249.Follow the requirement to compose a 350words essay using the given calculation.

Follow the requirement to compose a 350words essay using the given calculation.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.All the work has to be 100percent original.
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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.

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