that x equals 19.62 is (Use the Standard Normal Cumulative Probability Table.) Select one: a. 0.000 b. 0.0055 c. 0.4945 d. 0.9945

A

X is a normally distributed random variable with a mean of 5 and a variance of 4. The probability that x is greater than 10.52 is (Use the Standard Normal Cumulative Probability Table.) Select one: a. 0.0029 b. 0.0838 c. 0.4971 d. 0.9971

A

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 Select one: a. 0.1359 b. 0.8185 c. 0.3413 d. 0.4772

A the z value for 183=(183-180)/(24/sqrt(64))=1, the z value for 186 is 2. On z table: Pr(z<1)=0.8413; Pr(z<2)=0.9772. The difference is the probability between 183 and 186. The correct answer is: 0.1359

A standard normal distribution is a normal distribution with Select one: a. a mean of 1 and a standard deviation of 0 b. a mean of 0 and a standard deviation of 1 c. any mean and a standard deviation of 1 d. any mean and any standard deviation

B

Given that z is a standard normal random variable, what is the value of z if the area to the right of z is 0.1112? Use the Standard Normal Cumulative Probability Table. Select one: a. 0.3888 b. 1.22 c. 2.22 d. 3.22

B

The following data was collected from a simple random sample from a process (an infinite population). 13 15 14 16 12 Refer to Exhibit 7-1. The point estimate of the population standard deviation is (Hint: page 16 of ppt file) Select one: a. 2.500 b. 1.581 c. 2.000 d. 1.414

B

The standard deviation of all possible values is called the Select one: a. standard error of proportion b. standard error of the mean c. mean deviation d. central variation

B

A sample of 25 observations is taken from a process (an infinite population). The sampling distribution of is Select one: a. not normal since n < 30 b. approximately normal because is always normally distributed c. approximately normal if np ≥ 5 and n(1-p) ≥ 5 d. approximately normal if np > 30 and n(1-p) > 30

C

A sample of 51 observations will be taken from a process (an infinite population). The population proportion equals 0.85. The probability that the sample proportion will be between 0.9115 and 0.946 is Select one: a. 0.8633 b. 0.6900 c. 0.0819 d. 0.0345

C

A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. Refer to Exhibit 8-6. The 95% confidence interval for the average hourly wage of all information system managers is Select one: a. 40.75 to 42.36 b. 39.14 to 40.75 c. 39.14 to 42.36 d. 30 to 50

C

Exhibit 7-2 Four hundred registered voters were randomly selected asked whether gun laws should be changed. Three hundred said "yes," and one hundred said "no." Refer to Exhibit 7-2. The point estimate of the proportion in the population who will respond "yes" is (Hint: page 16 of ppt file) Select one: a. 300 b. approximately 300 c. 0.75 d. 0.25

C

In order to determine an interval for the mean of a population with unknown standard deviation a sample of 61 items is selected. The mean of the sample is determined to be 23. The number of degrees of freedom for reading the t value is Select one: a. 22 b. 23 c. 60 d. 61

C

X is a normally distributed random variable with a mean of 22 and a standard deviation of 5. The probability that x is less than 9.7 is (Use the Standard Normal Cumulative Probability Table.) Select one: a. 0.000 b. 0.4931 c. 0.0069 d. 0.9931

C It's type A question. z-score for 9.7 is (9.7-22)/5=-2.46, Using the tale, at z-score =-2.46, the area to the left is .9332; at z-score=2, the area to the left is 0.0069. So the probability is 0.0069

A random sample of 150 people was taken from a very large population. Ninety of the people in the sample were females. The standard error of the proportion of females is Select one: a. 0.0016 b. 0.2400 c. 0.1600 d. 0.0400

D

The following data was collected from a simple random sample from a process (an infinite population). 13 15 14 16 12 Refer to Exhibit 7-1. The mean of the population (Hint: page 16 of ppt file) Select one: a. is 14 b. is 15 c. is 15.1581 d. could be any value

D

The standard deviation of is referred to as the Select one: a. standard proportion b. sample proportion c. average proportion d. standard error of the proportion

D

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What is the probability that a randomly selected item weighs exactly 8 ounces? Use the Standard Normal Cumulative Probability Table. Select one: a. 0.5 b. 1.0 c. 0.3413 d. None of the alternative answers is correct.

D

X is a normally distributed random variable with a mean of 8 and a standard deviation of 4. The probability that x is between 1.48 and 15.56 is (Use the Standard Normal Cumulative Probability Table.) Select one: a. 0.0222 b. 0.4190 c. 0.5222 d. 0.9190

D

__________ is a property of a point estimator that is present when the expected​ value of the point estimator is equal to the population parameter it estimates. Select one: a. ​Predictable b. ​Precise c. ​Symmetric d. Unbiased

D

A population has a mean of 80 and a standard deviation of 7. A sample of 49 observations will be taken. The probability that the mean from that sample will be larger than 82 is Select one: a. 0.5228 b. 0.9772 c. 0.4772 d. 0.0228

D the z score of 82=(82-80)/(7/square root of 49)=2, On z table Pr(z<2)=0.9772, so the probability bigger than 82=1-0.9772. The correct answer is: 0.0228

A population of size 1,000 has a proportion of 0.5. Therefore, the proportion and the standard deviation of the sample proportion for samples of size 100 are Select one: a. 500 and 0.047 b. 500 and 0.050 c. 0.5 and 0.047 d. 0.5 and 0.050

D The standard deviation of sample proportion, aka standard error of proportion is: sqrt(0.5*0.5/100） The correct answer is: 0.5 and 0.050

Imagine we are trying to sell to a customer who demands that the mean of a random sample of 64 bulbs lasts at least 2,050 hours before they will buy. The population mean = 2,000 hours, and the population standard deviation is 100 hours. What mean length of bulb life could you be 90% confident that the sample mean will be at least that long? (Use the Standard Normal Cumulative Probability Table. Round to nearest integer)

First, it is a type D question since it asks you the cutoff point where 90% of time that a random variable is bigger than a number, ie. 10% to the left. Using the z-table, we can see that the z-value is -1.285, or -1.282 using computer. Giving z-value, we use cutoff value=mean+std.dev*z-value to find out the cutoff. Here mean is 2000, the std. dev of 64 samples is std.dev of the whole population divided by square root of sample size, which is 100/square root of 16. You will get the answer of 1984. The correct answer is: 1984

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What percentage of items will weigh at least 11.7 ounces? Use the Standard Normal Cumulative Probability Table. Select one: a. 46.78% b. 96.78% c. 3.22% d. 53.22%

It's type B question. z-score for 11.7 is (11.7-8)/2=1.85. Using the tale, at z-score =1.85, the area to the left is .9678; . So the are to the right is 1-.9678 The correct answer is: 3.22%

Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What is the probability that a randomly selected item will weigh between 11 and 12 ounces? Use the Standard Normal Cumulative Probability Table.

It's type C question. z-score for 11 is (11-8)/2=1.5, and z-score for 12 is (12-8)/2=2. Using the tale, at z-score =1.5, the area to the left is .9332; at z-score=2, the area to the left is 0.9772. So the difference is the probability between 11 and 12. The correct answer is: 0.0440

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What percentage of items will weigh between 6.4 and 8.9 ounces? Select one: a. 0.1145 b. 0.2881 c. 0.1736 d. 0.4617

The correct answer is: 0.4617

Random samples of size 17 are taken from a population that has 200 elements, a mean of 36, and a standard deviation of 8. Refer to Exhibit 7-5. The mean and the standard deviation of the sampling distribution of the sample means are (Hint: this is a finite population question, we need the finite population correction factor, see page 21 of the ppt file) Select one: a. 8.7 and 1.94 b. 36 and 1.94 c. 36 and 1.86 d. 36 and 8

This is a finite population question. We need to time the finite population correct factor (see page 21 on ppt file for details). So the stand deivation is: sqrt(200-17/200-1)*(8/sqrt(17). This part will not be required in the exam, i.e. you don't need to worry about the finite population issue in the exam. The correct answer is: 36 and 1.86

A 95% confidence interval for a population mean is determined to be 100 to 120. If the confidence coefficient is reduced to 0.90, the interval for μ Select one: a. becomes narrower b. becomes wider c. does not change

a

If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect Select one: a. the size of the confidence interval to increase b. the size of the confidence interval to decrease c. the size of the confidence interval to remain the same d. the sample size to increase

a

In a random sample of 144 observations, = 0.6. The 95% confidence interval for P is Select one: a. 0.52 to 0.68 b. 0.144 to 0.200 c. 0.60 to 0.70 d. 0.50 to 0.70

a

A random sample of 1000 people was taken. Four hundred fifty of the people in the sample favored Candidate A. The 95% confidence interval for the true proportion of people who favors Candidate A is Select one: a. 0.419 to 0.481 b. 0.40 to 0.50 c. 0.45 to 0.55

a =0.45+1.96*SQRT((0.45)*(1-0.45)/1000)

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. Refer to Exhibit 8-1. With a 0.95 probability, the margin of error is approximately Select one: a. 0.39 b. 1.96 c. 0.20 d. 1.64

a =1.96*1.8/SQRT(81) z value The correct answer is: 0.39

a. With 95% confidence, what is the margin of error? (Specify to 2 decimal places. Round the final answer, no rounding on intermediate steps) Answer 3.8 Correct b. Compute the 95% confidence interval for the population mean (Specify to 2 decimal places). ( Answer 76.20 Correct , Answer 83.80 Correct ) c. Assume that the same sample mean was obtained from a sample of 120 items. With 95% confidence, what is the margin of error? (Specify to 2 decimal places. Round the final answer, no rounding on intermediate steps) Answer 2.68 Correct d. Compute the 95% confidence interval for the population mean (Specify to 2 decimal places). ( Answer 77.32 Correct , Answer 82.68 Correct )

a =1.96*15/SQRT(60) z value b =80+1.96*15/SQRT(60) c =1.96*15/SQRT(120) d =80+ 1.96*15/SQRT(120)

A sample of 400 observations will be taken from a process (an infinite population). The population proportion equals 0.8. The probability that the sample proportion will be greater than 0.83 is Select one: a. 0.4332 b. 0.9332 c. 0.0668 d. 0.5668

the z-score is (.83-.8)/sqrt(0.8*(1-0.8)/400)=1.5. sqrt(0.8*(1-0.8)/400) is the standard error of proportion. From the table, it is .9332, so the area to the right is 1-.9332 The correct answer is: 0.0668