Calculations to Write Four Essays

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Question Description

Use the given calculations to write 4 new essays. 1 page for each.

Please note that they are 4 individual essay, you need to finish them one by one using the given calculation on each.

And also you need to submit 4 separate files to me in order.

You cant put all 4 answers into one file.

All the work must be original

Turnitin report is required

The calculation is provided below, please write a new essay using this calculation. Which means except the calculation, every sentences else needs to be writing on your own words. Essay 1. Use the following background information to complete the essay. Case Study Scenario Land and Agua Insurance Company has a call center in Tempe, Arizona. The business was originally established in Phoenix, Arizona in 1972 as a small business, and it has grown with the population of the area. The insurance company specia lizes in bundling insurance for cars, off-road vehicles, and watercraft (e.g., jet skis and boats). The company has 150,000 clients in Arizona. Marjorie Jones, Vice President of Operations, is concerned about customer complaints and the amount of time representatives are taking to resolve the calls. She has established a team to investigate the call center data and identify areas where the team should focus on making improvements. 2. Answer the questions below in essay format. Your essay must include an introd uction, a body, and a conclusion. It must address all relevant parts of each question. Your response should be a minimum of 500 words in length, and it should include your analysis of the probability calculations. Make sure to cite any source you use. Prop er citation format for a source includes the name of the author(s), the title of the work, the date of the publication, and the page number if you directly quote the source. Essay: Call Errors Using the Quality Summary and Call Center Data, provide a summary report for the vice president including the following information in an essay with a minimum of 500 words: 1. List the relative frequency for overall type of calls, call quality, and call errors. 2. Provide descriptive statistics for call time to include the mean, median, mode, variance, standard deviation, and range. 3. Provide the following probabilities: 1. CLM Error and AM Shift 2. COV Error and PM Shift 3. SAV Error or AM Shift 4. Given that the call comes in the morning, what is the probability of a CLM Error? 4. Based on the data, which error(s) should the team focus on? Does shift matter? Explain your answers. 5. Evaluate the current call time if customers expect to have their calls handled within approximately 7.5 minutes on the phone. Answer Associations in the present business world face a lot of vulnerability when they are deciding. Likelihood gives an important device to measure this vulnerability so directors can settle on better choices (Donnelly, 2015). There are two ideas we have to ace, one is the meaning of subjective likelihood - depends on the experience of the specialist and instinct to appraise the probabilities (Donnelly, 2015), and the other one is the meaning of Conditional likelihood - the likelihood of Event A happening, knowing, or given the condition that Event B has happened (Donnelly, 2015). The equation of the added substance run is P (A or B) = P (A) +P (B) - P (A+B). The equation of restrictive likelihood is P (A|B) =P (A and B)/P (B). Base on the informational collection, the aggregate number of value control set is 1000, so the relative recurrence is 1. The recurrence of right call quality is 850, at that point we have the relative recurrence 850/1000=0.850. The recurrence of inaccurate call quality is 150, at that point the relative recurrence is 150/1000=0.15. The aggregate number of general approaching calls by type is 95669, so the relative recurrence is 1. The recurrence of Check scope of strategy is 25625, the relative recurrence is 25625/95669=0.2679. The recurrence of Check status of claim is 22654, the relative recurrence is 22654/95669 = 0.2368. The recurrence of Update address is 18723, the relative recurrence is 18723/95669 = 0.1957. The recurrence of File a claim 13499, the relative recurrence is 13499/95669 = 0.1411. The recurrence of Update Information on assert 6200, the relative recurrence is 6200/95669 = 0.0648. The recurrence of Update Policy 5126, the relative recurrence is 5126/95669 = 0.0536. The recurrence of Cancel Policy 3842, the relative recurrence is 3842/95669 = 0.0402. The aggregate number of call blunders is 150, so the relative recurrence is 1. The recurrence of Incorrect scope quote (COV) is 54, the relative recurrence is 54/150 = 0.36. The recurrence of Incorrect catch of claim (CLM) is 42, the relative recurrence is 42/150 = 0.28. The recurrence of did not exchange to "Spare a Policy" (SAV) is 32, the relative recurrence is 32/150 = 0.21. The recurrence of Incorrect claim status gave (STAT) is 22, the relative recurrence is 22/150 = 0.15. The mean, middle, mode, difference, standard deviation, and range can without much of a stretch measure by figuring in exceed expectations. The mean is 11.07, the middle is 10.63, the mode is 12, the difference is 11.37, the standard deviation is 3.38, and the range is 14.25. The condition of computing the likelihood of CLM Error and AM Shift is P= (probability of CLM Error and AM)/ (likelihood of AM) =16/75=0.213. The condition of figuring the likelihood of COV Error and PM Shift is P= (likelihood of COV Error and PM)/ (likelihood of PM) =21/75=0.28. The condition of computing the likelihood of SAV Error or AM Shift is P= (likelihood of SAV) + (likelihood of AM) - (likelihood of SAV Error and AM)=0.21+0.519/150=0.21+0.5-0.127=0.583. The condition of ascertaining the likelihood of a CLM Error is P= (probability of CLM blunder and AM)/ (likelihood of CLM error) = (16/150)/0.28=0.381. • Provide the following probabilities: o CLM Error and AM Shift =19/150=.1267 o COV Error and PM Shift=28/150=.1867 o SAV Error or AM Shift =(32+75-20)/150=.580 o Given that the call comes in the morning, what is the probability of a CLM Error=.1267/.5000=.2533 Accordingly, I would state we should center around the SAV Error or AM move, since they have the greatest likelihood. The move does make a difference. Reference: Donnelly, R. A. (2015). Business statistics (2nd ed.). Upper Saddle River, NJ: Pearson.
Essay 1. Use the background information below to create the essay. Case Study Scenario Land and Agua Insurance Company has a call center in Tempe, Arizona. The business was originally established in Phoenix, Arizona in 1972 as a small business, and it has grown with the population of the area. The insurance company specializes in bundling insurance for cars, off-road vehicles, and watercraft (e.g., jet skis and boats). The company has 150,000 clients in Arizona. Marjorie Jones, Vice President of Operations, is concerned about customer complaints and the amount of time representatives are taking to resolve the calls. You are part of the team investigating the data to determine the probabilities of errors and call times. Ms. Jones also wants to understand the approximate range around the average for call times. 2. Answer the questions below in essay format. Your essay must include an introduction, a body, and a conclusion. It must address all relevant parts of each question. Your response should be a minimum of 500 words in length, and it should include your analysis of the probability calculations. Make sure to cite any source you use. Proper citation format for a source includes the name of the author(s), the title of the work, the date of the publication, and the page number if you directly quote the source. Essay: Probability Using the Quality Summary and Call Center Data, provide a summary report for the vice president including the following information in an essay with a minimum of 500 words: 1. Based on the probability of an error provided in the quality summary under call quality using a sample size of 15, predict the probability of both < 2 errors or errors using the correct discrete probability distribution. Assume calls are either correct or incorrect. 2. Using the call time mean and standard deviation from the quality sample, find the probability of a call time < 7min, between 7 and 9 min, and > 9 min. 3. Calculate and evaluate the 95% confidence interval for the mean from the call time data. Answer The central limit theorem and the typicality of the disseminations are basic components for insights and utilized for the rest of the lessons. "The Central Limit Theorem expresses that the example methods for huge estimated tests will be ordinarily disseminated paying little heed to the state of their populace appropriations" (Donnelly, 2015, p. 301). For this situation, we have to know the room for mistakes and the standard blunder are two unique ideas. The room for give and take, or the width of the interim, is the basic z-score duplicated by the standard mistake for the mean. Be that as it may, a bigger example lessens the standard mistake and, along these lines, the room for give and take. Typical conveyances utilize the z-score estimation, which you found out about in Lesson 2, to recognize the likelihood. To additionally develop this idea, there are times when you should discover the probabilities that are > < or some place in the middle of two distinctive z-scores (Donnelly, 2015). In light of the likelihood of a blunder gave in the quality synopsis under call quality utilizing an example size of 15, we have n = 15 calls. The consequent phase is to determine the likelihood of <2 blunders. P (mistake) = 0.15. At that point we make utilization of binomial likelihood recipe: (x < = 2) = P(x=0) + P(x=1) + P(x=2); P(x=0) = 15C0 * 0.15^0 * 0.85^15 = 0.08735; P(x=1) = 15C1 * 0.15^1 * 0.85^14 = 0.23123; P(x=2) = 15C2 * 0.15^2 * 0.85^13 = 0.28564. In this manner, we have P(X<=2) = 0.28564 + 0.23123 + 0.08735. The last answer will be P(x < = 2) = 0.604225. The calls are perfect. Binomial Probabilities Data Sample size Probability of an event of interest 15 0.15 Statistics Mean Variance Standard deviation 2.25 1.9125 1.3829 Binomial Probabilities Table X P(X) 0.0874 0.2312 0.2856 0.2184 0.1156 0.0449 0 1 2 3 4 5 <2 0.3186 >=5 0.0617 6 7 8 9 10 11 12 13 14 15 0.0132 0.0030 0.0005 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 We have the quantities of test size, mean and standard deviation figured through exceed expectations. The example estimate is 15, Mean is 12.05, Standard Deviation is 4.502. The second inquiry is to discover P(X<7). Since (x-mean)/standard deviation=2, P(2<(7-12.05)/4.502)=P(z<-1.12)=P(z>1.12)=1-P(z>1.12). With a z appropriation table, the last answer will be P(x<7) =1-0.869=0.131. The subsequent stage is to ascertain the likelihood in the vicinity of 7 and 9. P(70.68)- 0.131=1-P(z<0.68)- 0.131=0.1169. P(x>9) =P (z> (9-12.05)/4.502) =P (z>-0.68) =P (z<0.68) =0.752. Calculate and evaluate the 95% confidence interval for the mean from the call time data. Problem uses t critical value since you do not know population standard deviation and you are using the sample standard deviation. 11.07 +/- (1.98)(3.38/sqrt(150)= 10.52, 11.61 All in all, the likelihood of either < 2 mistakes or ≥ 5 blunders is 0.604. The likelihood of a call time < 7 min is 0.131, the likelihood of a call time in the vicinity of 7 and 9 min is 0.1169, and the likelihood of a call time > 9 min is 0.752. Moreover, we should know about the meaning of the certainty interim for the mean - an interim gauge around the example imply that gives a range of where the genuine populace means falsehoods. Reference: Donnelly, R. A. (2015). Business statistics (2nd ed.). Upper Saddle River, NJ: Pearson.
The calculation is provided below, please write a new essay using this calculation. Which means except the calculation, every sentences else needs to be writing on your own words. Essay 1. Use the information below to complete the essay. Case Study Scenario Land and Agua Insurance Company has a call center in Tempe, Arizona. The business was originally established in Phoenix, Arizona in 1972 as a small business, and it has grown with the population of the area. The insurance company specializes in bu ndling insurance for cars, off-road vehicles, and watercraft (e.g., jet skis and boats). The company has 150,000 clients in Arizona. Marjorie Jones, Vice President of Operations, is concerned about customer complaints and the amount of time representatives are taking to resolve the calls. The team must focus on two areas from the preliminary data collection including call time and error type by shift. 2. Answer the questions below in essay format. Your essay must include an introduction, a body, and a conclusion. It must address all relevant parts of each question. Your response should be a minimum of 500 words in length, and it should include both the appropriate statistics and support for the analysis and interpretation. Make sure to cite any source you use. Proper citation format for a source includes the name of the author(s), the title of the work, the date of the publication, and the page number if you directly quote the source. Essay Questions: Hypothesis Tests Using the Sample Hypothesis Test Data and Chi-Square Data, provide a summary report for the vice president including the following information in an essay with a minimum of 500 words (the level of significance is .05): • Two-sample hypothesis test ‒ Discuss the hypothesis test assumptions and test you used. Provide the test statistic and critical-value in your response. Evaluate the results • of the hypothesis test with the scenario. Provide recommendations for the vice president. Chi-square hypothesis test ‒ Discuss the hypothesis test assumptions and test you used. Provide the test statistic and critical value in your response. Evaluate the results of the hypothesis test with the scenario. Provide recommendations for the vice president. Sample Essay: Introduction Comparison of two population means is very common. Hypothesis testing uses statistics to determine between hypotheses whether the data collected is statistically significant or occurred by chance. A comparison between two population samples depends on both the means and the standard deviations. Using student t distribution, two population means with unknown population standard population and independent sample student t distribution in the comparison can be used. This paper looks at two types of hypothesis tests; the two-sample hypothesis test and the chi-square data test. Two Sample Hypothesis Test According to (Donnelly, 2015) there are steps to follow in hypothesis testing: step one is stating the hypothesis to be tested; null and alternative hypothesis. Step two is a selection of the significance level value which in this case is 5% level of significance. And the third and last step is the calculation of the appropriate test statistic. The test assumes that the data sampled are independently from a normal distribution. The dependent samples have a relationship with each other. As (Donnelly, 2015) states, a match-pair test compares two means or proportions. For the above case, for solving two-sample test, we have our null hypothesis as u1=u2, and our alternative hypothesis is u1 is not equal to u2, with u1 representing the mean call time in am shift and u2 representing the mean call time in pm shift. The formula of t-test is T= (u1-u2)/[Sp^2* (sqrt1/n1+1/n2)], where Sp^2=(n-1) S1^2+(n2-1) S2^2/(n1+n2-2). From this formula, we have the mean 1 being the sample mean call time in am shift. Mean 2 the sample mean call time in pm shift, S1^2 is the sample variance call time of am shift, S2^2 is the sample variance of call time of pm shift, n1 is the sample size of am shift, n2 is the sample size of pm shift. With the data provided in the excel, we get 75 for the sample size of am shift, 10 for the sample mean of am shift and 2.86 for the standard deviation of the am shift; 75 for the sample size of pm shift, 12.15 for the sample mean of pm shift, 3.54 for the standard deviation of the pm shift. Then we can get 3.2152 for the pooled standard deviation and 4.1 for the test statistic. Now we find critical t (alpha, n1+n2-2) as 1.976. Since t-test is greater than t critical, we reject the null hypothesis. Chi-Square Data With the Chi-Square Data, the tests seek to examine whether the frequency of the given certain values is different from the frequency distribution expected (Sharpe, 2015). The categorical values or the observed frequency refers to the integer values which are countable since they are collected from the population of interest. While the expected frequency distribution is that likely to occur randomly by chance if the null hypothesis is true. The Chi-Square Data assumes that the null hypothesis is all dependent and the alternative hypothesis is not all dependent (Lowry, 2014). The Chi-Square tests with known, observed frequency and expected frequency are easily solved using a calculator. By entering all the data we have into the calculator, we will get the result for chi-square. Therefore Chi-squared equals 2.637 with 7 degrees of freedom. The two-tailed P value equals 0. 9164. By conventional criteria, this difference is not statistically significant. Hence we accept the alternative hypothesis. Conclusion The vice president of the Land and Agua Insurance Company finds that there is no difference in the two population means since from the two test hypothesis the t statistic is higher than the t critical hence. The vice president can conclude that the call time and the shift have no difference therefore all the clients should be served regardless of the call time. Since from the chi-square data the alternative hypothesis was accepted, the vice president can conclude that the call time and the shift are significantly related. References Donnelly, R. A. (2015). Business statistics (2nd ed.). Upper Saddle River, NJ: Pearson. Lowry, R. (2014). Concepts and applications of inferential statistics. Sharpe, D. (2015). Your chi-square test is statistically significant: Now what?. Practical Assessment, Research & Evaluation, 20.
The calculation is provided below, please write a new essay using this calculation. Which means except the calculation, every sentences else needs to be writing on your own words. Essay 1. Complete the essay using the information below. Case Study Background: Land and Agua Insurance Company has a call center in Tempe, Arizona. The business was originally established in Phoenix, Arizona in 1972 as a small business, and it has grown with the population of the area. The insurance company specializes in bundling insurance for cars, off-road vehicles, and watercraft (e.g., jet skis and boats). The company has 150,000 clients in Arizona. Marjorie Jones, Vice President of Operations, is concerned about customer complaints and amount of time representatives are taking to resolve the calls. After studying the issue, the company implemented improvements to operations, and a new data sample was collected after three months. You are responsible for analyzing the results to determine whether the improvements made a difference, and determining if the new processes are meeting customer specifications. Additionally, you must review the data and predict the customer call time if a defect occurs. 2. Answer the question below in essay format. Your essay must include an intr oduction, a body, and a conclusion. It must address all relevant parts of each question. Your response should be a minimum of 500 words in length, and it should include both the appropriate statistics and support for the analysis and interpretation. Make s ure to cite any source you use. Proper citation format for a source includes the name of the author(s), the title of the work, the date of the publication, and the page number if you directly quote the source. Essay Questions: New Call Time Using the New Call Center Data, provide a summary report for the vice president including the following information in an essay with a minimum of 500 words: • Using the two-sample hypothesis test comparing old call time to new call time at a .05 level of significance, discuss the hypothesis test assumptions and tests used. Provide the test statistic and p -value in your response. Evaluate the results of the hypothesis test with the scenario. Provide recommendations for the vice president. • Using new call time and coded quality, develop a prediction equation for new call time. Evaluate the model and discuss the coefficient of determination, signi ficance, and use the prediction equation to predict a call time if there is a defect. • Evaluate whether the new call time meets customer specification. As stated in a previous lesson, customers indicated they did not want a call time longer than 7.5 minutes. Assume a standard deviation of 2 min is acceptable. Is the call center now meeting the customer specifications? If not where is the specification not being met? Explain your answers. Answer: In solving two-sample hypothesis test, z-test is used. Basing on this case, there are two different assumptions listed. Considering the null hypothesis, there is similarity of both sample average of old time and the new time. According to the alternative hypothesis, the sample average of the old and new call is completely different. During the calculation, there is need of sample size in the analysis. The assumption summation of the old call is equal to 11.066.In the new calls, the same function calculation is applied, and the sample size of the new call is assumed to be equal to 8.742.In the calculation of the sample variance, the value of the old call (11.066) is assumed to be y while the value of new calls (8.853) is assumed to be x. The statistical test formula is defined as t= (mean x-mean y) or [s*sqrt (1/mean y+1/mean x), where s=sqrt (mean x*s^2x + mean y*s^2y) (John, 2014). In calculation of the values, statistical test is given by [(170*11.066 +170*8.853)/(11.066+8.853)=sqrt 11.02=4.04. After that, all the data is feed to the formula,(11.066-8.853)/[4.04*(1/170+1/170)], 5.296 is the final answer of the test statistics. According to the analysis of the critical value at 0.05 levels of the significant are 3.985. Given the fact that 5.296 is higher than 3.985, the alternative hypothesis is applied since the sample average of the old call, and the new request is not identical. In accordance to test statistics, verification for the advancement in the original call, based on the average test and recommend to the vice president (Wargocki, 2014). "A correlation coefficient indicated both the strength and direction of their linear relationship between a dependent and independent variable" (John, 2014). Hypothesis test is carried out to examine if the population association coefficient is considerably from the zero-based correspondence coefficient. To fit a prediction equation for a new call data with code quality, we can apply hypothesis text. Regression formula is given by Y=a+bz where z is the coded quality where the value for z is either 0 or 1. "The regression formula assists in approximating the dependent value given an independent variable"(John, 2014). According to the data, the sum of z*y is 142.8, the sum of z is 16, the sum of z^2 is 16.Through feeding the given data in the formula b=(n*sum of z*ysum of y*sum of z)/n*sum of z^2-mean of z^2=(170*142.81328*16)/[170*16(0.106)^2]=0.0698;a=sum of y/n-b*(sum of z/n)=1328/1700.0698(16/50)=7.986. Therefore, the prediction equation for the new call is Y=7.986+1*0.0698. To evaluate customer's approval, when the average call duration is 7 min with a standard deviation of 2 min is acceptable. The z test is used to ensure customers fulfillment by sample mean Y=7.978.A null hypothesis is assumed to be uy=7 while the alternative theory is considered to be uy not equal to 7. In calculating the z value we use formula (mean y-uy)/s*sqrt (ny). After feeding the data, the answer is 8.02. In connection with 0.05 levels, if crucial, the critical p-value is 2.00. From the result it is clear that the z observed is higher than z recorded, an alternative hypothesis is put into consideration since the sample average of the old call is not the same (Blair, 2013). About recommendation, it is very crucial to verify the advancement in the new request on the bases of an average test. Example Process Capability Net weight specification = 9.0 oz  0.5 oz Process mean = 8.80 oz Process standard deviation = 0.12 oz Cp= (Upper Specification-Lower Specification)/6(standard deviation) Cp=(9.5-8.5)/(6*0.12)=1.39 References Wargocki, P., Wyon, D. P., & Fanger, P. O. (2014). The performance and subjective responsesof call‐center operators with new and used supply air filters at two outdoor air supplyrates. Indoor air, 14(s8), 7-16 .Blair, C. D. (2013). U.S. Patent N. 7,203,285. Washington, DC: U.S. Patent and Trademark Office. Jenkins, P., Olivier, M., Hodges, R., Parker, H. S., Farris, M., Mosby, K., ... & Cot, A. (2014). Preparing for Storms in Louisiana Take Home Guide, (English/Spanish) John Walker, S. (2014). Big data: A revolution that will transform how we live, work, and think.

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School: Rice University

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Running Head: New Call Time

1

Name
Course
University

New Call Time

2
New call time

Introduction
A two-sample hypothesis test is used to compare to means in order to retain or reject a null
hypothesis. A z-test is used when solving a two-sample hypothesis test. Based on the New Call
Center Data, two assumptions can be made. One is that the data follows a normal distribution
and that the values are independent.
Hypothesis
H0: Old time=New time
H1: Old time ≠New time
For the z-statistic to be computed from the data, there must be a sample size. Summation of the
assumption shows that the sample sizes for the old and new call are 11.066 and 8.742
respectively. We assume the sample variances for the new and old calls to be x and y
respectively. In calculation we use the statistical test formula of t= (mean x-mean y) or [s*sqrt
(1/mean y+1/mean x), where S=sqrt (mean x*s^2x + mean y*s^2y) (John, 2014).
Substituting the values to find S with the data given we have [(170*11.066
+170*8.853)/(11.066+8.853)= 11.02-2=4.04. Substituting to the rest of the formula we get
[(11.066-8.853)/[4.04*(1/170+1/170)]= 5.296.
The critical value of from the N-tables with 5% significance levels is 3.985. 5.296 is greater than
3.985, therefore, we reject Ho. We recommend to the vice president that the advancements made
in call time are significant and should continue.

New Call Time

2

According to John (2014), in a linear relationship of a dependent and independent variable, we
can estimate its strength and direction using a correlation coefficient. Correlation coefficients lie
between -1 and +1 (John, 2014). In order t...

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