STATS Ch.8, Applied Statistics Midterm 2

Lakukan tugas rumah & ujian kamu dengan baik sekarang menggunakan Quizwiz!

The __________ is the population we want to make inferences about.

Target Population

=NORM.S.DIST is used to compute...

the Cumulative Probability given a Z-score

=NORM.S.INV is used to compute...

the Z-value given a Cumulative Probability

Where is the highest point on a normal curve?(Normal Probability Distribution)

the mean

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.

(Type B question) The correct answer is: .0495

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? a. 0.3413 b. 0.8413 c. 0.1587 d. 0.5000

(Type B Question) --> z-score=1 --> Using the table, at z-score =1, the area 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

"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? (Keep four decimal places) (Use Stand. Normal Cumulative Probability Table) Check page 32 on ppt file for an example

(#1) find .0500 (#2) between -1.64 (.0505) & -1.65 (.0495) (#3) z value is (-1.64 + -1.65) / 2 = -1.645; 5% area to the left. (keep neg. sign) (#4): 6 - (1.645 x 0.3) = 5.5065 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? (Keep two decimal places) (Use Stand. Normal Cumulative Probability Table)

(#1) find .1000 in table (#2) between -1.28 (.1003) & -1.29 (.0985) (#2) z value is (1.28 + 1.29) /2 = 1.285; where 10% area to the right (turn -1.2.. into +) (#3) 20 + (1.285 x 8) = 30.28 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? (Keep two decimal places.) (Use Stand. Normal Cumulative Probability Table)

(#1) prob. 0.0250 is found at z-value -1.96 (#2) z value is 1.96 where 2.5% area to the right. (#3) = 20 + (1.96 x 8) The correct answer is: 35.68

(STEPS TO SOLVE???) 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 a. 0.5228 b. 0.9772 c. 0.4772 d. 0.0228

(#1): calculate the z score of 82 (#2): ((82-80)/7√49))=2 (#3): using z-table prob. (z<2)=0.9772 (#4): prob. BIGGER than 82=1-0.9772. (#5): 1- 0.9772 = 0.0228 The correct answer is: 0.0228

Replacing each sampled element before selecting subsequent elements is called:

sampling with replacement

standard deviation

=NORM.S.DIST

z-score

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) a. 300 b. approximately 300 c. 0.75 d. 0.25

0.75

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) a. is 14 b. is 15 c. is 15.1581 d. could be any value

could be any value

The standard deviation of all possible [X-bar] values is called the a. standard error of proportion b. standard error of the mean c. mean deviation d. central variation

standard error of the mean

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

standard error of the mean decreases

we know that 90% of the time, the sample mean is within +/-1.645 ________ ______ of the _______

standard error of the mean.

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 a. 0.5228 b. 0.9772 c. 0.4772 d. 0.0228

(Type B Question) answer: 0.0228

The standard deviation of [P-bar] is referred to as the a. standard proportion b. sample proportion c. average proportion d. standard error of the proportion

standard error of the proportion

The distribution of a Normal Probability Distribution is _____________

symmetric

the Normal Probability Distribution is __________

symmetric

(STEPS TO SOLVE???) 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

(#1): calculate z value for 183 a} ((183-180)/24√(64))= 1 (#2) calculate the z value for 186 a} ((186-180)/24√(64))= 2 (#3) Using z table: Prob. (z<1) a} find prob.≤1 on table b} prob. ≤1 on table ---> 0.8413 (#4):Using z table: Prob. (z<2) a} find prob.≤2 on table b} prob. ≤2 on table ---> 0.9772 (#5): difference is prob. between 183 & 186 a} 183 --> 1 ---> 0.8413 b} 186 --> 2 ---> 0.9772 c} 0.9772 - 0.8413 = 0.1359 The correct answer is: 0.1359

(STEPS TO SOLVE???) 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.

(#1): find the z score of 57.95 (#2): ((57.95-53)/21√ 49))=1.65 (#3): using z-table find prob.(z < 1.65) (#4): 1.65 in table is 0.9505 (#5): prob. BIGGER than 57.95=1- .9505 (#6) 1 - .9505 = .0495

(STEPS TO SOLVE???) The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the probability that a randomly selected item will weigh between 11 and 12 ounces? a. 0.4772 b. 0.4332 c. 0.9104 d. 0.0440

(#1): z-score for 11 is (11-8) / 2 = 1.5 (#2): z-score for 12 is (12-8) / 2 = 2 (use Stand. Normal Cumulative Prob. Table) (#3): using table at z-score =1.5 the area to the left is .9332 (#4): using table at z-score=2, the area to the left is 0.9772. (difference is probability between 11 and 12) (#5): 0.9772 - 0.9332 = 0.044

(STEPS TO SOLVE???) 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) a. 8.7 and 1.94 b. 36 and 1.94 c. 36 and 1.86 d. 36 and 8

(Finite Population Question) (#1) see page 21 on ppt file for details So the stand deviation is: √((200-17)÷(200-1))×(8÷√(17). The correct answer is: 36 and 1.86

(STEPS TO SOLVE???) 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 a. 0.4332 b. 0.9332 c. 0.0668 d. 0.5668

(Proportion Question) (#1): calculate the z-score (0.83-0.8)÷√((0.8×(1-0.8))÷400) a} (0.83-0.8)= 0.03 b}√0.8 × (1-0.8) / 400)= 0.02 b} 0.03 ÷ 0.02= 1.5 c} (0.03)÷(0.02)= 1.5 (#2): Using the table, find prob.≤ z-value 1.5 a} in table for 1.5 prob. is 0.9332 (prob.≤1.5) b} the area to the right (prob.≥1.5) is 1-0.9332 c} 1-0.9332=0.0668 The correct answer is: 0.0668

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 Stand. Normal Cumulative Probability Table) a. 0.000 b. 0.4931 c. 0.0069 d. 0.9931

(Type A Question) (#1) z-score for 9.7 is (9.7-22) / 5 = -2.46 (#2) Using table, at z-score= 2, the area to the left is 0.0069 (#3) So the probability is 0.0069 The correct answer is: 0.0069

(STEPS TO SOLVE???) In the 2016 election, Candidate A took 46.1% of the popular vote. A few days early, a poll conducted by the Economist magazine showed that among around 3700 interviewed voters, 45% of them prefer to vote for Candidate A. Another poll conducted by CBS showed that among around 1500 interviewed voters, 43% of them prefer to vote for Candidate A. We assume that the voters' preference didn't change, i.e. 46.1% is the population proportion when the polls were conducted. (Data source: relaclearpolitics.com) What is the probability that the Economist magazine can get a polling result that is equal to or lower than 45%? a. .0901 b. .0808 c. .0951 d. .1112 e. 0

(Type A Question) (#1): calculate z value (0.45-0.461)÷√((0.461)×(1-0.461)÷3700) a} (0.45-0.461)= -0.011 b} √((0.461)×(1-0.461)÷3700) = 0.0081949 c} (-0.011)÷(-1.342297255...) = -1.342297... (#2) Use table to find "probability less than" or prob. ≤ -1.3 a} locate -1.3 and 0.04 using table (-1.34) b} -1.34 on table ---> 0.0901 The correct answer is: .0901

In the 2016 election, Candidate A took 46.1% of the popular vote. A few days early, a poll conducted by the Economist magazine showed that among around 3700 interviewed voters, 45% of them prefer to vote for Candidate A. Another poll conducted by CBS showed that among around 1500 interviewed voters, 43% of them prefer to vote for Candidate A. We assume that the voters' preference didn't change, i.e. 46.1% is the population proportion when the polls were conducted. (Data source: relaclearpolitics.com) What is the probability that the Economist magazine can get a polling result that is equal to or lower than 45%? a. .0901 b. .0808 c. .0951 d. .1112 e. 0

(Type A Question) .0901

Imagine a business is worried about one of its salespeople's performance. The population proportion of contacts leading to sales = 0.20. Imagine Charlie has contacted 100 customers this week but only made 10 sales (assume these contacts are a simple random sample of those who could have been called upon). What is the probability that a random sample of 100 customers made less than or equal to 10 sales, i.e. the sample proportion ≤ 10/100)? a. 0.0228 b. 0.1234 c. 0.0062 d. None of these answers is correct e. 0.2554

(Type A Question) 0.0062

(STEPS TO SOLVE???) Imagine a business is worried about one of its salespeople's performance. The population proportion of contacts leading to sales = 0.20. Imagine Charlie has contacted 100 customers this week but only made 10 sales (assume these contacts are a simple random sample of those who could have been called upon). What is the probability that a random sample of 100 customers made less than or equal to 10 sales, i.e. the sample proportion ≤ 10/100)? a. 0.0228 b. 0.1234 c. 0.0062 d. None of these answers is correct e. 0.2554

(Type A Question) (#1): calculate z value (0.1-0.2)÷√((0.2)×(1-0.2)÷100) a} (0.1-0.2)= -0.1 b} √((0.2)×(1-0.2)÷100)= 0.04 c} (-0.1)÷(0.04)= -2.5 (#2) Use table to find "probability less than" or prob. ≤ -2.5 a} found at -2.5 and 0.00 on table headers b} -2.5 on table --> 0.0062 The correct answer is: 0.0062

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What percentage of items will weigh at least 11.7 ounces? (Stand. Normal Cumulative Prob. Table) a. 46.78% b. 96.78% c. 3.22% d. 53.22%

(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 -0.9678 --> So the area to the right is 1 - 0.9678 =0.0322 --> 0.0322 x 100 = 3.22% The correct answer is: 3.22%

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the probability that a randomly selected item will weigh between 11 and 12 ounces? (Use Stand. Normal Cumulative Probability Table) a. 0.4772 b. 0.4332 c. 0.9104 d. 0.0440

(Type C Question) (#1): z-score for 11 is (11-8) / 2 = 1.5 (#2): z-score for 12 is (12-8) / 2 = 2 (#3): using table at z-score =1.5 the area to the left is .9332 (#4): using table at z-score=2, the area to the left is 0.9772. (#5): 0.9772 - 0.9332 = 0.044 The correct answer is: 0.0440

a Random Sample from an Infinite Population is a Sample selected that satisfies the following (2) conditions:

- Each element selected comes from the population of interest. - Each element is selected independently.

Finite Populations are defined by lists such as:

- Organization membership roster - Credit card account numbers - Inventory product numbers

Stratified Random Sampling

- The population is first divided into groups of elements called strata. - Each element in the population belongs to one and only one stratum. - A simple random sample is taken from each stratum.

Cluster Sampling

- The population is first divided into separate groups of elements called clusters. - A simple random sample of the clusters is then taken.

=NORM.DIST

- Value - Mean of Normal Distribution; - Stand. Dev. of Normal Distribution (Cumulative Prob. or not)

Systematic Sampling

- We randomly select one of the first n/N elements from the population list. - has the properties of a Simple Random Sample, especially if the list of the population elements is a Random Ordering.

Convenience Sampling

- a Non-Probability Sampling Technique. - Items are included in the sample without known probabilities of being selected.

Examples of on-going processes, with Infinite Populations:

- parts manufactured on a production line - transactions occurring at a bank - phone calls arriving at a technical help desk - customers entering a store

Assume that our sample data of height in inches data represents the population of students at UMD and that this population is approximately normal with mean = 70 inches and standard deviation = 3.5 inches. Let x = height in inches (a continuous, normally distributed random variable) What is P(x = 74), ie. probability that x equals to 74? a. None of these is correct. b. 0.8729 c. 0 d. 0.3729 e. 0.1271

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 Stand. Normal Cumulative Probability Table) a. 0.000 b. 0.0055 c. 0.4945 d. 0.9945

0.000

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 Stand. Normal Cumulative Probability Table) a. 0.0029 b. 0.0838 c. 0.4971 d. 0.9971

0.0029

Imagine we are trying to sell to a customer who demands that the mean of a random sample of 16 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. Assume that it is known bulb life is normally distributed. What is the probability we get the sale, i.e. the probability that the sample mean is long enough? (Type B) a. 0.0228 b. None of these answers is correct c. 0.0000 d. 0.3085 e. 0.9772

0.0228

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 a. 0.0016 b. 0.2400 c. 0.1600 d. 0.0400

0.0400

In a sample of 400 voters, 360 indicated they favor the incumbent governor. The 95% confidence interval of voters not favoring the incumbent is

0.071 to 0.129 z-value =0.1+1.96*SQRT((0.1)*(0.9)/400)

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 a. 0.8633 b. 0.6900 c. 0.0819 d. 0.0345

0.0819

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

0.1359

Exhibit 8-1 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

0.20

0.39 z-value =1.69*1.8/SQRT(81)

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

0.419 to 0.481 z-value =0.45+1.96*√((0.45)*(1-0.45)/1000)

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? a. 0.1145 b. 0.2881 c. 0.1736 d. 0.4617

0.4617

(Complete the Sentence) the probabilities that the random variable smaller than the mean is

0.5

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

0.5 and 0.050

In a random sample of 144 observations, (p bar) = 0.6. The 95% confidence interval for P is

0.52 to 0.68 z-value =0.6+1.96*SQRT((0.6)*(0.4)/144)

Assume that our sample data of height in inches data represents the population of students at UMD and that this population is approximately normal with mean = 70 inches and standard deviation = 3.5 inches. Let x = height in inches (a continuous, normally distributed random variable) What is P(63 < x< 74)? a. 0.9772 b. 0.9931 c. 0.8501 d. None of these is correct e. 0.8729

0.8501

Assume that our sample data of height in inches data represents the population of students at UMD and that this population is approximately normal with mean = 70 inches and standard deviation = 3.5 inches. Let x = height in inches (a continuous, normally distributed random variable) What is P(x < 74), ie. probability that x less than 74? Select one: a. 0.3729 b. 0.5000 c. 0.8729 d. 1.0000 e. None of them is correct

0.8729

If an interval estimate is said to be constructed at the 90% confidence level, the confidence coefficient would be

0.9

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 Stand. Normal Cumulative Probability Table) a. 0.0222 b. 0.4190 c. 0.5222 d. 0.9190

0.9190

A random variable X follows a normal distribution with mean = 2000 and standard deviation = 100. For a randomly selected sample of 16 observations, what is the probability that the sample mean is less than 2050? a. 0.0000 b. 0.9772 c. 0.0228 d. 0.3085

0.9772

Assume that our sample data of height in inches data represents the population of students at UMD and that this population is approximately normal with mean = 70 inches and standard deviation = 3.5 inches. Let x = height in inches (a continuous, normally distributed random variable) What is P(x > 63), ie. probability that x is bigger than 63? a. 1 b. 0.5 c. 0.0228 d. 0.9772 e. None of these is correct.

0.9772

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 Stand. Normal Cumulative Probability Table) a. 1.08 b. 0.1401 c. 2.16 d. -1.08

1.08

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 Stand. Normal Cumulative Probability Table.) a. 0.3888 b. 1.22 c. 2.22 d. 3.22

1.22

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) a. 2.500 b. 1.581 c. 2.000 d. 1.414

1.581

Exhibit 8-6 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

1.611 =1.993*7/SQRT(75)

Exhibit 8-6 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

1.993

Office workers receive an average of 15.0 faxes per day with a sample size of 80 and sample standard deviation of 3.5. Based on this information construct and interpret a 95% confidence interval for the mean. What is the lower bound of the 95% confidence interval? Select one: a. 13.97 b. 14.22 c. 14.36 d. 15.00 e. 14.23

14.22

Office workers receive an average of 15.0 faxes per day with a sample size of 80 and sample standard deviation of 3.5. Based on this information construct and interpret a 95% confidence interval for the mean. What is the upper bound of the 95% confidence interval? Select one: a. 15.00 b. 15.29 c. 15.78 d. 15.77 e. 16.5

15.78

Office workers receive an average of 15.0 faxes per day with a sample size of 80 and sample standard deviation of 3.5. Based on this information construct and interpret a 95% confidence interval for the mean. What is the upper bound of the 95% confidence interval? a. 15.00 b. 15.29 c. 15.78 d. 15.77 e. 16.5

15.78

Imagine we are trying to sell to a customer who demands that the mean of a random sample of 16 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. Imagine we are not sure of the distribution so the customer says we should increase the sample size to 64. What mean length of bulb life could you be 95% confident that the sample mean will be at least that long? (Type D) Select one: a. None of these answers is correct. b. 1979 c. 1929 d. 2021 e. 1836

1929

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

24.4 to 26.5 t-value =25+2.387*2/SQRT(64)

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) a. 8.7 and 1.94 b. 36 and 1.94 c. 36 and 1.86 d. 36 and 8

36 and 1.86

Exhibit 8-6 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

39.14 to 42.36

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

60 degrees of freedom =n-1

Assume that our sample data of height in inches data represents the population of students at UMD and that this population is approximately normal with mean = 70 inches and standard deviation = 3.5 inches. Let x = height in inches (a continuous, normally distributed random variable) If you want to set a bar that at least 95% of UMD students should be taller than that bar, what's the height of the bar?

64.24

As degrees of freedom increase, the difference between the ____________ and the standard normal probability distribution, z distribution, becomes smaller and smaller.

t distribution

8.61 to 9.39 hours

(STEPS TO SOLVE???) Office workers receive an average of 15.0 faxes per day with a sample size of 80 and sample standard deviation of 3.5. Based on this information construct and interpret a 95% confidence interval for the mean. What is the lower bound of the 95% confidence interval? a. 13.97 b. 14.22 c. 14.36 d. 15.00 e. 14.23

=(15-1.99)×(3.5÷√(80)) (NEED TO KNOW: where -1.99 came from) The correct answer is: 14.22

(STEPS TO SOLVE???) Office workers receive an average of 15.0 faxes per day with a sample size of 80 and sample standard deviation of 3.5. Based on this information construct and interpret a 95% confidence interval for the mean. What is the upper bound of the 95% confidence interval? a. 15.00 b. 15.29 c. 15.78 d. 15.77 e. 16.5

=15+1.99×(3.5÷√(80)) (NEED TO KNOW: where +1.99 came from) The correct answer is: 15.78

=15-1.99×(3.5÷(80)) The correct answer is: 14.22

For the past 10 years, the average high temperature on September 1 is 80° with a standard deviation of 5°. If the temperature is normally distributed, what is the probability that the high temperature on September 1 of next year will be between 80° and 95°? a. Approximately 25%. b. Approximately 33%. c. Approximately 50%. d. Approximately 100%.

Approximately 50%.

Imagine a business is worried about one of its salespeople's performance. The population proportion of contacts leading to sales = 0.20. Imagine Charlie has contacted 100 customers this week but only made 10 sales (assume these contacts are a simple random sample of those who could have been called upon). Charlie says he just had a bad week. You think Charlie is probably not very good at his job. Which of the following explanations is more likely to be true? a. Better chance he had a bad week b. Better chance he is not good at selling the product

Better chance he is not good at selling the product

(Name the Term) "In selecting random samples of size n from a population, the sampling distribution of the sample mean x ̅ can be approximated by a normal distribution as the sample size becomes large"

Central Limit Theorem

When the population from which we are selecting a random sample does not have a normal distribution, the _____ ______ ______ is helpful in identifying the shape of the sampling distribution of x ̅.

Central Limit Theorem

Example of a Non-Probability Sampling Technique:

Convenience Sampling

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? (Round to nearest integer) (Use Stand. Normal Cumulative Probability Table)

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 64. You will get the answer of 1984.

(Type of Sampling???) "The person most knowledgeable on the subject of the study selects elements of the population that he or she feels are most representative of the population."

Judgment Sampling

Given your answer to 1 and 2, is it likely another sample would have a mean of at least 16.5 faxes/day? a. Yes, it would likely have a sample mean of at least 16.5 faxes/day b. No, it would not be likely to have a sample mean of at least 16.5 faxes/day.

No, it would not be likely to have a sample mean of at least 16.5 faxes/day.

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the probability that a randomly selected item weighs exactly 8 ounces? (Use Stand. Normal Cumulative Probability Table) a. 0.5 b. 1.0 c. 0.3413 d. None of the alternative answers is correct.

None of the alternative answers is correct.

Interval Estimate =

Point Estimate +/- Margin of Error

In __________ we use the data from the sample to compute a value of a sample statistic that serves as an estimate of a population parameter.

Point Estimation

With __________________, the sample results can provide "good" estimates of the population characteristics

Proper Sampling Methods

In large sampling projects, computer-generated _______________are often used to automate the sample selection process.

Random #'s

The _____________ is the population from which the sample is actually taken.

Sampled Population

When the population from which we are selecting a random sample does not have a normal distribution, the central limit theorem is helpful in identifying the _______ of the sampling distribution of x ̅.

Shape

A " ____________________________" is a sample selected such that each possible sample of size n has the same probability of being selected

Simple Random Sample of size n from a finite population of size N

As degrees of freedom increase , the difference between the t distribution and the _______ __________ _________ ___________, z distribution, becomes smaller and smaller.

Standard Normal Probability Distribution

Judgment Sampling

The person most knowledgeable on the subject of the study selects elements of the population that he or she feels are most representative of the population.

(STEPS TO SOLVE???) 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) a. 2.500 b. 1.581 c. 2.000 d. 1.414

The point estimate of population standard deviation is sample standard deviation. We need to use the equation we used in midterm 1. For more information about sample standard deviation, please check notes from chapter 3.

Proportion Question (example)

The probability that the sample proportion will be greater than 0.83 is

(STEPS TO SOLVE???) A population of size 1,000 has a proportion of 0.5. Therefore, the expected value and the standard deviation of the sample proportion for samples of size 100 are a. 500 and 0.047 b. 500 and 0.050 c. 0.5 and 0.047 d. 0.5 and 0.050

The standard deviation of sample proportion, AKA standard error of proportion is: √((0.5 x 0.5/100)) The correct answer is: 0.5 and 0.050

Suppose we try to estimate the average money spend by customers. Two different random samples of 100 data values are taken from the large population. One sample has a larger sample standard deviation than the other. Each of the samples is used to construct a 95% confidence interval. How do you think these two confidence intervals would compare?

This is the population standard deviation unknown case. We have to use the sample standard deviation as the estimator for the population standard deviation. With a larger sample standard deviation, the margin of error will be larger, and the confidence interval also will be larger. The correct answer is: None of these options is correct

(True or False) It is not possible to talk about the probability of the random variable assuming a particular value.

True

(True or False) Sampling without replacement is the procedure used most often.

True

__________ 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. a. Predictable b. Precise c. Symmetric d. Unbiased

Unbiased

Type C Question (example)

What is the probability that a randomly selected item will weigh between 11 and 12 ounces?

Type B Question (example)

What is the probability that a randomly selected item will weigh more than 10 ounces?

Type A Question (example)

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

A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is 15. a.) With 95% confidence, what is the margin of error? (Specify to 2 decimal places. Round the final answer, no rounding on intermediate steps) b.) Compute the 95% confidence interval for the population mean (Specify to 2 decimal places). 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) d.) Compute the 95% confidence interval for the population mean (Specify to 2 decimal places).

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 standard normal distribution is a normal distribution with 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

a mean of 0 and a standard deviation of 1

In the case of an infinite population, we must select _______________ in order to make valid statistical inferences about the population from which the sample is taken.

a random sample

An interval estimate can be computed by...

adding and subtracting a margin of error to the point estimate

A continuous random variable may assume 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

all values in an interval or collection of intervals

a Random Sample from _____ Population is a Sample selected that satisfies __ conditions.

an Infinite; 2

Populations are often generated by ___________ where there is no upper limit on the number of units that can be generated

an ongoing process

A sample of 25 observations is taken from a process (an infinite population). The sampling distribution of is 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

approximately normal if np≥5 & n(1-p)≥5

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

becomes narrower

As the number of degrees of freedom for a t distribution increases, the difference between the t distribution and the standard normal distribution

becomes smaller

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, α, is decreased, the interval for the population proportion (Hint: check textbook section 8.1 for the definition of level of significance)

becomes wider the level of significance is alpha, smaller alpha has bigger confidence coefficient. So, the interval should be wider.

Continuous Random Variable

can assume any value in an interval on the real line or in a collection of intervals

CONFIDENCE.NORM (function)

can return the margin of errors for a giving population standard deviation

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 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.

central limit theorem

Excel's NORM.DIST function can be used to compute 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

cumulative probabilities for a normally distributed x value

A specific t distribution depends on a parameter known as the ________ __ ________.

degrees of freedom

In point estimation we use the data from the sample to compute a value of a sample statistic that serves as an ____________ of a ___________ _________.

estimate of a population parameter

The sample results provide only ________ of the values of the population characteristics

estimates

(Complete the Sentence) The t distribution is a ...

family of similar probability distributions

(STEPS TO SOLVE ???) P(z > -1.08)

find the interception of row -1.0 and column 0.08. The number 0.1401 is the probability that z<=-1.08. Therefore, 1-0.1401 is the probability that z>-1.08

(STEPS TO SOLVE ???) Prob. (-1.33 ≤ z ≤ 1.67)

find the interception of row -1.3 and column 0.03. The number 0.0918 is the probability that z<=-1.33. Use the answer from the question a, 0.9525 is the probability that z<=1.67. 0.9525-0.0918 is the answer for this question.

(STEPS TO SOLVE ???) Prob. (z ≤ 1.67)

find the interception of row 1.6 and column 0.07 on Standard Normal Cumulative Probability Table, the number in the interception cell is the answer.

(Complete the Sentence) The purpose of an interval estimate is to provide information about...

how close the point estimate is to the value of the parameter

As degrees of freedom (increase/decrease) , the difference between the t distribution and the standard normal probability distribution, z distribution, becomes smaller and smaller.

increase

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) a. is 5 b. is 14 c. is 4 d. cannot be determined because the population is infinite

is 14

Population

is a collection of all the elements of interest

A normal probability distribution 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

is a continuous probability distribution

Frame

is a list of the elements that the sample will be selected from

Sample

is a subset of the population

Element

is the entity on which data are collected

Normal Probability Distribution

is the most important distribution for describing a continuous random variable.

Sampled Population

is the population from which the sample is drawn

The sampling distribution of the sample mean 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

is the probability distribution showing all possible values of the sample mean

=NORM.DIST is used to compute...

is used to compute the Cumulative Probability given a X-value for any Normal Distribution

What is the "skewness measure" of Normal Probability Distributions

its skewness measure is zero

mean

The entire family of normal probability distributions is defined by its ________ & __________

mean & standard deviation

The highest point on the normal curve is at the _____, which is A.K.A. the _____ and _____.

mean; median; mode

Mean is A.K.A. (Normal Prob. Distribution)

median & mode

In selecting random samples of size n from a population, the sampling distribution of the sample mean x ̅ can be approximated by a _______ distribution as the sample size becomes large

normal

The highest point on the ______ curve is at the mean, which is also the median and mode.

normal curve

When the population from which we are selecting a random sample does not have a _______ __________, the central limit theorem is helpful in identifying the shape of the sampling distribution of x ̅.

normal distribution

If a sample size of n is desired from a population containing N elements, we might sample ___ ____ for every _____ ______ in the population.

one element for every n/N element

The general form of an interval estimate of a population proportion is:

p ̅± Margin of Error

=NORM.S.INV

probability

=NORM.INV

probability; mean of normal distribution

When the population from which we are selecting a _______ ________ does not have a normal distribution, the central limit theorem is helpful in identifying the shape of the sampling distribution of x ̅.

random sample

(Complete the Sentence) The sample results provide only estimates of the values of the population characteristics because...

sample contains only a portion of the population

we know that 90% of the time, the _______ ______ is within +/-1.645 standard error of the mean.

sample mean

In point estimation we use the data from the ______ to compute a ________________________ that serves as an estimate of a population parameter

sample; value of a sample statistic

Excel's NORM.INV function can be used to compute 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

the normally distributed x value given a cumulative probability

Sampling Distribution is..

the probability distribution of all possible values of the sample proportion p ̅

In interval estimation, the t distribution is applicable only when

the sample standard deviation is used to estimate the population standard deviation

If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect

the size of the confidence interval to increase

σp ̅ is referred to as..

the standard error of the proportion

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the random variable in this experiment? a. the weight of items produced by a machine b. 8 ounces Incorrect c. 2 ounces d. the normal distribution

the weight of items produced by a machine

=NORM.INV is used to compute...

the x-value given a Cumulative Probability

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

STATS Ch.8, Applied Statistics Midterm 2

Set pelajaran terkait

Possible Solutions to Climate Change

BIO 1090 -- Unit 1

Physics ch 6

PUBLIC HEALTH INFORMATICS: AN INTRODUCTION

Exit hesi- Pharm, RN Exit Hesi

SAEM Combined Deck

Function Anatomy 5-6

g t 7

Korean Verbs (loading...)

LEJA Practice Final

1. CURRENT ELECTRICITY AND ETC

OB Exam #3

CompTIA Security+ Final Assessment (40)

Ch. 5: Foundations of Employee Motivation (power points)

Test 2 Macro Study Guide

South America & Caribbean

quiz 12 psychiatric disorders

Lipids

Google Analytics

Nutrition Quiz 6