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I have attached my original problem in word format and my answers in excel format.
Could the OTA please check my work and offer suggestions where mistakes are
obvious as this is the premise for me to build on in my coursework.
Also, I am looking for detailed help in understanding what this data means/implies.
For example, how would I read the data to myself in my head gleaning meaning from
it? Or, the big, so what does this really say or imply? In six weeks for my final exam I
will have to interpret problems as well as do the formulas and I need to begin
practicing on work that is relevant for me. I decided to start with some role modeling
from you. I hope I have offered payment compensatory for this added necessary
feedback.
P9.43
The SEC requires companies to file annual reports concerning their financial status. It is
impossible to audit every account receivable. Suppose we audit a random sample of 49
accounts receivable invoices and find a sample average of $128 and a sample standard
deviation of $53.
a) Find a 99% confidence interval for the mean size of an accounts receivable
invoice. Does your answer require that the sizes of the accounts receivable
invoices follow a normal distribution?
Using the formula:
SE = SD/SQRT(N) = 53/ SQRT(49) = 7.57
Two-tailed t (0.01, 48) = 2.7
Then the Confidence intervals: lower and upper bounds of the range of values that
contain the population parameter with a given probability.
The confidence interval = Mean +- t*SE = 128 +- 2.7*7.57 = (107.69, 148.31)
So your calculation is correct in this part.
 It is true that normal distribution is required to derive the confidence interval
 Moreover, the data should be sampled randomly and are independent
However, the central limit theorem states that no matter what the shape of the original
distribution, the sampling distribution of the mean approaches a normal distribution.
Furthermore, for most distributions, a normal distribution is approached very quickly as
N increases. (An detailed explanation is attached.)
Therefore, as the sample size is large enough, the distribution is approaching normal
itself.
b) How large a sample do we need if we want to be 99% sure that we can estimate
the mean invoice size within $5?
Determining sample size. If one has a preference about the size of the standard error, and
an estimate of the standard deviation in the sample, then N may be estimated by working
the formula for the SE backwards.
The sample size formula uses three key factors:
(1) the alpha level, the level of acceptable risk the researcher is willing to accept that the
true margin of error exceeds the acceptable margin of error; i.e., the probability that
differences revealed by statistical analyses really do not exist; also known as Type I error.
The alpha level used in determining sample size in this survey is 0.01. In the formula, the
alpha level is incorporated into the formula by utilizing the t-value for the alpha level
selected (In this case, the two-tailed t-value for alpha level of .05 is 2.617 for sample
sizes above 120).
(2) Acceptable Margin of Error: the risk the researcher is willing to accept in the survey,
or the error the researcher is willing to accept.
In this case, the error is +- $2.5.
(3) A critical component of sample size formulas is the estimation of variance in the
primary variables of interest in the study.
Var=SD^2= 53^2=2809 in this question.
Now we can calculate the required sample size to be:
N = Var * (t/e)^2 = 2809*(2.617/2.5) ^2 = 3078
Therefore, we should use sample size of at least 3078, if we want to be 99% sure that we
can estimate the mean invoice size within $5
c) Take new sample size of 100, with sample mean equal to $210 and sample
standard deviation of $24. Compute 99.8% confidence interval. Then, how large a
sample size do we need to be sure of that 99.8% mean to within 5$?
Compare and contrast the results of a,b) with results of c).
N=100, M=210, SD=24 and df = 100-1=99
SE = SD/SQRT(N) = 24/ SQRT(100) = 2.4
Two-tailed t(0.2%, 99) = 3.175
Then the confidence interval = M +- t*SE = 210 +- 3.175*2.4 = (202.38, 217.62)
Var=SD^2= 24^2=576 in this question.
We can calculate the required sample size to be:
N = Var * (t/e)^2 = 576*(3.175/2.5) ^2 = 929
Therefore, we should use sample size of at least 929, if we want to be 99.8% sure that we
can estimate the mean invoice size within $5
Although our confidence level is higher in part (c), the required sample size is smaller
than part (b). Why does this happen?
The key point here is the standard deviation (or variance). We notice that the standard
deviation in part (c) is much smaller than part (b), which indicate that this new sample
behaves better and is of higher quality. Therefore, the sample is very close to the real
population and we will only require fewer observations to reach 99.8% confidence level
Appendix
The central limit theorem states that given a distribution with a mean μ and variance σ2,
the sampling distribution of the mean approaches a normal distribution with a mean (μ)
and a variance σ2/N as N, the sample size, increases.
The amazing and counter- intuitive thing about the central limit theorem is that no matter
what the shape of the original distribution, the sampling distribution of the mean
approaches a normal distribution. Furthermore, for most distributions, a normal
distribution is approached very quickly as N increases. Keep in mind that N is the sample
size for each mean and not the number of samples. Remember in a sampling distribution
the number of samples is assumed to be infinite. The sample size is the number of scores
in each sample; it is the number of scores that goes into the computation of each mean.
Let’s show this property by the results of a simulation exercise to demonstrate the central
limit theorem. The computer sampled N scores from a uniform distribution and computed
the mean. This procedure was performed 500 times for each of the sample sizes 1, 4, 7,
and 10.
The resulting frequency distributions each based on 500 means. For N = 4, 4 scores were
sampled from a uniform distribution 500 times and the mean computed each time. The
same method was followed with means of 7 scores for N = 7 and 10 scores for N = 10.
Two things should be noted about the effect of increasing N:
1. The distributions becomes more and more normal.
2. The spread of the distributions decreases.