GEN305 Mathematical Concepts And Statistical Applications

Question:

Basic Computation – Confidence Interval for P Consider n = 100 Binomial Trials with r = 30, successes

a)Check Requirements Should a normal distribution be used to approximate p?

b.Find a 90% confidence interval to determine the percentage of successes in the population.

c.Interpretation Explain what the confidence interval that you calculated means to you.

Basic Computation: Confidence Interval to p Take into account n = 200 Binomial Trials with r =80 successes

a)Check Requirements Should a normal distribution be used to approximate p?

b.Find a 95% confidence interval to determine the population’s success rate p.

c.Interpretation Explain what the confidence interval that you calculated means to you.

Basic Computation: How large is the minimum sample size required to give a 95% confidence interval a maximum margin error of 0.1?

If the preliminary estimate of p is 0.25

If there is no preliminary estimate for the value of p?, then b.

Basic Computation: How large is the minimum sample size required to achieve a 99% confidence interval with a maximum margin of error of 0.06?

If the preliminary estimate for p = 0.8, what is a?

If there is no preliminary estimate for the value of p?, then b.

Myers-Briggs – Actor Isabel Briggs Myers was an early pioneer in the study and application of personality types.

The following information is adapted from MBTI Manual: A Guide for the Development and Use Of the Myers-Briggs Type Indiator by Myers & McCaulley (Consulting Psychologists Press).

A random sample of 62 actors was used to determine that 39 extroverts were present.

Let p stand for the percentage of actors who exhibit personality.

Get a point estimate of p.

b. Find a 95% confidence interval to p. Give a brief description of the confidence interval that you have discovered.

c. Check Requirements. Do you believe that the conditions of np >5 and nq>5 have been met for this problem?

Explain why this would be important.

Trick or treating was discovered in a survey of 35 Cherry Creek households. It was found that 11 households lit up the lights and made pretend they were not home on Halloween.

a)Calculate a confidence interval of 90% for p, the percentage of Cherry Creek houses that pretend they are not home on Halloween.

b. What assumptions are necessary for the calculation of the confidence interval for part a?

c.Interpretation The national percentage of American households that go out on Halloween and pretend they are not home is 0.28.

Is 0.28 the confidence interval that you calculated?

Your answer suggests that Cherry Creek is different from all U.S. neighborhoods in terms of the percentage of residents living there.

Marketing: Customer loyalty

a) Let p stand for the percentage of women shoppers who are loyal to their supermarket.

Here’s a point estimate for p.

b. Find a 95% confidence interval to calculate p. Provide a brief explanation about the meaning of this interval.

c.Interpretation What would your reporting style be if you were a journalist and wanted to report on the results from a survey about the percentage of female supermarket shoppers who have remained loyal over the past 12 months?

Is there a margin of error for a 95% confidence range?

Answer:

Basic Computation: Confidence Interval to pConsider that n = 100 binomial trials have r = 30, with success.

Review Requirements

The normal distribution is not appropriate in the case of the given experiment, which has a result that can be called success or failure.

The confidence interval for success in the population is 90%

Given that there are n=100 Binomial Trials and r=30 Success;

The 90 % confidence interval (for p) is given by;

+ E

Where is p

E =? and = = = 0.33

[ ] and E = 0.3.

is the critical value to determine the confidence interval c in the normal standard distribution. u is the average of the total population under consideration.

E =?

The confidence interval found proves that deed success percentage is 0.375. This gives us a good estimate of this proportion.

Basic Computation: Confidence Interval to pConsider that n = 200 binomial trials are possible with r = 80 success.

The normal distribution is not appropriate in the case of the given experiment, which has a result that can be called success or failure.

A 95% confidence interval is needed for the population’s success rate p.

Given that n=200 Binomial Trials were performed and r=80 success rates;

The 95 percent confidence interval for p can be found at: p

+ E

Where is p

= = = 0.8, and E =?

[ ] and E = 0.8.

is the critical value to determine the confidence interval c in the normal standard distribution. u is the average of the total population under consideration.

E =?

The confidence interval found proves that deed success percentage is 0.8. This gives a good estimate for this proportion.

Basic Computation: Sample size

A preliminary estimate for pis 0.25

To have 95 % confidence in the value of, we need to know that?

Which is the critical value for C under the standard normal distribution? It is 1.96. Estimate of p= 0.25, error margin of 0.0.1; We get the sample size n

Sample size n to get a preliminary estimate for 0.25 = = 72

If there is not a preliminary estimate for p

With a 95% confidence, the value?

This is the critical value for C under the standard normal distribution. It is 1.96. For no preliminary estimate and error margins of 0.1, we obtain the sample size: n

Assume equal probability in the experiment, n is the sample size for a preliminary estimation. Therefore, p=0.5

Thus, the sample size would be == 96

Basic Computation: Sample size

A preliminary estimate for pis 0.8?

With a confidence of 99%, the value?

This is the critical value of C under the standard normal distribution. It is 2.58, an estimate of P= 0.8, and an error margin de 0.1. The sample size is n.

Sample size n to get a preliminary estimate for 0.8 = =107

If there is not a preliminary estimate for p

With a confidence of 99%, the value?

This is the critical value of C under the normal distribution. It is 2.58, an estimate of P= 0.8, and an error margin of 0.0.1. The sample size is n.

Sample size n, for no preliminary estimate. The experiment assumes a double chance. Therefore, the value of the n becomes = =166

Myers-Briggs: ActorsIsabel Briggs Myers was an early pioneer in the study and application of personality types.

The following information is adapted from MBTI Manual: A Guide for the Development and Use Of the Myers-Briggs Type Indiatorby McCauley (Consulting Psychologists Press).

A random sample of 62 actors was used to determine that 39 extroverts were present.

Let us prepresent the extrovert proportion of all actors.

Get a point estimate of p.

Sample size = 62.

The proportion p (extroverts). ==0.6290. Q=0.371

Point estimation for the p is equal to the extravert proportion, thus point estimate 39/62 = 0,371.

For p, find a 95% confidence interval. Give a brief description of the meaning of your confidence interval.

The 95 percent confidence interval for p can be found at: p

+ E

Where is p

= = = 0.62290 and E =?

[ ] and E = 0.6290.

is the critical value to determine the confidence interval c in the normal standard distribution. u is the average of the total population under consideration.

E =?

The confidence interval shown above includes the point estimate, therefore the proportion point estimate is accurate.

Check RequirementsDoes this problem satisfy the conditions np>5 and nq>5?

Explain why you think this is important.

To ensure that the conditions np>5 & nq>5 are satisfied, it is important to have a large number of trials.

The problem can be solved by r=39, which is a high value and fulfills both conditions.

Trick or TreatIn a survey that included 35 households from the Cherry Creek neighborhood in Denver, 11 households lit up the lights and claimed not to be there on Halloween.

For p, calculate a 90% confidence interval. This is the percentage of Cherry Creek homes that pretend to not be home for Halloween.

Given that n= 35 is greater than r= 11, the confidence interval of p at a 90 % level for p can be calculated by: p

+ E

Where is p

= = = 0.3143, and E =?

[ ] and E =????.

is the critical value to determine the confidence interval c in the normal standard distribution. u is the average of the total population under consideration.

E =?

What assumptions must be made to calculate the confidence range of part (a).

When calculating the confidence interval, you can make these assumptions:

This sample data was taken from a very small population.

The population distribution was roughly normal.

Interpretation: 0.28 is the national proportion of all American households that pretend they are not at home on Halloween.

Is that 0.28 in your confidence interval?

Your answer suggests that Cherry Creek is different from all U.S. neighborhoods in terms of the percentage of residents living there.

The confidence interval was calculated to give 0.28.

It is clear that Cherry Creek’s population is significantly different than the entire United States.

This is because 0.28 does not represent a higher or lower percentage of all households in the United States.

Marketing: Customer loyalty

Let’s consider the proportion of all women shoppers who are loyal to their supermarket.

Get a point estimate of p.

The P estimate is p.

which is equal = 628/730= 0.8602.

Provide a short explanation of the 95% confidence interval (for p)

The 95 percent confidence interval for p can be found at: p

+ E

Where is p

= = = 0.8602; E =?

[ ] and E =?

is the critical value to determine the confidence interval c in the normal standard distribution. u is the average of the total population under consideration.

E =?

The confidence interval as calculated above shows that the point estimator includes and falls within the confidence range. This can be used to conclude that customers have remained loyal.

Interpretation

Is there a margin of error for a 95% confidence range?

According to a market survey, 86% of the customers remained loyal to the supermarket over the past year.

The error term margin = +- [(0.8853)/ (0.8602)]/ [2]

The error margin = +- (0.5545).

Refer to

Statistics Science.

Cambridge University Press.

Illustration of the Binomial distribution’s Confidence Interval and fiducially limit usage

A., Pavlov, V.I (1999).

Statistical Reliability Engineering.

Wiley and Sons.

Clinical Trials Data and Safety Monitoring Committees.

CRC Press.

Introduction to Statistical Quality Control.

Wiley and Sons.

Concerning the Problem of Confidence intervals.

The Annals of Mathematical Statistics.

Wiley and Sons.

Radolsky L. (2014) Probability-Based Structural Fire Load.

Cambridge University Press.

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