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DeVry MATH 221 Week 2 Homework - Latest

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DeVry MATH 221 Week 2 Homework - Latest 2015

MATH 221 Homework Week 2

Two variables have a positive linear correlation. Does the dependent variable increase or decrease as the independent variable increases?

Choose the correct answer below.

The dependent variable decreases

The dependent variable increases

Discuss the difference between r and p

Choose the correct answers below.

R represents the sample correlation coefficient.

P represents the population correlation coefficient

The scatter plot of a paired data set is shown. Determine whether there is a perfect positive linear correlation, a strong positive linear correlation, a perfect negative linear correlation, a strong negative linear correlation, or no linear correlation between the variables.

Choose the correct answer below.

no linear correlation

strong positive linear correlation

strong negative linear correlation

perfect negative linear correlation

perfect positive linear correlation

3 The scatter plot of a paired data set is shown. Determine whether there is a perfect positive linear correlation, a strong positive linear correlation, a perfect negative linear correlation, a strong negative linear correlation, or no linear correlation between the variables.

no linear correlation

strong positive linear correlation

strong negative linear correlation

perfect negative linear correlation

perfect positive linear correlation

Identify the explanatory variable and the response variable.

A golfer wants to determine if the amount of practice every year can be used to predict the amount of improvement in his game.

The explanatory variable is the amount of practice

The response variable is the amount of improvement in his game

4 Identify the explanatory variable and the response variable.

A teacher wants to determine if the amount of textbook used by her students can be used to predict the students’ test scores

The explanatory variable is the type of text book

The response variable is the students’ test scores

Two variables have a positive linear correlation. Is the slope of the regression line for the variables positive or negative?

The slope is positive. As the independent variable increases the dependent variable also tends to increase

The slope is negative. As the independent variable increases the dependent variable tends to decrease

The slope is negative. As the independent variable increases the dependent variable tends to increase.

The slope is positive. As the independent variable increases the dependent variable tends to decrease.

Given a set of data and a corresponding regression line, describe all values of x that provide meaningful predictions for y.

Prediction values are meaningful for all x-values that are realistic in the context of the original data set.

Prediction values are meaningful for all x-values that are not included in the original data set.

Prediction values are meaningful for all x-values in (or close to) the range of the original data.

Match this description with a description below.

The y-value of a data point corresponding to

Choose the correct answer below.

correct answer

Match this description with a description below.

The y-value for a point on the regression line corresponding to

Choose the correct answer below.

correct answer

Match the description below with its symbol(s).

The mean of the y-values

Select the correct choice below.

correct answer

Match the regression equation with the appropriate graph.

Choose the correct answer below.

C is the correct answer.

Match the regression equation with the appropriate graph.

D is the correct answer

Use the value of the linear correlation coefficient to calculate the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R= -0.312

Calculate the coefficient of determination

.097 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

9.7% of the variation can be explained by the regression line. (Round to three decimal places as needed)

About the unexplained variation?

90.3% of the variation is unexplained and is due to other factors or to sampling error. (Round to three decimal places as needed)

11 Use the value of the linear correlation coefficient to calculate the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R= -0.324

Calculate the coefficient of determination

0.105 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

10.5 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

About the unexplained variation?

89.5 % of the variation is unexplained and is due to other factors or to sampling error. (Round to three decimal places as needed)

R = 0.481

Calculate the coefficient of determination

.231 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

2.1 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

76.9 % of the variation is unexplained and is due to other factors or to sampling error. (Round to three decimal places as needed)

12 Use the value of the linear correlation coefficient to calculate the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R = 0.224

Calculate the coefficient of determination

0.050 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

5 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

95 % of the variation is unexplained and is due to other factors or to sampling error. (Round to three decimal places as needed)

The equation used to predict college GPA (range 0-4.0) is is high school GPA (range 0-4.0) and x2 is college board score (range 200-800). Use the multiple regression equation to predict college GPA for a high school GPA of 3.5 and college board score of 400.

The predicted college GOA for a high school GPA of 3.5 and college board of 400 is 2.9. (Round to the nearest tenth as needed).

13 Use the value of the linear correlation coefficient to calculate the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R = 0.909

Calculate the coefficient of determination

.826 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

82.6 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

17.4 % of the variation is unexplained and is due to other factors or to sampling error. (Round to three decimal places as needed)

The equation used to predict the total body weight (in pounds) of a female athlete at a certain school is is the female athlete’s height (in inches) and x2 is the female athlete’s percent body fat. Use the multiple regression equation to predict the total body weight for a female athlete who is 64 inches tall and has 17% body fat.

The predicted total body weight for a female athlete who is 64 inches tall and has 17% body fat is 140.9 pounds. (Round to the nearest tenth as needed).

The equation used to predict college GPA (range 0-4.0) is is high school GPA (range 0-4.0) and X2 is college board score (range 200-800). Use the multiple regression equation to predict college GPA for a high school GPA of 3.2 and a college board score of 500.

The predicted college GPA for a high school GPA of 3.2 and college board score of 500 is 2.9. (Round to the nearest tenth as needed).

The equation used to predict the total body weight (in pounds) of a female athlete at a certain school is is the female athlete’s height (in inches) and X2 is the female athlete’s percent body fat. Use the multiple regression equation to predict the total body weight for a female athlete who is 67 inches tall and has 24% body fat.

The predicted total body weight for a female athlete who is 67 inches tall and has 24% body fat is 137.8 pounds. (Round to the nearest tenth as needed).

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DeVry MATH 221 Week 2 Discussions Regression - Latest

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MATH 221 Week 2 Discussions Regression

Suppose you are given data from a survey showing the IQ of each person interviewed and the IQ of his or her mother. That is all the information that you have. Your boss has asked you to put together a report showing the relationship between these two variables. What could you present and why?

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DeVry MATH 221 Week 1 DQ Descriptive Statistics - latest

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MATH 221 Week 1 DQ Descriptive Statistics

If you were given a large data set such as the sales over the last year of our top 1,000 customers, what might you be able to do with this data? What might be the benefits of describing the data?

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DeVry MATH 221 Quiz Week 5 - Latest

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MATH 221 Quiz Week 5 DeVry

Sixty percent of households say they would feel secure if they had $50,000 in savings. You randomly select 8 households and ask them if they would feel secure if they had $50,000 in savings. Find the probability that the number that say they would feel secure is (a) exactly five, (b) more than five, and (c) at most five.

Find the probability that the number that say they would feel secure is exactly five.

P(5) = 0.279 (Round to three decimal places as needed)

Find the probability that the number that say they would feel secure is more than five.

P(x>5) = 0.315 (Round to three decimal places as needed)

Find the probability that the number that say they would feel secure is at most five.

P(x≤5) = 0.685 (Round to three decimal places as needed)

Hint:

Let X be the number of households say they would feel secure if they had $50,000 in savings. Clearly X is binomial with n = 8 and p = 0.60. The probability mass function binomial variable is given by .The probability for different values of x are given below.

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

00.00065536

0.00065536

00.99934464

0.00786432

0.00851968

0.00065536

0.99148032

0.99934464

0.04128768

0.04980736

0.00851968

0.95019264

0.99148032

0.12386304

0.1736704

0.04980736

0.8263296

0.95019264

0.2322432

0.4059136

0.1736704

0.5940864

0.8263296

0.279

0.685

0.4059136

0.315

0.5940864

0.20901888

0.89362432

0.68460544

0.10637568

0.31539456

0.08957952

0.98320384

0.89362432

0.01679616

0.10637568

0.01679616

0.98320384

00.01679616

(a) P (5) = = 0.279

(b) P (x > 5) = P (x = 6) + P (x = 7) + P (x = 8)

= 0.315

Suppose 80% of kids who visit a doctor have a fever, and 25% of kids with a fever have sore throats. What’s the probability that a kid who goes to the doctor has a fever and a sore throat?

The probability is 0.200. (Round to three decimal places as needed)

Hint:

Let A and B respectively be the events that the kids have fever and sore throat.

Given that P (A) = 80% = 0.80

P (B|A) = 25% = 0.25

P (a kid who goes to the doctor has a fever and a sore throat) = P (A∩B)

= P (B|A) * P (A)

= 0.25 * 0.80

= 0.200

Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n = 90, p = 0.8

The mean, µ is 72.0 (Round to the nearest tenth as needed)

The variance, σ2, is 14.4 (Round to the nearest tenth as needed)

The standard deviation, σ is 3.8 (Round to the nearest tenth as needed)

Hint:

Mean = np = 90 * 0.8 = 72.0

Variance, σ2 = np(1 – p) = 90 * 0.8 * (1 – 0.8) = 14.4

Standard deviation, σ = √14.4 = 3.8

Use the bar graph below, which shows the highest level of education received by employees of a company, to find the probability that the highest level of education for an employee chosen at random is E.

The probability that the highest level of education for an employee chosen at random is E is 0.069. (Round to the nearest thousandth as needed)

Hint:

P (the highest level of education for an employee chosen at random is E)

= 6/(6 + 22 + 36 + 16 + 6 + 1)

= 0.069

A company that makes cartons finds that the probability of producing a carton with a puncture is 0.05, the probability that a carton has a smashed corner is 0.09, and the probability that a carton has a puncture and has a smashed corner is 0.005. Answer parts (a) and (b) below.

Are the events “selecting a carton with a puncture” and “selecting a carton with a smashed corner” mutually exclusive?

No, a carton can have a puncture and a smashed corner.

Yes, a carton can have a puncture and a smashed corner

Yes, a carton cannot have a puncture and a smashed corner

Mo, a carton cannot have a puncture and a smashed corner

If a quality inspector randomly selects a carton, find the probability that the carton has a puncture or has a smashed corner.

The probability that a carton has a puncture or a smashed corner is 0.135. (Type an integer or a decimal. Do not round)

Given that x has a Poisson distribution with �� = 8, what is the probability that x = 3?

P(3) ≈ 0.0286 (Round to four decimal places as needed)

Hint:

The probability density function of Poisson random variable is given by, where λ = 8, x = 3

P (3) = = 0.0286

Perform the indicated calculation.

= 0.0017 (Round to four decimal places as needed)

Hint:

5P2 = 5!/(5 – 2)! = 20

12P4 = 12!/(12 – 4)! = 11880

5P2/12P4 = 20/11880 = 0.0017

A frequency distribution is shown below. Complete parts (a) through (d)

The number of televisions per household in a small town

Televisions 0 1 2 3

Households 26 448 730 1400

Use the frequency distribution to construct a probability distribution

X P(x)

0 0.010

1 0.172

2 0.280

3 0.538

(Round to the nearest thousandth as needed)

Hint: Please see the excel spreadsheet.

Graph the probability distribution using a histogram. Choose the correct graph of the distribution below.

Describe the histogram’s shape. Choose the correct answer below.

Skewed right

Skewed left

Symmetric

Find the mean of the probability distribution

µ = 2.3 (round to the nearest tenth as needed)

Find the variance of the probability distribution

σ2 = 0.6 (round to the nearest tenth as needed)

Find the standard deviation of the probability distribution

σ = 0.8 (round to the nearest tenth as needed)

Interpret the results in the context of the real-life situation.

The mean is 2.3, so the average household has about 3 television. The standard deviation is 0.6 of the households differ from the mean by no more that about 1 television

The mean is 0.6, so the average household has about 1 television. The standard deviation is 0.8 of the households differ from the mean by no more that about 1 television

The mean is 2.3, so the average household has about 2television. The standard deviation is 0.8 of the households differ from the mean by no more that about 1 television

The mean is 0.6, so the average household has about 1 television. The standard deviation is 2.3 of the households differ from the mean by no more that about 3 television

Hint:

Households

P (X)

X^2

X * P (X)

X^2 * P (X)

026

0.010

448

0.172

0.17204301

0.172043011

730

0.280

0.56067588

1.121351767

1400

0.538

1.61290323

4.838709677

Total

2604

2.34562212

6.132104455

μ = ∑ X * P (X) = 2.3

σ2 = = 6.1321 – (2.3456)^2 = 0.6

σ = √0.6 = 0.8

In the general population, one woman in eight will develop breast cancer. Research has shown that 1 woman is 650 carries a mutation of the BRCA gene. Nine out of 10 women with this mutation develop breast cancer.

Find the probability that a randomly selected woman will develop breast cancer given that she has a mutation of the BRCA gene.

The probability that a randomly selected woman will develop breast cancer given that she has a mutation of the BRCA gene is 0.9. (Round to one decimal place as needed)

Hint:

P (a randomly selected woman will develop breast cancer given that she has a mutation of the BRCA gene)

= 9/10

= 0.9

Find the probability that a randomly selected woman will carry the mutation of the BRCA gene and will develop breast cancer.

The probability that a randomly selected woman will carry the gene nutation and develop breast cancer is 0.0014. (Round to four decimal places as needed)

Hint:

P (A∩B) = P (A|B) * P (B)

= 0.9 * (1/650)

= 0.0014

Are the events of carrying this mutation and developing breast cancer independent or dependent?

Dependent

Independent

Hint:

Here P (A|B) ≠ P (A). Hence the events are dependent.

Students in a class take a quiz with eight questions. The number x of questions answered correctly can be approximated by the following probability distribution. Complete parts (a) through (e)

X 0 1 2 3 4 5 6 7 8

P(x) 0.04 0.04 0.06 0.06 0.12 0.24 0.23 0.14 0.07

Use the probability distribution to find the mean of the probability distribution

µ= 4.9 (Round to the nearest tenth as needed)

Use the probability distribution to find the variance of the probability distribution

σ2=4.0 (Round to the nearest tenth as needed)

Use the probability distribution to find the standard deviation of the probability distribution

2.0 (Round to the nearest tenth as needed)

Use the probability distribution to find the expected value of the probability distribution

4.9 (Round to the nearest tenth as needed)

Interpret the results

The expected number of questions answered correctly is 2.0 with a standard deviation of 4.9 questions.

The expected number of questions answered correctly is 4 with a standard deviation of 2.0 questions.

The expected number of questions answered correctly is 4.9 with a standard deviation of 0.04 questions.

The expected number of questions answered correctly is 4.9 with a standard deviation of 2.0 questions.

Hint:

P(x)

X^2

X * P (X)

X^2 * P (X)

00.04

0.04

0.06

0.12

0.24

0.06

0.18

0.54

0.12

0.48

1.92

0.24

1.2

0.23

1.38

8.28

0.14

0.98

6.86

0.07

0.56

4.48

Total

4.94

28.36

μ = ∑ X * P (X) = 4.9

σ2 = = 28.36 – (4.94)^2 = 4.0

σ = √4 = 2.0

Expected value = μ = 4.9

Identify the sample space of the probability experiment and determine the number of outcomes in the sample space. Randomly choosing a multiple of 5 between 21 and 49

The sample space is {25,30,35,40,45} (Use a comma to separate answers as needed. Use ascending order)

There are 5 outcome(s) in the sample space.

Decide if the events shown in the Venn diagram are mutually exclusive.

Are the events mutually exclusive?

Yes

Hint:

Since the events are overlapping, they have common elements. Hence events are not mutually exclusive.

Determine whether the random variable is discrete or continuous.

The number of free-throw attempts before the first shot is made

The weight of a T-bone steak

The number of bald eagles in the country

The number of points scored during a basketball game

The number of hits to a website in a day

Is the number of free-throw attempts before the first shot is made discrete or continuous?

The random variable is continuous

The random variable is discrete

Hint:

Since the number of free-throw attempts before the first shot can only be expressed as a whole number, it is discrete.

Is the weight of a T-bone steak discrete or continuous?

The random variable is discrete

The random variable is continuous

Hint:

The weight can take any value and is not restricted to whole numbers. Hence it is continuous.

Is the number of bald eagles in the country discrete or continuous?

The random variable is discrete

The random variable is continuous

Hint:

Since the number of bald eagles in the country can only be expressed as a whole number, it is discrete.

Is the number of points scored during a basketball game discrete or continuous?

The random variable is discrete

The random variable is continuous

Hint:

Since the number of points scored during a basketball game can only be expressed as a whole number, it is discrete.

Is the number of hits to a website in a day discrete or continuous?

The random variable is discrete

The random variable is continuous

Hint:

Since the number of hits to a website in a day can only be expressed as a whole number, it is discrete.

A survey asks 1100 workers, :Has the economy forced you to reduce the amount of vacation you plan to take this year?” Fifty-six percent of those surveyed say they are reducing the amount of vacation. Twenty workers participating in the survey are randomly selected. The random variable represents the number of workers who are reducing the amount of vacation. Decide whether the experiment is a binomial experiment. If it is, identify a success, specify the values of n, p, and q, and list the possible values of the random variable x.

Is the experiment a binomial experiment?

Yes

What is a success in this experiment?

Selecting a worker who is reducing the amount of vacation

Selecting a worker who is not reducing the amount of vacation

This is not a binomial experiment

Specify the value of n. Select the correct choice and fill in any answer boxes in your choice

N=20

This is not a binomial experiment

Specify the value of p. Select the correct choice below and fill in any answer boxes in your choice.

P=0.56 (Type an integer or a decimal)

This is not a binomial experiment

Specify the value of q. Select the correct choice below and fill in any answer boxes in your choice.

Q=0.44(Type an integer r a decimal)

This is not a binomial experiment

List the possible values of the random variable x

X=0, 1, 2,…, 20

X=1, 2, 3,…, 1100

1, 2,…, 20

This is not a binomial experiment

Determine whether the distribution is a discrete probability distribution.

Is the distribution a discrete probability distribution? Why? Choose the correct answer below.

Yes, because the probabilities sum to 1 and are all between 0 and 1, inclusive

No, because the total probability is not equal to 1

Yes, because the distribution is symmetric

No, because some of the probabilities have values greater than 1 or less than 0

The table below shows the results of a survey that asked 2872 people whether they are involved in any type of charity work. A person is selected at random from the sample. Complete parts (a) through (e).

Frequency Occasionally Not at all Total

Male 226 455 793 1474

Female 206 450 742 1398

Total 432 905 1535 2872

Find the probability that the person is frequently or occasionally involved in charity work

P(being frequently involved or being occasionally involved) = 0.466 (Round to the nearest thousandth as needed)

Hint:

P(being frequently involved or being occasionally involved) =

= (432 + 905)/2872

= 0.466

Find the probability that the person is male or frequently involved in charity work

P(being male or being frequently involved) = 0.585

Hint:

P(being male or being frequently involved) = P (male) + p (being frequently involved) – P (male and being frequently involved)

= (1474/2872) + (432/2872) – (226/2872)

= 0.585

Find the probability that the person is female or not involved in charity work at all

P(being female or not being involved) = 0.763 (Round to the nearest thousandth as needed)

Hint:

P(being female or not being involved) = P (female) + P (not being involved) – P (female and not being involved)

= (1398/2872) + (1535/2872) – (742/2872)

= 0.763

Find the probability that the person is female or not frequently involved in charity work

P(being female or not being frequently involved) = 0.921 (Round to the nearest thousandth as needed)

Hint:

P(being female or not being frequently involved) = P (female) + P (not being frequently involved) – P (being female AND not being frequently involved)

= (1398/2872) + [(905 + 1535)/2872] – [(450 + 742)/2872)

= 0.921

Are the events “being female” and “being frequently involved in charity work” mutually exclusive?

No, because 206 females are frequently involved in charity work.

Yes, because no females are frequently involved in charity work.

Yes, because 206 females are frequently involved in charity work.

No, because no females are frequently involved in charity work.

For the given pair of events, classify the two events as independent or dependent.

Swimming all day at the beach

Getting a sunburn

Choose the correct answer below.

The two events are independent because the occurrence on one does not affect the probability of the occurrence of the other.

The two events are dependent because the occurrence of one does not affect the probability of the occurrence of the other.

The two events are independent because the occurrence of one affects the probability of the occurrence of the other.

The two events are dependent because the occurrence of one affects the probability of the occurrence of the other.

Outside a home, there is a 9-key keypad with letters A, B, C, D, E, F, G, H, and I that can be used to open the garage if the correct nine-letter code is entered. Each key may be used only once. How many codes are possible?

The number of possible codes is 362880.

Hint:

Here each letter can be used only once. Therefore, the first letter can be set in 9 different ways using these 9 letters, 2nd letter in 8 different letters and so on.

Hence the number of possible codes = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

= 362880

Determine the number of outcomes in the event. Decide whether the event is a simple event or not.

A computer is used to select randomly a number between 1 and 9, inclusive. Event C is selecting a number less than 5.

Event C has 4 outcome(s)

Is the event a simple event?

No, because event C has more than one outcome.

Hint:

C = {1, 2, 3, 4}

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DeVry MATH 221 Final Exam - Latest 2015

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DeVry MATH 221 Final Exam - Latest

The table below shows the number of male and female students enrolled in nursing at a university for a certain semester. A student is selected at random. Complete parts (a) through (d).

Find the probability that the student is male or a nursing major.

P (being male or being nursing major) = 0.513

Find the probability that the student is female or not a nursing major.

P(being female or not being a nursing major) = 0.972

Find the probability that the student is not female or a nursing major

P(not being female or not being a nursing major) = 0.513

Are the events “being male” and “being a nursing major” mutually exclusive? Explain.

Answer: No, because there are 99 males majoring in nursing.

Hint:

(a) P (male or a nursing major) = P (male) + P (nursing major) – P (male AND a nursing major)

= (1115/3539) + (799/3539) – (99/3539)

= 0.513

(b) P (female or not a nursing major) = P (female) + P (not a nursing major) – P (female AND not a nursing major)

= (2424/3539) + (2740/3539) – (1724/3539)

= 0.972

= (1115/3539) + (799/3539) – (99/3539)

= 0.513

An employment information service claims the mean annual pay for full-time male workers over age 25 without a high school diploma is $22,325. The annual pay for a random sample of 10 full-time male workers over age 25 without a high school diploma is listed. At a = 0.10, test the claim that the mean salary is $22,325. Assume the population is normally distributed.

20,660 – 21,134 – 22,359 – 21,398 – 22,974, - 16,919 – 19,152 – 23,193 – 24,181 – 26,281

Write the claim mathematically and identify

Which of the following correctly states ?

Answer:

Find the critical value(s) and identify the rejection region(s).

What are the critical values? Answer: -1.833, 1.833

Which of the following graphs best depicts the rejection region for this problem?

Find the standardized test statistics. t = -0.60

Decide whether to reject or fail to reject the null hypothesis. reject because the test statistics is in the rejection region.

fail to reject because the test statistic is not in the rejection region. c. reject because the test statistic is not in the rejection region. d. fail to reject because the test statistic is in the rejection region.

(e) Interpret the decision in the context of the original claim. a. there is sufficient evidence to reject the claim that the mean salary is $22,325. b. there is not sufficient evidence to reject the claim that the mean salary is not $22,325. c. there is sufficient evidence to reject the claim that the mean salary is not $22,325. d. there is not sufficient evidence to reject the claim that the mean salary is $22,325.

Hint:

Test statistic, = -0.60

Critical values are obtained from the Student’s t distribution table with d.f. 9 at the significance level 0.10.

Critical values = -1.833, 1.833

Since the test statistic does not fall in the critical region, we do not reject the null hypothesis.

The times per week a student uses a lab computer are normally distributed, with a mean of 6.1 hours and a standard deviation of 1.2 hours. A student is randomly selected. Find the following probabilities. (a) the probability that the student uses a lab computer less than 5hrs a week. (b) the probability that the student uses a lab computer between 6-8 hrs a week.

= 0.180

= 0.477

= 0.008

Hint:

P (X <5) == P (Z < -0.9167) = 0.180

P (6<X <8) == P (-0.0833 <Z <1.5833) = 0.477

P (X >9) == P (Z >2.4167) = 0.008

Write the null and alternative hypotheses. Identify which is the claim. A study claims that the mean survival time for certain cancer patients treated immediately with chemo and radiation is 13 months.

Find the indicated probability using the standard normal distribution. P(z>-1.58) = 0.9429

Hint:

P (Z > -1.58) = 1 – P (Z < -1.58) = 1 – 0.0571 = 0.9429

The Gallup Organization contacts 1323 men who are 40-60 years of age and live in the US and asks whether or not they have seen their family doctor.

What is the population in the study? Answer: Adult men who are 40-60 years old and live in the US.

What is the sample in the study? Answer: The 1323 adult men who are 40-60 years old and live in the US.

The ages of 10 brides at their first marriage are given below. 4 32.2 33.6 41.2 43.4 37.1 22.7 29.9 30.6 30.8

(a) find the range of the data set. Range = 20.7 (b) change 43.4 to 58.6 and find the range of the new date set. Range = 35.9 (c) compare your answer to part (a) with your answer to part (b) Answer: Changing the maximum value of the date set greatly affects the range.

Hint:

Range = maximum value – minimum value = 43.4 – 22.7 = 20.7

Range = maximum value – minimum value = 58.6 – 22.7 = 35.9

The following appear on a physician’s intake form. Identify the level of measurement of the data. (a) Martial Status (b) Pain Level (0-10) (c) Year of Birth (d) Height

(a) what is the level of measurement for marital status Nominal

(b)what is the level of measurement for pain level Ordinal

(d) What is the level of measurement for height Ratio

To determine her air quality, Miranda divides up her day into 3 parts; morning, afternoon, and evening. She then measures her air quality at 3 randomly selected times during each part of the day. What type of sampling is used? Answer: Stratified

Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line. Then use the regression equation to predict the value of y for each of the given x-values, if meaningful. The caloric content and the sodium content (in milligrams) for 6 beef hot dogs are shown in the table below.

X= 150 calories

X= 100 calories

X = 120 calories

X = 60 calories

Find the regression equation. = 2.509x + (37.832) Choose the correct graph below.

(a) predict the value of y for x = 150. Answer: 414.182 (b) predict the value of y for x = 100. Answer: 188.372 (c) predict the value of y for x = 120. Answer: 338.912 (d) predict the value of y for x = 60. Answer: not meaningful.

Hint:

Regression equation is y = 2.509x + 37.832 (Please see the excel spreadsheet)

(a) When x = 150, y = (2.509 * 150) + 37.832 = 414.182

(b) When x = 100, y = (2.509 * 100) + 37.832 = 288.732

(d) since the value of x is outside the range of values of x, not meaningful

A restaurant association says the typical household spends a mean of $4072 per year on food away from home. You are a consumer reporter for a national publication and want to test this claim. You randomly select 12 households and find out how much each spent on food away from home per year. Can you reject the restaurant association’s claim at a = 0.10? Complete parts a through d.

Write the claim mathematically and identify. Choose the correct the answer below.

Use technology to find the P-value. P = 0.03 Decide whether to reject or fail the null hypothesis.

Answer: Reject Interpret the decision in the context of the original claim. Assume the population is normally distributed. Choose the correct answer below. Answer: At the 10% significance level, there is a sufficient evidence to reject the claim.

Hint: Please see the excel spreadsheet

The table below shows the results of a survey in which 147 families were asked if they own a computer and if they will be taking a summer vacation this year.

(a) find the probability that a randomly selected family is not taking a summer vacation year. Probability = 0.299 (b) find the probability that a randomly selected family owns a computer Probability = 0.388 (c) find the probability that a randomly selected family is taking a summer vacation this year and owns a computer Probability = 0.825 (d) find the probability a randomly selected family is taking a summer vacation this year and owns a computer. Probability = 0.320

Are the events of owning a computer and taking a summer vacation this year independent or dependent events? Answer: Dependent

Hint:

(a) P (a randomly selected family is not taking a summer vacation year)

= 44/147

= 0.299

(b) P (randomly selected family owns a computer) =

= 57/147

= 0.388

= (47/147)/(57/147)

= 47/57

= 0.825

(d) P (a randomly selected family is taking a summer vacation this year and owns a computer)

= 47/147

= 0.320

Assume the Poisson distribution applies. Use the given mean to find the indicated probability. Find P(5) when ᶙ = 4

P(5) = 0.156

Hint:

P (5) = , Where λ = 4, x = 5

Therefore, P (5) = = 0.156

In a survey of 7000 women, 4431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week.

A 99% confidence interval for the population proportion is 618, 0.648

Hint:

90% Confidence Interval for proportion is given by

where p = x/n = 4431/7000 = 0.633, = 2.5758, n = 7000

That is,

= (0.618, 0.648)

A random sample of 53 200-meter swims has a mean time of 3.32 minutes and the population standard deviation is 0.06 minutes. Construct a 90% confidence interval for the population mean time. Interpret the results.

The 90% confidence interval is 3.31, 3.33.

Interpret these results. Choose the correct answer: Answer: With 90% confidence, it can be said that the population mean time is between the end points of the given confidence interval.

Hint:

The confidence interval is given by, where = 3.32, σ = 0.06, n = 53, = 1.6449

That is,

= (3.31, 3.33)

Determine whether the variable is qualitative or quantitative: Weight

Quantitative Qualitative

32% of college students say that they use credit cards because of the reward program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the reward program is (a) exactly two, (b), more than two, and (c), between two and five inclusive.

(a) P(2) = 0.211 (b) P(X>2) = 0.669 (c) P(2<x<5) = 0.816

Hint:

Let X be the number of college students say that they use credit cards because of the reward program. Clearly X is binomial with n = 10 and p = 0.32. The probability mass function binomial variable is given by .The probability for different values of x are given below.

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

00.021139228

0.021139228

00.978860772

0.099478721

0.120617949

0.021139228

0.879382051

0.978860772

0.210660821

0.33127877

0.120617949

0.66872123

0.879382051

0.264358677

0.595637447

0.33127877

0.404362553

0.66872123

0.217707146

0.813344593

0.595637447

0.186655407

0.404362553

0.122940506

0.936285099

0.813344593

0.063714901

0.186655407

0.048211963

0.984497062

0.936285099

0.015502938

0.063714901

0.012964562

0.997461623

0.984497062

0.002538377

0.015502938

0.002287864

0.999749487

0.997461623

0.000250513

0.002538377

0.000239254

0.999988741

0.999749487

1.1259E-05

0.000250513

1.1259E-05

0.999988741

01.1259E-05

(a) P (2) = = 0.211

(b) P (X > 2) = 1 – P (X ≤ 2) = 1 – [P (X = 0) + P (X = 1) + P (X = 2)] = 0.669

= 0.816

A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 950 hours. A random sample of 74 light bulbs has a mean life of 943 hours with a standard deviation of 90 hours. Do you have enough evidence to reject the manufacturer’s claim? Use ᶏ = 0.04

Identify the critical value(s). -1.75 (c) identify the standardized test statistic. z = -0.67 (d) decide whether to reject or fail to reject the null hypothesis.

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