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DeVry MATH 221 Week 7 Quiz with Formulas – Latest

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A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 35,000 miles and a standard deviation of 2800 miles. He wants to give a guarantee for free replacement of tires that don’t wear well. How should he word his guarantee if he is willing to replace approximately 10% of the tires?

Tires that wear out by 31,412 miles will be replaced free of charge.

Hint:

Here we need a value of X say K such that P (X < K) = 0.10

We know P (Z < -1.282) = 0.10

Thus from Z score of K we have,

Solving for K we have K = 31412

Find the indicated z-score shown in the graph to the right.

The z-score is -1.18

Hint:

From standard normal table, P (Z < -1.18) = 0.1190

Therefore, z = -1.18

A researcher wishes to estimate, with 95% confidence, the amount of adults who have high-speed internet access. Her estimate must be accurate within 4% of the true proportion. a) find the minimum sample size needed, using a prior study that found that 32% of the respondents said they have high-speed internet access.

b) no preliminary estimate is available. Find the minimum sample size needed.a) = 523 b) = 601

Hint:

(a)

The sample size is given by where p = 0.32

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >522.4384

Hence the minimum sample size required is n = 523

(b)

The sample size is given by where p = 0.50

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >600.2279

Hence the minimum sample size required is n = 601

The total cholesterol levels of a sample of men aged 35-44 are normally distributed with a mean of 221 milligrams per deciliter and a standard deviation of 37.7 milligrams per deciliter. (a) what percent of men have a total cholesterol level less than 228 milligrams per deciliter of blood? (b) if 251 men in the 35-44 age group are randomly selected, about how many would you expect to have a total cholesterol level greater than 264 milligrams per deciliter of blood?a) = 57.37% b) = 32

Hint:

(a) P (X < 228) = = P (Z <0.1857) = 0.5737 = 57.37%

(b) P (X > 264) = = P (Z >1.1406) = 0.127

Hence the number = 251 * 0.127 = 32

Find the z-score that has a 12.1% of the distribution’s area to it’s left. Answer = -1.17

Hint:

We need the k such that P (Z < k) = 0.121

From standard normal table, P (Z < -1.17) = 0.121

Therefore k = -1.17

A doctor wants to estimate the HDL cholesterol of all 20-29 year old females. How many subjects are needed to estimate the HDL cholesterol within 2 points with 99% confidence assuming σ = 18.1? suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?99% = 554 subjects 90% = 222 subjectshow does the decrease in confidence affect the sample size required? Answer: The lower the confidence% level the smaller the sample size.

Hint:

The minimum sample size is given by the formula

Please see the excel spreadsheet for calculations.

Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.054Answer: -1.607

Hint:

We need the k such that P (Z < k) = 0.054

From standard normal table, P (Z < -1.607) = 0.121

Therefore k = -1.607

In a survey of 3076 adults, 1492 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.Answers: 462, 0.508With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

Hint:

99% Confidence Interval for proportion is given by

Please see the excel spreadsheet for calculation.

Assume the random variable x is normally distributed with mean u = 89 and standard deviation o = 4. Find the indicated probability. P(76<x<82) Answer: 0395

Hint:

P (76 <X <82) = = P (-3.25 <Z < -1.75) = 0.0395

Find the margin of error for the given values of c,s, and n. c = .90, s = 3.1, n =49 Answer: 728

Hint:

Margin of error = = 1.645 * (3.1/√49) = 0.728

Find the critical value Tc for the confidence level c = .90 and sample size n = 29. Tc = 701

Hint:

Degrees of freedom = n – 1 = 29 – 1 = 28

From Student’s t distribution table, critical value Tc with d.f. 28 at the confidence level 0.90 is given by,

Tc = 1.701

The mean height of women in a country (ages 20-29) is 63.9 inches. A random sample of 65 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 inches? Assume o = 2.91 Answer = 0012

Hint:

P (>65) == P (Z >3.0476) = 0.0012

A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 age group, the heights were normally distributed with a mean of 67.9 inches and a standard deviation of 3.0 inches. A study participant is randomly selected. Complete parts (A) through (C). (a) the probability that his height is less than 68 inches. Answer: 5133 (b) the probability that his height is between 68-71 inches. Answer: 0.3360 (c) the probability that his height is more than 71 inches. Answer: 0.1507

Hint:

(a) P (X <68) = = P (Z < 0.03333) = 0.5133

(b) P (68 <X <71) = = P (0.0333 <Z < 1.0333) = 0.3360

For the standard normal distribution shown on the right, find the probability of z occurring in the region.Probability = .6950

Hint:

P (Z < 0.51) = 0.6950

Find the indicated probability using the standard normal distribution. (P – 1.35 < z < 1.35) = 8230

Hint:

(P – 1.35 < z < 1.35) = P (z < 1.35) – P (z < -1.35)

= 0.9115 – 0.0885

= 0.8230

Find the margin of error for the given values of c, s, n. c = 0.98, s = 5, n = 6 Answer: 9

Hint:

Margin of error = = 3.3649 * (5/√6) = 6.9

The systolic blood pressures of a sample of adults are normally distributed with a mean pressure of 115 millimeters of mercury and a standard deviation of 3.6 millimeters of mercury. The systolic blood pressures of four adults selected at random are 122 millimeters of mercury, 113 millimeters of mercury, 106 millimeters of mercury, and 128 millimeters of mercury. The graph of the standard normal distribution is below. Complete parts a – c.

Match the values with the letters a/b/c/d. A = 106 B = 113 C = 122 D = 128

Find the z-scores that corresponds to each value A = -2.50 B = -0.56 C = 1.94 D = 3.61

Hint:

A = (X – μ)/σ = (106 – 115)/3.6 = -2.50

B = (X – μ)/σ = (113 – 115)/3.6 = -0.56

C = (X – μ)/σ = (122 – 115)/3.6 = 1.94

D = (X – μ)/σ = (128 – 115)/3.6 = 3.61

Determine if any of the values are unusual, and classify them as either unusual or very unusual.

Answer: The unusual value(s) is/are 106. The very unusual value(s) is/are 128.

Hint:

106 is unusual, since the z-score is outside ±2. 128 is very unusual, since the z-score is outside ±3.

You are given a sample mean and standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 60 home theater systems has a mean price of $134.00 and a standard deviation is $19.60The 90% confidence interval is (84), (138.16). The 95% confidence interval is (129.04, (138.96).Interpret the results.

Answer: With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

The monthly incomes for 12 randomly select people, each with a bachelor’s degree in economics, are shown on the right. Assume the population is normally distributed.

Mean = 4263.3 Standard Deviation = 260.1 99% confidence interval = 4030.1, 4496.5

What is the total area under the normal curve? Answer = 1

A population has a mean u = 82 and a standard deviation o = 36. Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 81 82

The amounts of time employees at a large corporation work each day are normally distributed, with a mean of 7.4 hours and a standard deviation of 0.38 hour. Random sample of size 25 and 37 are drawn from the population and the mean of each sample is determined. What happens to the mean and the standard deviation of the distribution of sample means as the size of the sample increases?Mean of distribution = 4 Standard deviation of distribution = 0.08If the sample size is n = 37, find the mean and standard deviation.Mean = 7.4 Standard deviation = 0.06

What happens to the mean and standard deviation of the distribution of sample means as the size of the sample increases? Answer: The mean stays the same, the but standard deviation decreases.

Assume a member is selected at random from the population represented by the graph. Find the probability that the member selected at random is from the shaded area of the graph.

Answer = 0.3038

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letssuperstudentdinosaur-blog · 9 years ago

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DeVry MATH 221 Entire Course - LATEST 2015

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gotremendouspolicecollector-blog · 9 years ago

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MATH 221

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MATH 221

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fantasticblazedinosaur-blog · 10 years ago

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DeVry MATH 221 Week 7 Quiz with Formulas – Latest

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Tires that wear out by 31,412 miles will be replaced free of charge.

Hint:

Here we need a value of X say K such that P (X < K) = 0.10

We know P (Z < -1.282) = 0.10

Thus from Z score of K we have,

Solving for K we have K = 31412

Find the indicated z-score shown in the graph to the right.

The z-score is -1.18

Hint:

From standard normal table, P (Z < -1.18) = 0.1190

Therefore, z = -1.18

b) no preliminary estimate is available. Find the minimum sample size needed.

a) = 523 b) = 601

Hint:

(a)

The sample size is given by where p = 0.32

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >522.4384

Hence the minimum sample size required is n = 523

(b)

The sample size is given by where p = 0.50

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >600.2279

Hence the minimum sample size required is n = 601

a) = 57.37% b) = 32

Hint:

(a) P (X < 228) = = P (Z <0.1857) = 0.5737 = 57.37%

(b) P (X > 264) = = P (Z >1.1406) = 0.127

Hence the number = 251 * 0.127 = 32

Find the z-score that has a 12.1% of the distribution’s area to it’s left. Answer = -1.17

Hint:

We need the k such that P (Z < k) = 0.121

From standard normal table, P (Z < -1.17) = 0.121

Therefore k = -1.17

99% = 554 subjects 90% = 222 subjects

how does the decrease in confidence affect the sample size required? Answer: The lower the confidence% level the smaller the sample size.

Hint:

The minimum sample size is given by the formula

Please see the excel spreadsheet for calculations.

Answer: -1.607

Hint:

We need the k such that P (Z < k) = 0.054

From standard normal table, P (Z < -1.607) = 0.121

Therefore k = -1.607

In a survey of 3076 adults, 1492 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.

Answers: 462, 0.508

With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

Hint:

99% Confidence Interval for proportion is given by

Please see the excel spreadsheet for calculation.

Assume the random variable x is normally distributed with mean u = 89 and standard deviation o = 4. Find the indicated probability. P(76<x<82) Answer: 0395

Hint:

P (76 <X <82) = = P (-3.25 <Z < -1.75) = 0.0395

Find the margin of error for the given values of c,s, and n. c = .90, s = 3.1, n =49 Answer: 728

Hint:

Margin of error = = 1.645 * (3.1/√49) = 0.728

Find the critical value Tc for the confidence level c = .90 and sample size n = 29. Tc = 701

Hint:

Degrees of freedom = n – 1 = 29 – 1 = 28

From Student’s t distribution table, critical value Tc with d.f. 28 at the confidence level 0.90 is given by,

Tc = 1.701

Hint:

P (>65) == P (Z >3.0476) = 0.0012

Hint:

(a) P (X <68) = = P (Z < 0.03333) = 0.5133

(b) P (68 <X <71) = = P (0.0333 <Z < 1.0333) = 0.3360

For the standard normal distribution shown on the right, find the probability of z occurring in the region.

Probability = .6950

Hint:

P (Z < 0.51) = 0.6950

Find the indicated probability using the standard normal distribution. (P – 1.35 < z < 1.35) = 8230

Hint:

(P – 1.35 < z < 1.35) = P (z < 1.35) – P (z < -1.35)

= 0.9115 – 0.0885

= 0.8230

Find the margin of error for the given values of c, s, n. c = 0.98, s = 5, n = 6 Answer: 9

Hint:

Margin of error = = 3.3649 * (5/√6) = 6.9

Match the values with the letters a/b/c/d. A = 106 B = 113 C = 122 D = 128

Find the z-scores that corresponds to each value A = -2.50 B = -0.56 C = 1.94 D = 3.61

Hint:

A = (X – μ)/σ = (106 – 115)/3.6 = -2.50

B = (X – μ)/σ = (113 – 115)/3.6 = -0.56

C = (X – μ)/σ = (122 – 115)/3.6 = 1.94

D = (X – μ)/σ = (128 – 115)/3.6 = 3.61

Determine if any of the values are unusual, and classify them as either unusual or very unusual.

Answer: The unusual value(s) is/are 106. The very unusual value(s) is/are 128.

Hint:

106 is unusual, since the z-score is outside ±2. 128 is very unusual, since the z-score is outside ±3.

The 90% confidence interval is (84), (138.16). The 95% confidence interval is (129.04, (138.96).

Interpret the results.

The monthly incomes for 12 randomly select people, each with a bachelor’s degree in economics, are shown on the right. Assume the population is normally distributed.

Mean = 4263.3 Standard Deviation = 260.1 99% confidence interval = 4030.1, 4496.5

What is the total area under the normal curve? Answer = 1

A population has a mean u = 82 and a standard deviation o = 36. Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 81 82

Mean of distribution = 4 Standard deviation of distribution = 0.08

If the sample size is n = 37, find the mean and standard deviation.

Mean = 7.4 Standard deviation = 0.06

What happens to the mean and standard deviation of the distribution of sample means as the size of the sample increases? Answer: The mean stays the same, the but standard deviation decreases.

Assume a member is selected at random from the population represented by the graph. Find the probability that the member selected at random is from the shaded area of the graph.

Answer = 0.3038

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loudunknownphilosopher-blog · 10 years ago

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MATH 221

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fancylightpenguin-blog · 10 years ago

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DeVry MATH 221 Week 3 DQ Statistics in the News - Latest

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MATH 221 Week 3 DQ Statistics in the News

Keep your eyes and ears open as you read or listen to the news this week. Find/discover an example of statistics in the news to discuss the following statement that represents one of the objectives of statistics analysis: “Statistics helps us make decisions based on data analysis.” Briefly discuss how the news item or article meets this objective. Cite your references.

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i-qualitycupcakestudent-blog · 10 years ago

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DeVry MATH 221 Week 7 Quiz with Formulas – Latest

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Tires that wear out by 31,412 miles will be replaced free of charge.

Hint:

Here we need a value of X say K such that P (X < K) = 0.10

We know P (Z < -1.282) = 0.10

Thus from Z score of K we have,

Solving for K we have K = 31412

Find the indicated z-score shown in the graph to the right.

The z-score is -1.18

Hint:

From standard normal table, P (Z < -1.18) = 0.1190

Therefore, z = -1.18

b) no preliminary estimate is available. Find the minimum sample size needed.

a) = 523 b) = 601

Hint:

(a)

The sample size is given by where p = 0.32

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >522.4384

Hence the minimum sample size required is n = 523

(b)

The sample size is given by where p = 0.50

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >600.2279

Hence the minimum sample size required is n = 601

a) = 57.37% b) = 32

Hint:

(a) P (X < 228) = = P (Z <0.1857) = 0.5737 = 57.37%

(b) P (X > 264) = = P (Z >1.1406) = 0.127

Hence the number = 251 * 0.127 = 32

Find the z-score that has a 12.1% of the distribution’s area to it’s left. Answer = -1.17

Hint:

We need the k such that P (Z < k) = 0.121

From standard normal table, P (Z < -1.17) = 0.121

Therefore k = -1.17

99% = 554 subjects 90% = 222 subjects

how does the decrease in confidence affect the sample size required? Answer: The lower the confidence% level the smaller the sample size.

Hint:

The minimum sample size is given by the formula

Please see the excel spreadsheet for calculations.

Answer: -1.607

Hint:

We need the k such that P (Z < k) = 0.054

From standard normal table, P (Z < -1.607) = 0.121

Therefore k = -1.607

In a survey of 3076 adults, 1492 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.

Answers: 462, 0.508

With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

Hint:

99% Confidence Interval for proportion is given by

Please see the excel spreadsheet for calculation.

Assume the random variable x is normally distributed with mean u = 89 and standard deviation o = 4. Find the indicated probability. P(76<x<82) Answer: 0395

Hint:

P (76 <X <82) = = P (-3.25 <Z < -1.75) = 0.0395

Find the margin of error for the given values of c,s, and n. c = .90, s = 3.1, n =49 Answer: 728

Hint:

Margin of error = = 1.645 * (3.1/√49) = 0.728

Find the critical value Tc for the confidence level c = .90 and sample size n = 29. Tc = 701

Hint:

Degrees of freedom = n – 1 = 29 – 1 = 28

From Student’s t distribution table, critical value Tc with d.f. 28 at the confidence level 0.90 is given by,

Tc = 1.701

Hint:

P (>65) == P (Z >3.0476) = 0.0012

Hint:

(a) P (X <68) = = P (Z < 0.03333) = 0.5133

(b) P (68 <X <71) = = P (0.0333 <Z < 1.0333) = 0.3360

For the standard normal distribution shown on the right, find the probability of z occurring in the region.

Probability = .6950

Hint:

P (Z < 0.51) = 0.6950

Find the indicated probability using the standard normal distribution. (P – 1.35 < z < 1.35) = 8230

Hint:

(P – 1.35 < z < 1.35) = P (z < 1.35) – P (z < -1.35)

= 0.9115 – 0.0885

= 0.8230

Find the margin of error for the given values of c, s, n. c = 0.98, s = 5, n = 6 Answer: 9

Hint:

Margin of error = = 3.3649 * (5/√6) = 6.9

Match the values with the letters a/b/c/d. A = 106 B = 113 C = 122 D = 128

Find the z-scores that corresponds to each value A = -2.50 B = -0.56 C = 1.94 D = 3.61

Hint:

A = (X – μ)/σ = (106 – 115)/3.6 = -2.50

B = (X – μ)/σ = (113 – 115)/3.6 = -0.56

C = (X – μ)/σ = (122 – 115)/3.6 = 1.94

D = (X – μ)/σ = (128 – 115)/3.6 = 3.61

Determine if any of the values are unusual, and classify them as either unusual or very unusual.

Answer: The unusual value(s) is/are 106. The very unusual value(s) is/are 128.

Hint:

106 is unusual, since the z-score is outside ±2. 128 is very unusual, since the z-score is outside ±3.

The 90% confidence interval is (84), (138.16). The 95% confidence interval is (129.04, (138.96).

Interpret the results.

The monthly incomes for 12 randomly select people, each with a bachelor’s degree in economics, are shown on the right. Assume the population is normally distributed.

Mean = 4263.3 Standard Deviation = 260.1 99% confidence interval = 4030.1, 4496.5

What is the total area under the normal curve? Answer = 1

A population has a mean u = 82 and a standard deviation o = 36. Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 81 82

Mean of distribution = 4 Standard deviation of distribution = 0.08

If the sample size is n = 37, find the mean and standard deviation.

Mean = 7.4 Standard deviation = 0.06

What happens to the mean and standard deviation of the distribution of sample means as the size of the sample increases? Answer: The mean stays the same, the but standard deviation decreases.

Assume a member is selected at random from the population represented by the graph. Find the probability that the member selected at random is from the shaded area of the graph.

Answer = 0.3038

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superwisebouquetblazeposts-blog · 10 years ago

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DeVry MATH 221 Week 7 DQ Rejection Region - Latest

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MATH 221 Week 7 DQ Rejection Region

How is the rejection region defined and how is that related to the z-score and the p value? When do you reject or fail to reject the null hypothesis? Why do you think statisticians are asked to complete hypothesis testing? Can you think of examples in courts, in medicine, or in your area?

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gosuperbcollectionwizarduni-blog · 10 years ago

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MATH 221 Week 7 DQ Rejection Region - Latest

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MATH 221 Week 7 DQ Rejection Region

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yourmaximumcollectiondestin-blog · 10 years ago

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MKT 421 (Marketing Plan) Complete Course - New

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MKT 421 (Marketing Plan) Entire Course - New

MKT 421 Week 1 Individual Favorite Brand Paper

MKT 421 Week 1 DQ 1

MKT 421 Week 1 DQ 2

MKT 421 Week 1 DQ 3

MKT 421 Week 2 Individual Marketing Mix Presentation

MKT 421 Week 2 Individual Personal Branding Plan Paper

MKT 421 Week 2 DQ 1

MKT 421 Week 2 DQ 2

MKT 421 Week 2 DQ 3

MKT 421Week 3 Individual Perpetual Map Presentation

MKT 421 Week 3 Learning Team Strategy and Positioning Paper

MKT 421 Week 3 DQ 1

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MKT 421 Week 4 Individual Blue Ocean Strategy Paper

MKT 421 Week 4 Learning Team - Product, Pricing, and Channels Paper Paper

MKT 421 Week 4 DQ 1

MKT 421 Week 4 DQ 2

MKT 421 Week 4 DQ 3

MKT 421 Week 5 Individual - Map the Supply Chain Paper

MKT 421 Week 5 Learning Team Final Marketing Plan Paper

MKT 421 Week 5 Learning Team Final Marketing Plan Presentation

MKT 421 Week 5 DQ 1

MKT 421 Week 5 DQ 2

MKT 421 Week 5 DQ 3

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screechingcollectiontragedyblog · 9 years ago

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DeVry MATH 221 Week 7 DQ Rejection Region – Latest

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MATH 221 Week 7 DQ Rejection Region

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omglazycoffeecollection-blog · 10 years ago

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MKT 421 week 3 Learning Team Strategy and Positioning Paper - NEW

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MKT 421 week 3 Learning Team Strategy and Positioning Paper - NEW

Select a new product or service that will be launched by either an existing organization or one you will create. This product or service will serve as the basis for the Marketing Plan you will write throughout the course.

Obtain your instructor's approval of your product or service before beginning this project.

Write a 2,800- to 3,500-word paper that includes the following:

An overview of the organization

A description of the product or service

A SWOT analysis of the organization and offering

A competitive analysis of the organization and offering using Porter's five competitive forces model

The criteria you will use to segment your market and select your target market including geographic, demographic, psychographic, and behavioral factors

A description of your target market

Needs that cause your target market to buy including emotional and logical drivers

A written positioning statement that identifies:

Your target market

The needs that drive purchase

Your organization's industry category

How your organization solves the target's needs

Your organization's competition

What makes your organization different from its competition

Include sources of secondary research that support your analysis.

Format your paper consistent with APA guidelines.

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screechingcollectiontragedyblog · 9 years ago

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DeVry MATH 221 Week 6 DQ Confidence Interval Concepts – Latest

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MATH 221 Week 6 DQ Confidence Interval Concepts

Consider the formula used for any confidence interval and the elements included in that formula. What happens to the confidence interval if you (a) increase the confidence level, (b) increase the sample size, or (c) increase the margin of error? Only consider one of these changes at a time. Explain your answer with words and by referencing the formula.

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screechingcollectiontragedyblog · 9 years ago

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DeVry MATH 221 Week 5 DQs Interpreting Normal Distributions – Latest

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MATH 221 Week 5 DQs Interpreting Normal Distributions – 2015

Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?

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screechingcollectiontragedyblog · 9 years ago

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DeVry MATH 221 Week 4 DQ Discrete Probability Variables – latest

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MATH 221 Week 4 DQ Discrete Probability Variables

What are examples of variables that follow a binomial probability distribution? What are examples of variables that follow a Poisson distribution? When might you use a geometric probability?

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DeVry MATH 221 Week 3 quiz with Formulas – Latest

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MATH 221 Week 3 Quiz – 2015

Use the Venn diagram to identify the population and the sample.

Choose the correct description of the population.

The number of home owners in the state

The income of home owners in the state who own a car

The income of home owners in the state

The number of home owners in the state who own a car

Choose the correct description of the sample

The income of home owners in the state who own a car

The income of home owners in the state

The number of home owners in the state who own a car

The number of home owners in the state

Hint: Population is the entire group about which information is needed.

A sample is a smaller group of members of a population selected to represent the population.

Determine whether the variable is qualitative or quantitative.

Favorite sport

Is the variable qualitative or quantitative?

Qualitative

Quantitative

Hint: Since the favorite sport cannot be expressed in numerical terms, it is qualitative.

Students in an experimental psychology class did research on depression as a sign of stress. A test was administered to a sample of 30 students. The scores are shown below.

43 50 10 91 76 35 64 36 42 72 53 62 35 74 50

72 36 28 38 61 48 63 35 41 22 36 50 46 85 13

To find the 10% trimmed mean of a data set, order the data, delete the lowest 10% of the entries and highest 10% of the entries, and find the mean of the remaining entries. Complete parts (a) through (c).

Find the 10% trimmed mean for the data.

The 10% trimmed mean is 48.8.(Round to the nearest tenth as needed.)

Hint: Please see the excel spreadsheet.

Compare the four measures of central tendency, including the midrange.

Mean = 48.9 (Round to the nearest tenth as needed.)

Median = 47

Mode = 35, 36, 50 (Use a comma to separate answers as needed.)

The midrange is found by the following formula.

(Maximum data entry) + (Minimum data entry)

Midrange = 50.5 (Round to the nearest tenth as needed.)

Hint:

Mean = = 48.9

Median = item when the items are arranged in ascending or descending order of magnitude.

= item = 15.5th item

= (46 + 48)/2

= 47

Mode = most frequently occurring item

Here the observations 35, 36 and 50 occur 3 times and hence all these are modes.

Midrange = (Maximum data entry) + (Minimum data entry)

= (10 + 91)/2 = 50.5

It simply decreases the number of computations in finding the mean.

It permits the comparison of the measures of central tendency.

It permits finding the mean of a data set more exactly.

It eliminates potential outliers that could affect the mean of the entries.

Construct a frequency distribution for the given data set using 6 classes. In the table, include the midpoints, relative frequencies, and cumulative frequencies. Which class has the greatest frequency and which has the least frequency?

Amount (in dollars) spent on books for a semester

457 146 287 535 442 543 46 405 496 385 517 56 33 132 64

99 378 145 30 419 336 228 376 227 262 340 172 116 285

Complete the table, starting with the lowest class limit. Use the minimum data entry as the lower limit of the first class. (Type integers or decimals rounded to the nearest thousandth as needed.)

Which class has the greatest frequency?

The class with the greatest frequency is from 374 to 459.

Which class has the least frequency?

The class with the least frequency is from 288 to 373.

Hint:

Minimum value = 30

Maximum value = 543

Number of classes required = 6

Now (543 – 30)/6 = 85.5

Hence the upper limit of the first class is 30 + 85 = 115

2nd class lower limit = 116

2nd class upper limit = 116 + 85 = 201

3rd class lower limit = 202

3rd class upper limit = 202 + 85 = 287

4th class lower limit = 288

4th class upper limit = 288 + 85 = 373

5th class lower limit = 374

5th class upper limit = 374 + 85 = 459

6th class lower limit = 460

6th class upper limit = 460 + 85 = 545

For the calculations of frequency, midpoint, etc. please see the excel spreadsheet.

Identify the data set’s level of measurement.

The nationalities listed in a recent survey (for example, American, German, or Brazilian)

Nominal

Ordinal

Interval

Ratio

Hint: Since the nationalities do not have any numerical value or order, it is nominal.

Explain the relationship between variance and standard deviation. Can either of these measures be negative?

Choose the correct answer below.

The standard deviation is the negative square root of the variance. The standard deviation can be negative but the variance can never be negative.

The standard deviation is the positive square root of the variance. The standard deviation and variance can never be negative. Squared deviations can never be negative.

The variance is the negative square root of the standard deviation. The variance can be negative but the standard deviation can never be negative.

The variance is the positive square root of the standard deviation. The standard deviation and variance can never be negative. Squared deviations can never be negative.

Hint:

Variance =

Standard deviation =

That is, standard deviation is the positive square root of variance.

For the following data (a) display the data in a scatter plot, (b) calculate the correlation coefficient r, and (c) make a conclusion about the type of correlation.

The number of hours 6 students watched television during the weekend and the scores of each student who took a test the following Monday.

Hours spent watching TV, x 0 1 2 3 3 5

Test score, y 98 90 84 74 93 65

Choose the correct scatter plot below.

The correlation coefficient r is -0.844 (Round to three decimal places as needed)

Which of the following best describes the type of correlation that exists between number of hours spent watching television and test scores?

Strong negative linear correlation

No linear correlation

Weak negative linear correlation

Strong positive linear correlation

Weak positive linear correlation

Hint:

For scatter plot and calculation of correlation coefficient, please see the excel spreadsheet.

Since the absolute value of correlation coefficient close to 1, it is strong. Since the correlation coefficient is negative, there is negative correlation. Hence there is a strong negative linear correlation exists between the variables.

Suppose a survey of 526 women in the United States found that more than 70% are the primary investor in their household. Which part of the survey represents the descriptive branch of statistics?

Choose the best statement of the descriptive statistic in the problem.

There is an association between the 526 women and being the primary investor in their household.

526 women were surveyed.

70% of women in the sample are the primary investor in their household.

There is an association between U.S. women and being the primary investor in their household.

Choose the best inference from the given information.

There is an association between the 526 women and being the primary investor in their household.

There is an association between U.S. women and being the primary investor in their household.

70% of women in the sample are the primary investor in their household

526 women were surveyed.

Hint:

Since 70% of women in the sample are the primary investor in their household, we can infer that there is an association between U.S. women and being the primary investor in their household.

Identify the sampling technique used.

A community college student interviews everyone in a particular statistics class to determine the percentage of students that own a car.

Radom

Cluster

Convenience

Stratified

Systematic

Hint: A convenience sample is made up of people who are easy to reach. Here the community college student interviews everyone in a particular statistics class which is easy for him. Hence it is a convenient sampling.

Use the frequency polygon to identify the class with the greatest, and the class with the least frequency.

What are the boundaries of the class with the greatest frequency?

5-30.5

25-31

5-29.5

28-31

What are the boundaries of the class with the least frequency?

10-13

5-11.5

7-13

5-12.5

Hint:

Clearly the greatest frequency is 14 and is in the boundary 26.5-29.5.

Least frequency is 1 and is in the boundary 8.5-11.5.

Determine whether the given value is a statistic or a parameter

In a study of all 2377 students at a college, it is found that 35% own a computer

Choose the correct statement below.

Parameter because the value is a numerical measurement describing a characteristic of a population.

Statistic because the value is a numerical measurement describing a characteristic of a population.

Statistic because the value is a numerical measurement describing a characteristic of a sample.

Parameter because the value is a numerical measurement describing a characteristic of a sample.

Compare the three data sets

Which data set has the greatest sample standard deviation?

Data set (iii), because it has more entries that are farther away from the mean

Data set (ii), because it has two entries that are far away from the mean.

Data set (i), because it has more entries that are close to the mean.

Which data set has the least sample standard deviation?

Data set (i), because it has more entries that are close to the mean.

Data set (ii), because it has less entries that are farther away from the mean.

Data set (iii), because it has more entries that are farther away from the mean.

How are the data sets the same? How do they differ?

The three data sets have the same standard deviations but have different means.

The three data sets have the same mean, median and mode but have different standard deviation.

The three data sets have the same mean and mode but have different medians and standard deviations.

The three data sets have the same mode but have different standard deviations and means

Decide which method of data collection you would use to collect data for the study.

A study of the effect on the human digestive system of a popular soda made with a caffeine substitute.

Choose the correct answer below.

Observational Study

Simulation

Survey

Experiment

Use the given frequency distribution to find the:

Class width

Class midpoint of the first class

Class boundaries of the first class

(a) 4 (b) 137.5 (c) 134.5-139.5

(a) 5 (b) 137 (c) 135-139

(a) 5 (b) 137 (c) 134.5-139.5

(a) 4 (b) 137.5 (c) 135-139

Hint:

Class width = 139.5 – 134.5 = 5

Class midpoint of the first class = (135 + 139)/2 = 137

Class boundaries of the first class = 134.5-139.5

Consider the following sample data values.

5 14 15 21 16 13 9 19

(a) Calculate the range

(b) Calculate the variance

The range is 16. (Type an integer or a decimal)

The sample variance is 57. (Type an integer or decimal rounded to two decimal places as needed)

The sample standard deviation is 15. (Type an integer or decimal rounded to two decimal places as needed)

Hint:

X – X̅

(X – X̅)^2

-9

-1

-5

112

186

Range = Highest observed value – Lowest observed value

= 21 – 5

= 16

Variance = = 26.57

Standard deviation = = 5.15

Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line. (the pair of variables have a significant correlation.) Then use the regression equation to predict the value of y for each of the given x-values, if meaningful. The number of hours 6 students spent for a test and their scores on that test are shown below.

Find the regression equation.

Y = 6.643 x + (31.238) (Round to three decimal places as needed)

Choose the correct graph below.

Predict the value of y for x = 4. Choose the correct answer below.

Not meaningful

Predict the value of y for x = 4.5. Choose the correct answer below.

Not meaningful

Predict �� the value of y for x = 2.5. Choose the correct answer below.

57.8

47.8

111.0

Not meaningful

Hint:

Scatter plot is given by,

Regression equation is given by,

Y = 6.643 x + 31.238 (Please see the excel sheet for workings)

(a) When x = 4,

Y = (6.643 * 4) + 31.238 = 57.8

(b) When x = 4.5,

Y = (6.643 * 4.5) + 31.238 = 61.1

Clearly x = 12 is outside the range of values of independent variable. Hence not meaningful.

(d) When x = 2.5,

Y = (6.643 * 2.5) + 31.238 = 47.8

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