MAT135 Homework #4: Basic Business Statistics Solution, Analysis

This document presents a complete solution to a basic business statistics homework assignment, addressing questions related to hypothesis testing, confidence intervals, and statistical analysis. The solution includes detailed calculations, interpretations, and conclusions for each problem, covering topics such as the probability of a sample mean, confidence intervals for the mean and proportion, and hypothesis testing for various scenarios. The assignment also explores the application of statistical concepts to real-world business problems, providing a comprehensive understanding of the subject matter. The document includes answers to questions related to the mean amount of fruit in yogurt containers, confidence intervals for proportions, and tests of hypotheses regarding weight loss and waiting times. The student has provided a detailed step-by-step approach to the solution.

Running head: BASIC BUSINESS STATISTICS
Basic Business Statistics
Name of the student:
Name of the University:
Author note:

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BASIC BUSINESS STATISTICS

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BASIC BUSINESS STATISTICS
Table of Contents
Answer to the question 1............................................................................................................3
Part (a)....................................................................................................................................3
Part (b)....................................................................................................................................3
Answer to the question 2............................................................................................................4
Part (a)....................................................................................................................................4
Part (b)....................................................................................................................................4
Answer to the question 3............................................................................................................5
Answer to the question 4............................................................................................................6
Part (a)....................................................................................................................................6
Part (b)....................................................................................................................................7
Part (c)....................................................................................................................................7
Answer to the question 5............................................................................................................7
Answer to the question 6............................................................................................................8
Answer to the question 7............................................................................................................9
Bibliography.............................................................................................................................11

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BASIC BUSINESS STATISTICS
Answer to the question 1
Part (a)
Given that
μ= 50
n= 25
X = 45
S= 11.51
P ( X < 45) = P (
X−μ
s
√ n
<
45−50
11.51
√ 25
)
= P (Z < -2.1720)
= 0.0149
The claim of the manufactures is the sample mean is less than 45. It has been seen that
the probability of the less than 45 mean is very poor that is 0.0149. Hence the claim of the
manufactures is not true.
Part (b)
The 95% confidence interval for mean amount of fruit is
X ± 1.96 * σ
√ n
= 45± 1.96 * 11.51
√25
= (40.49, 124.36)
40.49≤mean amount of fruit≤124.36

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BASIC BUSINESS STATISTICS
Answer to the question 2
Part (a)
Given that
n= 100
X= 23
P= 23/100
= 0.23
The 90% confidence interval for proportion is
P ±1.645 * √ p∗(1− p)
n
= 0.23±1.645 * √ 0.23∗(1−0.23)
100
= (0.16, 0.29)
Part (b)
Sample size (n) = (Z/p1)2* (P(1-P))
= (1.645/0.05) * (0.23*(1-0.23))
= 5.83

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BASIC BUSINESS STATISTICS
Answer to the question 3
Hence the 90% confidence interval
0.8516 ≤ Mean weight loss ≤ 3.348
Yes the program is effective.
0.8 1 1.2 1.4 1.6 1.8 2
0
0.5
1
1.5
2
2.5
90% confidence interval plot

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BASIC BUSINESS STATISTICS
Answer to the question 4
Part (a)
The 95% confidence interval for population standard deviation (σ) is obtained by
using t statistic.
X ± tn-1 * sample standard deviation
√n
= 7.15 ± 1.83 * 0.477
√10
= (6.81, 7.49)
Hence 6.81 ≤ σ ≤ 7.49
0 2 4 6 8 10 12
0
2
4
6
8
10
12
95% co fidence interval

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BASIC BUSINESS STATISTICS
Part (b)
The 95% confidence interval for population standard deviation (σ) is obtained by using t
statistic.
X ± tn-1 * sample standard deviation
√n
= 7.15 ± 1.83 * 1.82
√10
= (5.85, 8.45)
Hence 5.85 ≤ σ ≤ 8.45
Part (c)
Yes it shows a variation of time. In the first interval the confidence limit is lesser than
the second one. The second arrangement that is bank of providence is better.
Answer to the question 5
Null hypothesis: The proportion of death before thanksgiving is not less than 0.5.
Alternative hypothesis: The proportion of death before thanksgiving is less than 0.5.

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BASIC BUSINESS STATISTICS
Test statistic= -7.70
P-value = 1
Alpha = 0.05
It has been seen that P-value > alpha. Therefore the null hypothesis is not significant.
Hence it may be concluded that the proportion of death before thanksgiving is not less than
0.5.
Answer to the question 6
Null hypothesis: The mean weight loss is not greater than 0.
Alternative hypothesis: The mean weight loss is greater than 0.

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BASIC BUSINESS STATISTICS
Test statistic= 3.387
P-value = 0.0002
Alpha = 0.01
It has been seen that P-value < alpha. Therefore the null hypothesis is significant.
Hence it may be concluded that the mean weight loss is greater than 0.
Answer to the question 7
Null hypothesis: The standard deviation of waiting time is not less than 1.8 minutes.
Alternative hypothesis: The standard deviation of waiting time is less than 1.8 minutes.

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BASIC BUSINESS STATISTICS
Test statistic= 35.49
P-value = 0.000
Alpha = 0.05
It has been seen that P-value < alpha. Therefore the null hypothesis is significant.
Hence it may be concluded that the standard deviation of waiting time is less than 1.8
minutes.

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BASIC BUSINESS STATISTICS
Bibliography
De Winter, J.C., 2013. Using the Student's t-test with extremely small sample sizes. Practical
Assessment, Research, and Evaluation, 18(1), p.10.
Moinester, M. and Gottfried, R., 2014. Sample size estimation for correlations with pre-
specified confidence interval. The quantitative methods for psychology, 10(2), pp.124-130.
Test, N., Plot, T.S. and Plot, I.V., 2013. One-sample t-test.