Comprehensive Solution: Basic Business Statistics Assignment Analysis

This document presents a comprehensive solution to a basic business statistics assignment. It includes detailed calculations and explanations for various statistical concepts. The solution covers hypothesis testing, including null and alternative hypotheses, test statistics, critical values, and p-values. It also demonstrates the calculation of confidence intervals, including the determination of the margin of error and the construction of the confidence interval for a proportion. Furthermore, the assignment addresses probability calculations, such as determining the probability of obtaining a specific proportion of alcohol in tequila bottles and analyzing the probability of customer arrivals using Poisson distribution. The document also explores concepts like covariance, correlation coefficients, and the relationship between variables. Finally, it provides solutions for binomial distribution problems. The student has contributed this solution to Desklib, a platform offering AI-powered study tools and resources for students.

BASIC BUSINESS STATISTICS
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Question 2
(a) Null and alternative hypothesis
Null hypothesis H0 p=0.4
Alterative hypothesis Ha: p ≠ 0.40
Sample size = 100 bottles
Sample proportion ^p=0.43
Standard deviation σ p= √ p(1− p)
n = √ 0.4 (1−0.4)
100 =0.049
The test statistic z= 0.43−0.40
0.049 =0.61
For significance level of 1%, the critical z value would be 2.58. (from z table)
It can be said that it is a two tailed hypothesis test which implies that null hypothesis would
be rejected when z calculated would not be fall within in the range of critical value. In present
case, the z calculated comes out to be 0.61 which does fall within -2.58 and +2.58 and hence,
null hypothesis would not be rejected.
Therefore, it can be concluded that insufficient evidence is present to conclude that
proportion is not same as 0.40.
(b) 99% confidence interval for proportion
The z value for 99% confidence interval = 2.58 (z table)

Standard error ¿ √ ^p ( 1− ^p )
n = √ 0.43 ( 1−0.43 )
100 =0.049508
Margin of error E=z value∗Standard error =2.58∗0.049508=0.12773 0.128
99 % confidence interval= ^p ± E
Lower limit of 99 % confidence interval= ^p−E=0.43−0.128=0.30227
Lower limit of 99 % confidence interval= ^p+ E=0.43+0.128=0.55773
99%confidence interval = [0.30227 0.55773]
(c) Probability of getting a bottle of tequila that would contains a proportion of alcohol equal
to 0.43 or more =?
Here,
Mean μ=p=0.40
Variance σ = p∗1− p
n = 0.4 ( 1−0.4 )
100 =0.0024
Standard deviation= √ ( variance ) = √ ( 0.0024 ) =0.04899
Now,
P ( p ≥ 0.43 ) =P ( p>0.43 ) =P ( z > 0.43−0.4
0.04899 ) =P ( z >0.61 )
Further,
P ( p ≥ 0.43 ) =P ( z >0.61 ) =1−P ( z<0.61 )
From standard normal table
P ( z< 0.61 )=0.7291

Hence
P ( p ≥ 0.43 ) =1−0.7291=0.2709
Therefore, there is a 27.09% probability of getting a bottle of tequila that would contains a
proportion of alcohol equal to 0.43 or more.
Question 3
(a) Average return
E ( x )= ( 28 %∗12.2 ) + ( 12 %∗15.1 ) + ( 24 %∗0 ) + ( 18 %∗(−1.5 ) ) + ( 18 %∗(−3 ) )
E ( x )=4.418 %
(b) Variance of the returns
E ( x2 )= ( 28 %∗(12.2)2 )+ (12 %∗(15.1)2 )+ ( 24 %∗0 ) + (18 %∗(−1.5 )2 )+ (18 %∗(−3 )2 )
E ( x2 )=71.0614
Now,
Variance Var ( x )=E ( x2 )−E ( x )=71.0614− ( 4.4182 )
Variance=51.543
(c) Standard deviation of the returns
Standard deviation = sqrt (variance) =sqrt (51.543) = 7.179
Question 4
(a) Covariance = 12
The dependency of the variables mainly depends on the value of the correlation. The
covariance value is positive which indicates that corn and temperature are dependent to each
other. It means as the temperature increases then the corn yield would also be increased.

(b) Correlation coefficient r = Cov ( x , y )
√V ( x ) V ( y ) = 12
√9∗36 =0.67
The value of correlation coefficient positive indicates that variables corn yield and
temperature are positively correlated. Further, the value is higher than 0.5 which indicates
that variables corn yield and temperature is strongly positively correlated.
(c) Yes, it can be said that relationship between the corn yield per acre and temperature is
casual because the corn yield not solely depends on the temperature and is a function of
rain, quality of the seeds, weather conditions and so forth. Also, it is not practically
possible that maximum corn yield would be obtained at maximum temperature only.
Question 5
(a) Null and alternative hypothesis
Null hypothesis H0: μ=12, Population mean¿ subway is12 inches .
Alterative hypothesis Ha: : μ<12 Populationmean ¿ subway is12 inches .
The test statistic t= 11.5−12
0.9
√ ( 10 )
=−1.757
Degree of freedom dpf =10−1=9
The p value left tailed = 0.0564
The critical t value = -0.883
Confidence level = 80%
Alpha (Significance level) =0.20
It can be said from the above that p value is lower than level of significance and hence,
sufficient evidence are present to reject the null hypothesis and to accept the alternative
hypothesis. Further, test statistic is lower than critical value and hence, null hypothesis would
be rejected. Therefore, it can be concluded that population mean size of the sandwiches is
lower than 12 inches. It means the subway is short changing customers and hence, one can
sue subway for this.

(b) Sample size = 20
Test statistic
statistic t= 11.5−12
0.9
√ ( 20 )
=−2.485
Degree of freedom dpf =20−1=19
The p value left tailed = 0.0112
The critical t value = -1.729
Confidence level = 95%
Alpha (Significance level) =0.05
It can be said from the above that p value is lower than level of significance and hence,
sufficient evidence are present to reject the null hypothesis and to accept the alternative
hypothesis. Further, test statistic is lower than critical value and hence, null hypothesis would
be rejected. Therefore, it can be concluded that population mean size of the sandwiches is
lower than 12 inches. It means the subway is short changing customers and hence, one can
sue subway for this.
Question 6
(a) Probability of success = 0.65
Number of trials = 3
Odds of having 0 or 1 quizzes is computed as shown below.
Probability of odds of having 0 quiz=P ( 0 ) = 3!
0 ! ( 3−0 ) ! ( 0.65 ) 0 ( 1−0.65 ) 3−0=0.042875
Probability of odds of having1 quiz=P ( 0 )= 3 !
1! (3−1 ) ! ( 0.65 )1 ( 1−0.65 )3−1=0.238875
Mean number of quizzes ¿ np=3∗0.65=1.95
Variances=np q=3∗0.65 (1−0.65 )=0.6825

Standard deviation= √ ( 0.6825 ) =0.8261
(b) Average = 3 customer per hours
A work-days is 8 hours
Mean = 3*8 = 24
Mean and variance is same in case of Poisson distribution and hence,
Variance = Mean = 24
Standard deviation ¿ √(24 )=4.8989
Probability that either 14 or 15 customers would come in on a typical day
P ( x ) = eΎ Ύ x
x !
Where, Ύ =24 , x=14 , 15
P ( 14 ) =e−24 2414
14 ! =0.009109
P ( 15 ) =¿ e−24 2415
15! =0.01457
Probability that either 14 or 15 customers would come in on a typical day =
0.009109+0.01457 = 0.023679