The manager knows from previous years that the company will provide an annual bonus if the average sales per customer is greater than $80.00. Based on this sample – with a 5% level of confidence – should they expect an annual bonus? t-Test: Paired Two Sample for Means Variable 1 Variable 2 Mean 77.6005 0 Variance 3098.585 0 Observations 100 100 Pearson Correlation #DIV/0! Hypothesized Mean Difference 80 df 99 t Stat -0.43106 P(T<=t) one-tail 0.33368 t Critical one-tail 1.660391 P(T<=t) two-tail 0.667361 t Critical two-tail 1.984217
The manager knows from previous years that the company will provide an annual bonus if the average sales per customer is greater than $80.00. Based on this sample – with a 5% level of confidence – should they expect an annual bonus? t-Test: Paired Two Sample for Means Variable 1 Variable 2 Mean 77.6005 0 Variance 3098.585 0 Observations 100 100 Pearson Correlation #DIV/0! Hypothesized Mean Difference 80 df 99 t Stat -0.43106 P(T<=t) one-tail 0.33368 t Critical one-tail 1.660391 P(T<=t) two-tail 0.667361 t Critical two-tail 1.984217
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
The manager knows from previous years that the company will provide an annual bonus if the
average sales per customer is greater than $80.00. Based on this sample – with a 5% level of
confidence – should they expect an annual bonus?
t-Test: Paired Two Sample for Means | ||
Variable 1 | Variable 2 | |
Mean | 77.6005 | 0 |
Variance | 3098.585 | 0 |
Observations | 100 | 100 |
Pearson |
#DIV/0! | |
Hypothesized Mean Difference | 80 | |
df | 99 | |
t Stat | -0.43106 | |
P(T<=t) one-tail | 0.33368 | |
t Critical one-tail | 1.660391 | |
P(T<=t) two-tail | 0.667361 | |
t Critical two-tail | 1.984217 |
Here is the data that I ran through excel. I am having a hard time with the hypothesis, decision, and summary.
Expert Solution
Step 1
The test hypotheses are:
Null hypothesis:
H0: μ ≤ $80.00.
Alternative hypothesis:
H1: μ > $80.00.
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