Basic Business Statistics

This document provides solutions for Basic Business Statistics assignments and essays. It covers topics such as pooled standard deviations, test statistics, estimated variance, sample proportions, and more.

Basic Business Statistics
Student Name:
Student Number:
Date: 17th February 2019

Question 1:
a) Pooled standard deviations
Answer
Sp= √ ( n1 −1 ) s1
2+ ( n2−1 ) s2
2
n1+n2−2
n1=10 ,n2=16 , s1
2=0.292 , s2
2=0.320
Sp= √ ( n1 −1 ) s1
2+ ( n2−1 ) s2
2
n1+ n2−2 = √ ( 10−1 ) 0.292+ ( 16−1 ) 0.320
10+16−2 = √ 0.3095=0.556327
b) Test Statistics
Answer
t= x1 −x2 − ( μ1−μ2 )
s p √ 1
n1
+ 1
n2
= ( 1.24−2.46 ) −0.60
0.556327 √ 1
10 + 1
16
= −1.8 2
0.22426 =−8.1156
c) The p-value for the test
Answer
p-value = 0.00000
d) Can we reject the null hypothesis?
Answer
Yes we reject the null hypothesis since p-value is less than 1% level of significance.
Question 2:

a) Estimated variance of x1−x2
Answer
Sp
2= ( n1 −1 ) s1
2+ ( n2−1 ) s2
2
n1+ n2−2
n1=9 , n2=15 , s1
2=1242 , s2
2 =1605
Sp
2= ( n1 −1 ) s1
2+ ( n2−1 ) s2
2
n1+ n2−2 = ( 9−1 ) 1242+ ( 15−1 ) 1605
9+15−2 =1473
b) Estimated standard error of x1−x2
Answer
sp √ 1
n1
+ 1
n2
= √ 1473∗
√ 1
9 + 1
15 =38.37968∗0.421637=16.18229
c) Test Statistics
Answer
t= ( x1−x2 ) − ( μ1−μ2 )
S . E(x1−x2) = ( 204.3−213.7 )−30
16.18229 = −3 9.4
16.18229 =−2.43476
d) How many degrees of freedom?
Answer
DF =n1 +n2−2 → 9+15−2=22
e) The p-value for the test
Answer
p-value = 0.02347244
f) Can we reject the null hypothesis?
Answer
Yes we reject the null hypothesis since p-value is less than 5% level of significance.
Question 3:

a) Sample proportions p1and p2
Answer
p1= x1
n1
= 40
200 =0.2
p2= x2
n2
= 24 0
5 00 =0.48
b) Estimated variance of p2− p1
Answer
p2 ( 1−p2 )
n2
+ p2 ( 1− p2 )
n1
= 0.48 ( 1−0.48 )
500 + 0.2 ( 1−0.2 )
200 =0.001299
c) Estimated standard error of p2− p1
Answer
S . E ( p2− p1 )= √ p2 ( 1−p2 )
n2
+ p2 ( 1− p2 )
n1
= √ 0.48 ( 1−0.48 )
500 + 0.2 ( 1−0.2 )
200 = √0.001299=0.036044
d) Test Statistics
Answer
t= ( p2− p1 )− ( π 2−π 1 )
S . E ( p2− p1 ) = ( 0.48−0.2 )−0.25
0.036044 = 0.03
0.036044 =0.8323
e) The p-value for the test
Answer
p-value = 0.20262
f) Can we reject the null hypothesis?
Answer
No we fail to reject the null hypothesis since p-value is greater than 5% level of
significance.