Basic Computation And Myers-Briggs Self-Assessment Study Results Essay.

Confidence Interval for p

1. Basic Computation: Confidence Interval for p Consider n = 100 binomial trials with r = 30 successes.
a)Check Requirements Is it appropriate to use a normal distribution to approximate the p? distribution?
b)Find a 90% confidence interval for the population proportion of successes p.
c)Interpretation Explain the meaning of the confidence interval you computed.
2. Basic Computation: Confidence Interval for p Consider n = 200 binomial trials with r = 80 successes.
a)Check Requirements Is it appropriate to use a normal distribution to approximate the p? distribution?
b)Find a 95% confidence interval for the population proportion of successes p.
c)Interpretation Explain the meaning of the confidence interval you computed.
3. Basic Computation: Sample Size What is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1:
a)if a preliminary estimate for p is 0.25?
b)if there is no preliminary estimate for p?
4. Basic Computation: Sample Size What is the minimal sample size needed for a 99% confidence interval to have a maximal margin of error of 0.06:
a)if a preliminary estimate for p is 0.8?
b)if there is no preliminary estimate for p?
5. Myers-Briggs: Actors Isabel Briggs Myers was a pioneer in the study of personality types. The following information is taken from MBTI Manual: A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts.
a)Let p represent the proportion of all actors who are extroverts. Find a point estimate for p.
b)Find a 95% confidence interval for p. Give a brief interpretation of the meaning of the confidence interval you have found.
c)Check Requirements Do you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration.
6. Trick or Treat In a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 11 households turned out the lights and pretended not to be home on Halloween.
a)Compute a 90% confidence interval for p, the proportion of all housholds in Cherry Creek that pretend not to be home on Halloween.
b)What assumptions are necessary to calculate the confidence interval of part (a)?
c)Interpretation The national proportion of all households in the United States that turn out the lights and pretend not to be home on Halloween is 0.28. Is 0.28 in the confidence interval you computed? Based on your answer, does it seem that the Cherry Creek neighborhood is much different (either a higher or a lower proportion) from the population of all U.S. households? Explain.
7. Marketing: Customer Loyalty In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute).
a)Let p represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for p.
b)Find a 95% confidence interval for p. Give a brief explanation of the meaning of the interval.
c)Interpretation As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a 95% confidence interval? 

1. Basic Computation: Confidence Interval for pConsider n = 100 binomial trials with r = 30 successes.
2. Check RequirementsIs it appropriate to use a normal distribution to approximate the p? distribution?

For case given, it would not be appropriate to use the normal distribution since the result or outcome of the experiment is specified to be either success or failure.

1. Find a 90% confidence interval for the population proportion of successes p.

Given that n=100 binomial trials, and r=30 success;

The 90 % confidence interval for p is given by; p? - E < µ < p? + E

Where p? =   =  = 0.3 and E = ? [√ ] but q= (1-0.3) = 0.7 and ? is the critical value for the confidence interval c under the normal standard distribution and µ is the mean of the total population under consideration.

E = ? (√) = 1.645 [√] = 0.075

0.3 – 0.075 < µ <0.3 + 0.075

0.225 < µ < 0.375

1. InterpretationExplain the meaning of the confidence interval you computed.

The confidence interval obtained proves that in deed the proportion of success is 0.375 and thus a good estimate for the proportion.

2. Basic Computation: Confidence Interval for pConsider n = 200 binomial trials with r = 80 successes.

For case given, it would not be appropriate to use the normal distribution since the result or outcome of the experiment is specified to be either success or failure.

1. Find a 95% confidence interval for the population proportion of successes p.

Given that n=200 binomial trials, and r=80 success;

The 95 % confidence interval for p is given by; p? - E < µ < p? + E

Where p? =   =  = 0.8 and E = ? [√ ] but q= (1-0.8) = 0.2 and ? is the critical value for the confidence interval c under the normal standard distribution and µ is the mean of the total population under consideration.

E = ? (√) = 1.96 [√] = 0.055

0.8 – 0.055 < µ <0.8+ 0.055

0.745 < µ < 0.885

1. InterpretationExplain the meaning of the confidence interval you computed.

The confidence interval obtained proves that in deed the proportion of success is 0.8 and thus a good estimate for the proportion.

3. Basic Computation: Sample Sizewhat is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1:
4. If a preliminary estimate for pis 0.25?

For a 95 % confidence, the value of ? which is the critical value for c under the standard normal distribution is 1.96, estimate of p= 0.25 and error margin of 0.1; we obtain the sample size n as;

Sample size n for a preliminary estimate of 0.25 =  = 72

1. If there is no preliminary estimate for p?

For a 95 % confidence, the value of ? which is the critical value for c under the standard normal distribution is 1.96, for no preliminary estimate and error margin of 0.1; we obtain the sample size n as;

Sample size n for a preliminary estimate assume equal probability in the experiment and hence the value of p= 0.5.

Hence the sample size n would be =  = 96

4. Basic Computation: Sample Sizewhat is the minimal sample size needed for a 99% confidence interval to have a maximal margin of error of 0.06:
5. If a preliminary estimate for pis 0.8?

For a 99% confidence, the value of ? which is the critical value for c under the standard normal distribution is 2.58, estimate of p= 0.8 and error margin of 0.1; we obtain the sample size n as;

Sample size n for a preliminary estimate of 0.8 =  = 107

1. If there is no preliminary estimate for p?
2. For a 99% confidence, the value of ? which is the critical value for c under the standard normal distribution is 2.58, estimate of p= 0.8 and error margin of 0.1; we obtain the sample size n as;
3. Sample size n for no preliminary estimate, the experiment assumes double chance and thus the value of n becomes  = = 166
4. Myers-Briggs: ActorsIsabel Briggs Myers was a pioneer in the study of personality types. The following information is taken from MBTI Manual: A Guide to the Development and Use of the Myers-Briggs Type Indicatorby Myers and McCauley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts.
5. Let prepresent the proportion of all actors who are extroverts. Find a point estimate for p.

Sample size = 62. The proportion p of the (extroverts) =  = 0.6290 and q= 0.371

Point estimate for p is equivalent to the proportion of p  of the extraverts thus point estimate 39/62 = 0.371

1. Find a 95% confidence interval for p. Give a brief interpretation of the meaning of the confidence interval you have found.

The 95 % confidence interval for p is given by; p? - E < µ < p? + E

Check Requirements

Where p? =   =  = 0.6290 and E = ? [√ ] but q= (1- 0.6290) = 0.3709 and ? is the critical value for the confidence interval c under the normal standard distribution and µ is the mean of the total population under consideration.

E = ? {√} = 1.96 [√] = 0.12024

0.6290 – 0.1202 < µ <0.6290+ 0.1202

0.5088< µ < 0.7492

The confidence interval above is inclusive of the point estimate thus the point estimate for the proportion is accurate.

1. Check RequirementsDo you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration.

For the conditions np>5 and nq> 5 to be sufficient, the number of trials r should be large. In case, r= 39 which is relatively large value hence the two conditions are satisfied in the problem.

6. Trick or TreatIn a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 11 households turned out the lights and pretended not to be home on Halloween.
7. Compute a 90% confidence interval for p, the proportion of all households in Cherry Creek that pretend not to be home on Halloween.

Given that n= 35 and r= 11, 90 % confidence interval for p is given by; p? - E < µ < p? + E

Where p? =   =  = 0.3143 and E = ? [√ ] but q= (1- 0.3142) = 0.6857 and ? is the critical value for the confidence interval c under the normal standard distribution and µ is the mean of the total population under consideration.

E = ? {√} = 1.64 [√] = 0.1287

0.3143 – 0.1287 < µ < 0.3143 + 0.1287

0.1856< µ < 0.443

1. What assumptions are necessary to calculate the confidence interval of part (a)?

The following assumptions can be made when calculating the confidence interval;

1. The sample data is was obtain from a small populations
2. The population distribution was approximately normal.
3. InterpretationThe national proportion of all households in the United States that turn out the lights and pretend not to be home on Halloween is 0.28. Is 0.28 in the confidence interval you computed? Based on your answer, does it seem that the Cherry Creek neighborhood is much different (either a higher or a lower proportion) from the population of all U.S. households? Explain.

The value 0.28 is in the confidence interval computed. From this it seems that the Cherry Creek neighborhood is much different form population of all United States. This is because the value 0.28 is neither a higher nor a lower proportion from the proportion of all the United States households.

7. Marketing: Customer LoyaltyIn a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute).
8. Let prepresent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for p.

The estimate for P is p? which is equal = 628/730 = 0.8602.

1. Find a 95% confidence interval for p. Give a brief explanation of the meaning of the interval. The 95 % confidence interval for p isgiven by; p? - E < µ < p? + E

Where p? =   =  = 0.8602 and E = ? [√ ] but q= (1- 0.8602) = 0.1397 and ? is the critical value for the confidence interval c under the normal standard distribution and µ is the mean of the total population under consideration.

E = ? [√ ] = 1.96 [√] = 0.02515

0.8602 – 0.02515 < µ < 0.8602 + 0.02515

0.8351< µ < 0.8853

Based on the confidence interval calculated above, it can be concluded that the point estimator is inclusive and falls within the confidence interval and hence it can as well be asserted that the customers remained loyal to the supermarket.

1. InterpretationAs a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a 95% confidence interval?

According to the market survey conducted on a particular supermarket, about 86% of the customers are found to remain loyal to the supermarket during the past years.

The margin of the error term= ± [(0.8853) ÷ (0.8602)] ÷ [2]

The error margin= ± (0.5145).

References

Brown, L. D., & Cai, T T., & DasGupta, A. (2001). Interval Estimation for a Binomial Proportion. Statistics Science. Cambridge University Press.

Clopper, C., & Pearson, S. (2012). The use of Confidence Interval or fiducially limits illustrated in the case of the Binomial distribution. Biometrika

Gnedenko, B. V., & Ushakov, I. A., & Pavlov, V. I (1999). Statistical Reliability Engineering. Wiley and Sons.

Herson, J. (2009). Data and Safety Monitoring Committees in Clinical Trials. CRC Press.

Montgrometry, D. (2014). Introduction to Statistical Quality Control. 4^th Editions. Wiley and Sons.

Neyman, J. On the Problem of Confidence intervals. The Annals of Mathematical Statistics. Wiley and Sons.

Radolsky, L. (2014) Probability –Based Structural Fire Load. Cambridge University Press.