Statistical Investigation: Confidence Intervals and Hypothesis Testing

This document presents a comprehensive solution to a statistical investigation assignment. It begins with an analysis of confidence intervals and z-scores, calculating sample sizes and evaluating the correctness of statistical statements related to normal distributions. The assignment then progresses to a two-part problem. The first part involves analyzing customer satisfaction data from two regions, including calculating unbiased estimates, conducting F-tests for variance, and performing t-tests for means. The second part focuses on hypothesis testing for proportions, determining whether the proportion of satisfied customers has decreased. The solution includes detailed calculations, test statistics, p-values, and critical values, along with interpretations of the results. The document concludes with a discussion on the impact of sample size on standard error and a comparison of critical values between t-tests and tests based on the central limit theorem. References and appendix with supporting statistical output are also provided.

Running head: STATISTICAL INVESTIGATION
STATISTICAL INVESTIGATION
Name of the student:
Name of the university:
Author’s note:

1STATISTICAL INVESTIGATION
Q1.
(a) Given, μ = population mean = 11.6 and σ = population standard deviation = 14.8. Here,
x̅ = μ and s = (σ/√n) since population follows normal distribution. Therefore,
Sample mean or x̅=11.6 and
Population standard deviation = σ = 14.8.
(b) 95% z - score value of normal distribution is 1.96 when n is large. Again, 90% z - score
value of normal distribution is 1.644 with large n (Ross 2014). Interval gets wider with the
increase in confidence level. Therefore, 90% confidence interval in narrower from 95%
confidence interval.
(c) Let n denote the required sample size. Then,
n = [(zα/2)2 * σ * (1- σ)]/(margin of error)2.
Where, zα/2 = z – score at α/2, σ = population standard deviation and margin of error is the
permitted error level. Given, σ = 14.8, margin of error is 0.01 and 99% Z – score is 2.576.
Therefore, n = mod [{2.5762 * 14.8 * (1-14.8)}/ (0.01)2] = [6.635776*(-13.8)*14.8]/(0.00001)
=13552909.
The required sample size is 13552909.

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(d) The statement is not correct.
In case of normal distribution, 95% confidence interval of μ is given by {μ ± zα/2*
(σ/√n)}. Standard error multiplied with confidence co-efficient is to be added or subtracted here
rather than only the standard deviation (Leon-Garcia 2017). Therefore, we cannot claim with
95% probability that μ will fall in between 11.6 and 14.8.
(e) The statement is not correct. We can say with confidence that 68% of the observation will fall
in the interval of 11.6 ± 14.8.
A normal curve is a bell shaped curve with its mean as the central line of it (Grimmett
2018). The curve has half of its area in the right tail of the curve and the rest of the half in the left
tail. Standard deviation denotes the range of dispersion from mean. It has been seen that 68% of
the value falls within μ ± σ i.e. P[μ- σ<X<μ+ σ] ≈ 0.068.
(f) Let μ1 be the calculated mean and μ be the hypothesized mean . We are to test:
H0: μ = 12.5 vs H1: μ > 12.5.
Testing statistic is : t = (μ1 - μ)/(σ/√n)
So, calculated t is (14.5 – 12.5)/(14.8/√30) = 0.740 and tabulated t is 1.96.
Therefore, calculated t < tabulated t and null hypothesis is accepted and it can be said that the
average alcohol content is 12.5%.

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Q2.
(a)
For region 1:
Unbiased estimate of average customer satisfaction = x̅1 = 54.06
Unbiased estimate of variance of customer satisfaction = s1 = 45.30
For region 2:
Unbiased estimate of average customer satisfaction = x̅2 = 45.30
Unbiased estimate of variance of customer satisfaction = s2 = 39.06
(b)
The assumption is both the populations follow normal distribution (Lyons and Peres
2016).
(c)
Test hypothesis: H0 : σ1 = σ2 vs H1: σ1 ≠ σ2 , where σ1 is population standard deviation of
region1 and σ2 is population standard deviation of region2.
Test statistics: F = s1/ s2, where s1 is standard deviation of region 1 and s2 is standard
deviation for region 2.
F ~ F (15, 12)
Calculation results are:

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F statistic = 1.1317.
F – Critical value = 2.6371.
It can be seen that tabulated F > calculated F. Therefore the null hypothesis is rejected (Aidara
2018.). We cannot conclude at 5% significance that the variances are equal.
(d)
Test hypothesis: H0: μ1 < μ2 vs. H1: μ1 >= μ2, where μ1 and μ2 are hypothesized mean of
region 1 and region 2 respectively.
Test statistics: (x̅1 - x̅2)/ [s/{√(1/N1) +(1/N2 )}] ~ t27 , where x̅1 and x̅2 are mean of the given
dataset for region 1 and region 2 respectively.
Calculation results are:
T – stat = 3.5901.
P(T<=t) = 0.0006
T Critical value = 2.47
It can be seen that tabulated t < calculated t. Therefore, the null hypothesis can be
accepted and it can be said that average satisfaction of region 2 is not significantly less than
region 1 at level of significance 1.
(e)
Let p denote the proportion of people satisfied from the services (Allen 2014).

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Test hypothesis: H0 : p0 = 0.3 vs H1 : p0 < 0.3 .
Test statistic : k = (p – p0)/ [√{p0*(1-p0)}/n], where p0 is the hypothesized proportion and n is
total sample size.
Calculation results are:
Test statistic = 0.117
P-value = 0.4538
It can be seen that p-value < 0.05. Therefore, null hypothesis will be rejected and it can
be said that the proportion has decreased and less than 30% are satisfied in the present scenario
with level of significance 5%.
(f)
(i). With the assumption of homoscadaticity, the standard error for the differences in mean is
given by: √{(s1/n1) + (s2/n2)}. The sample size is inversely proportional to sample size.
Therefore, as sample size increases, standard error decreases.
(ii). In a two sample mean test , with assumptions of homoscadasticity and same sample size,
critical value of student t test can be larger than the critical value of approximate test on CLT.
Reason for this is that degree of freedom has to be taken into consideration for calculation of t
statistic that is sample size has to be taken into account. Whereas, no sample size has to be taken
into account for computing test based on central limit theorem. This factor creates the difference.

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References:
Ross, S.M., 2014. Introduction to probability models. Academic press.
Dudley, R.M., 2018. Real analysis and probability. CRC Press.
Leon-Garcia, A., 2017. Probability, statistics, and random processes for electrical engineering.
Aidara, N., 2018. Introduction to probability and statistics.
Lyons, R. and Peres, Y., 2016. Probability on trees and networks (Vol. 42). Cambridge
University Press.
Allen, A.O., 2014. Probability, statistics, and queueing theory. Academic Press.

8STATISTICAL INVESTIGATION
Appendix:
F-Test Two-Sample for
Variances
Variable 1 Variable 2
Mean
54.0666666
7
45.3076923
1
Variance
44.2095238
1
39.0641025
6
Observations 15 13
df 14 12
F 1.13171738
P(F<=f) one-tail
0.41943413
4
F Critical one-tail
2.63712355
8
t-Test: Two-Sample Assuming Unequal
Variances
Variable 1
Variable
2
Mean
54.0666666
7
45.3076
9
Variance
44.2095238
1 39.0641
Observations 15 13
Hypothesized Mean Difference 0
df 26
t Stat
3.59015591
4
P(T<=t) one-tail
0.00067403
1
t Critical one-tail
2.47862981
7
P(T<=t) two-tail
0.00134806
1
t Critical two-tail 2.77871452

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