STA1610
Assignment 5
Unique No: 867182
Due 08 September 2025
, Student Name: STA1610 Assignment 5
STA1610 Assignment 5
Due date: 08 September 2025
Question 1
Problem Statement:
The results of ComputerWorld’s annual job satisfaction exercise showed that 28% of
information systems managers are very satisfied, 46% somewhat satisfied, 12% neither
satisfied nor dissatisfied, 10% somewhat dissatisfied, and 4% very dissatisfied. Suppose a
sample of 500 computer programmers yielded:
• Very satisfied: 105
• Somewhat satisfied: 235
• Neither: 55
• Somewhat dissatisfied: 90
• Very dissatisfied: 15
At α = 0.05, test whether the job satisfaction distribution for computer programmers is
different from that for information systems managers.
Step 1 — Choose the test and state hypotheses
This is a chi-square goodness-of-fit test comparing observed counts to expected counts.
H0 : (0.28, 0.46, 0.12, 0.10, 0.04) (programmers follow same distribution)
Ha : Programmers differ in distribution.
Number of categories k = 5, so df = k − 1 = 4. Significance level α = 0.05.
Page 1
Assignment 5
Unique No: 867182
Due 08 September 2025
, Student Name: STA1610 Assignment 5
STA1610 Assignment 5
Due date: 08 September 2025
Question 1
Problem Statement:
The results of ComputerWorld’s annual job satisfaction exercise showed that 28% of
information systems managers are very satisfied, 46% somewhat satisfied, 12% neither
satisfied nor dissatisfied, 10% somewhat dissatisfied, and 4% very dissatisfied. Suppose a
sample of 500 computer programmers yielded:
• Very satisfied: 105
• Somewhat satisfied: 235
• Neither: 55
• Somewhat dissatisfied: 90
• Very dissatisfied: 15
At α = 0.05, test whether the job satisfaction distribution for computer programmers is
different from that for information systems managers.
Step 1 — Choose the test and state hypotheses
This is a chi-square goodness-of-fit test comparing observed counts to expected counts.
H0 : (0.28, 0.46, 0.12, 0.10, 0.04) (programmers follow same distribution)
Ha : Programmers differ in distribution.
Number of categories k = 5, so df = k − 1 = 4. Significance level α = 0.05.
Page 1
