Statistics for Business (STA101) - Assessment 1 Assignment Solution

This document provides a comprehensive solution to a statistics assignment for a business course (STA101). The solution includes detailed calculations and interpretations for several statistical concepts. Question 1 involves calculating covariance and correlation coefficient using sample data, interpreting the relationship between variables, determining the least squares line, and creating a scatter diagram. Question 2 covers regression analysis, including hypothesis testing for the significance of the slope, determination of the coefficient of determination (R2), and interpretation of the slope coefficient. Question 3 focuses on probability calculations, specifically involving malignant and benign cysts, and the accuracy of a test. Question 4 addresses descriptive statistics, including the calculation of mean, median, mode, and standard deviation, along with an analysis of outliers and an empirical rule test for normal distribution. The solution utilizes appropriate statistical formulas and provides clear explanations throughout, supported by relevant references.

STATISTICS FOR BUSINESS
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Question 1
(a) Computation of covariance
x y x- x bar y - y bar (x- x bar)*(y - y bar)
420 2.8 -105 -0.4 42
610 3.6 85 0.4 34
625 3.75 100 0.55 55
500 3 -25 -0.2 5
400 2.5 -125 -0.7 87.5
450 2.7 -75 -0.5 37.5
550 3.5 25 0.3 7.5
650 3.9 125 0.7 87.5
480 2.95 -45 -0.25 11.25
565 3.3 40 0.1 4
Total
5250 32 371.25
Mean
N= 10, x=5250/10 = 525 ¿ y=32/10 = 3.2
Covariance (Covar x , y )=¿(371.25/10) = 37.13
Computation of correlation coefficient
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(b) The above correlation coefficient of 0.979 needs to be interpreted considering both the
magnitude as well as the sign. The positive sign is indicative of a directly proportional
relationship between the two variables. The magnitude is indicative of the relationship
being strong (Morien,, 2017).
(c) Regression line
y = mx + c
m = r ∗൬
σy
σx
൰= 0.979 ∗൬
0.45
83.67൰= 0.0053
c = y bar − b ∗x bar = 3.2 − ሺ0.0053 ∗525ሻ= 0.4156
y = 0.4156 + 0.0053 x
(d) Scatter displayof x and y
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Question 2
Regression Output Model
(a)Hypothesis test for checking the significance of slope
Null hypothesis H0 : β=0(Slopeis insignificant )
Alternative hypothesis Ha : β ≠ 0( Slope is significant )
The p value = 0.6019
Level of significance = 0.05
Observation: p value >>Level of significance
Result: Fails to reject null hypothesis (Hair et. al.,2015).
Slope is statistically not significant for regression model and will be considered as zero.
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(b)Coefficient of determination (R2)
The coefficient of determination indicates that mere 1.140% changes in sound quality of
stereo speaker can be described by changes in the speaker prices. It means a significant
portion i.e. 98.9% in stereo speaker sound quality is dependent on the other variables. The
extremely poor R square value is indicative that the overall utility of regression model is not
good (Berenson et. al., 2015).
(c) Slope coefficient is -0.00239. It refers presence of negative correlation of price and sound
quality of speaker.It means when the price of speaker increases (1 unit) then the sound
quality of speaker decreases by 0.00239 (units). Clearly the relationship is negative which
would mean higher prices would lead to poor sound quality (Flick, 2015).
Question 3
Total cysts (T) = 10,000
Malignant cysts (M) = 1500
Benign cysts (B) = 8500
Accuracy test = 80%
(a) P(M) = 1500/10000 = 0.15
P(B) = 8500/10000 = 0.85
(b) P(Positive test)= P (+ and Malignant) + P (+ and Benign) = {(0.8) *(0.15)} + {(0.20)
*(0.85)} = 0.29
(c) P (Negative test) = 1 – P (Positive test) = 0.71
(d) P (Benign / Positive test) = {(0.20) *(0.85)}/ (0.29) = 0.586
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(e) P (Malignant / Negative test) = {(0.20) *(0.15)}/ (0.71) = 0.42
Question 4
(a) Calculation of mean,median and mode
Data in ascending order
Mean = =21740/30 = 724.67
Median =1/2 {(30/2)thterm+(30/2+1)thterm}=1/2 (720+720)=720
Mode= 730 (Highest Frequency)
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(b) It is apparent from above that mean, median and mode are same It means measures of
central tendency are not considered to be in agreement or else no deviation would have
been observed.
(c) Computation of standard deviation
Standard deviation = ඨ 1
n − 1෍ ሺx − x barሻ2
Standard deviation = ඨ378746.67
30 − 1 = 114.28
(d) Outliers or unusual data
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(e) The empirical rule for testing the normal distribution of data
As all the three empirical rules are satisfied by given data and hence, it can be assumed as
normally distributed through empirical rule.
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References
Berenson, M., Levine, D., Szabat, K.A. and Krehbiel, T.C. (2015) Basic Business Statistcs:
Concepts and Applications. 2nd ed. Brisbane: Pearson Higher Education AU, pp. 198
Flick, U. (2015) Introducing research methodology: A beginner's guide to doing a research
project. 4th ed. New York: Sage Publications, pp. 156
Hair, J. F., Wolfinbarger, M., Money, A. H., Samouel, P., and Page, M. J. (2015) Essentials
of business research methods. 2nd ed. New York: Routledge, pp. 189
Morien, D. (2017) Business Statistics. 3rd ed. Melbourne: Cengage learning Australia, pp. 133
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