Basic Business Statistics Homework: Solutions for Hypothesis Testing

This document presents a solved homework assignment in basic business statistics, covering key concepts such as pooled standard deviations, test statistics, and p-values. The solutions address three distinct questions involving hypothesis testing. The first question examines pooled standard deviations, test statistics, and p-values to determine whether to reject the null hypothesis. The second question focuses on estimating variance and standard error, calculating test statistics, determining degrees of freedom, and interpreting p-values to assess the null hypothesis. The final question involves sample proportions, estimating variance and standard error, calculating test statistics, and interpreting the p-value to evaluate the null hypothesis. The assignment demonstrates statistical analysis techniques commonly used in business contexts.

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.