Business Statistics 1B Assignment: NUST, Semester 1, 2019

This assignment solution covers several key concepts in business statistics, including probability calculations, confidence intervals, and hypothesis testing. It addresses problems related to normal distributions, sample means, and population proportions, applying these concepts to real-world business scenarios. The solution includes detailed calculations and interpretations, such as determining probabilities related to student spending on cell phones, constructing confidence intervals for the proportion of students visiting a library, and conducting hypothesis tests to evaluate claims about spending habits. The assignment also explores the effects of changing confidence levels on interval estimates and provides conclusions based on the statistical findings, offering insights for stakeholders such as the Mayor of Walvis Bay and the Namibian Chamber of Commerce and Industry (NCCI). The assignment showcases the application of statistical methods to draw meaningful conclusions in a business context.

Business for Statistics
Student Name
University Affiliation

Question 1 [30]
1.2.1 What is the probability that a student studies for more than N$ 60‐00 per week.
(6)
Solution
Given that the μ=52:00 , x=60 :00 σ =6 :00
Now , P ( x >60 )= (1−P ) ( x−μ
σ )= 60−52
6
¿ 1−P(z > 8
6 )
therefore , Z=1.333
Now using z ¿ P value calculator
P ( z=1.333 )=0.0913
¿ 1−0.0913=0.9087
The probability that a students spents more than N $ 60.00 per week is 0.9087
1.2.2 Find the probability that the mean amount of money on cell phones for three
randomly
selected students is less than N$ 60‐66 per week. (8)
solution
Given that n=3 , x=60.66 , μ=52.00 , σ =6.00
Z= x −μ
σ
√n
¿ 1−P ¿
¿ 1−P ¿
¿ 1−P ( z <2.50 )
¿ 0.9876
¿ 1−0.9876
¿ 0.0124
The probability that a student holdsless yhan N $ 60.66 is0.0124

1.3.1)
Estimate and interpret the 90 % and 95 % confidence interval estimates for the true unknown
population proportion of students that visit the library facility on a regular.
Solution,
Given n=256 , x =48
First calculating the proportion;
^P= x
n = 48
256 =0.1875
We will calculatehe confidence intervalusing the formula ,
I = ^P+ Z √ ^P( 1− ^P)
n
At 90 % confidence level ,
α =100−90=10=0.1
Z α
2
=Z 0.1
2
=Z ( 0.05 ) =1.6449
¿ 0.1875 ±1.6449 √ 0.1875(1−0.1875)
256
¿ 0.1875 ± 0.04013
¿ 0.1474 , 0.2276
Therefore the population that visits the library for 90 %interval is 0.1474∧0.2276
At 95 %interval
Z α
2
=Z 0.05
2
=Z ( 0.025 ) =1.960
Calcalating the the confidence interval ;
I = ^P+ Z √ ^P( 1− ^P)
n
¿ 0.1875 ±1.960 √ 0.1875(1−0.1875)
256
¿ 0.1875 ± 0.0478
¿ 0.1397 , 0.2353

Therefore the population that visits the library for 95 % interval is 0.1397∧0.2353
1.3.2) Comment on the difference between the two interval estimates obtained in question 3.1.1
(3)
95 % confidence intervalhas a wider width than90 % cofidence interval
2.1.2) Comment on the difference in sizes obtained between the two interv al estimates
obtained in question 2.1.1 (2)
When the confidence level decreases, the width of the confidence level also decreases equally.
2.2.1) Formulate a suitable null and alternative hypothesis for this situation.(2)
Null hypothesis , H 0: p=0.15
Alternate hypothesis H 1 : p≠ 0.15
2.2.2) Compute the sample statistic (3)
2.2.3) Formulate the decision rule by using  = 10 % (3)
Given the α=0.1 , Zα=1.6449
Since1.94029>1.6449 , we now reject H 0
2.2.4) Based upon the outcome of this test, what conclusion can the Mayor of Walvis Bay make
about the claim? (2)
There is enough evidence ¿ reject the claimthat , thecurrent standing of unemployment is15 %
2.3.2) Compute the sample statistic
Usi the formula
t= ^x−μ
s
√ n
¿ 95.5−90
15
√ 25

¿ 5.5
3
¿ 1.8333
Therefore , sample statistics is1.8333
2.3.3) Formulate the decision rule by using  = 10 % (3)
Given α=0.1
df =25−1=1=24
We now use the right tail area = 0.10
From the t table we find that, t critical value is 1.318
We will reject null hypothesis H0, if the calculated t test statistic value > 1.318, otherwise we fail
to reject H0.
Since t test statistic value (1.8333) is greater critical value (1.318) , we reject null hypothesis H0.
2.3.4) What conclusion can the chairman of the Namibian Chamber of Commerce and Industry (
NCCI)
draw from the findings? (2)
There is enough evidence to support the claim that visitors to the mall spend more than 90
minutes in the mall on every occasion.