STA1610
Assignment 5
Unique No: 867182
DUE 8 September 2025
, ASSIGNMENT 05
Unique Nr.: 867182
Fixed closing date: 08 September 2025
Question 1: Goodness-of-Fit Test
To determine if the job satisfaction for computer programmers is different from that of
information systems managers, I'll use a chi-square goodness-of-fit test.
Step 1: State the hypotheses
• Null hypothesis (𝐻0 ): The job satisfaction distribution for computer programmers
is the same as for information systems managers.
• Alternative hypothesis (𝐻𝑎 ): The job satisfaction distribution for computer
programmers is different from that for information systems managers.
Step 2: Determine the expected frequencies (𝐸𝑖 )
The expected frequencies are calculated by multiplying the total sample size (𝑛 = 500 )
by the population proportions from the ComputerWorld survey.
• Very Satisfied: 𝐸1 = 500 × 0.28 = 140
• Somewhat Satisfied: 𝐸2 = 500 × 0.46 = 230
• Neither: 𝐸3 = 500 × 0.12 = 60
• Somewhat Dissatisfied: 𝐸4 = 500 × 0.10 = 50
• Very Dissatisfied: 𝐸5 = 500 × 0.04 = 20
Note: The sum of the expected frequencies is 140 + 230 + 60 + 50 + 20 = 500 , which is
equal to the sample size.
Step 3: Calculate the chi-square test statistic (𝜒 2 )
Assignment 5
Unique No: 867182
DUE 8 September 2025
, ASSIGNMENT 05
Unique Nr.: 867182
Fixed closing date: 08 September 2025
Question 1: Goodness-of-Fit Test
To determine if the job satisfaction for computer programmers is different from that of
information systems managers, I'll use a chi-square goodness-of-fit test.
Step 1: State the hypotheses
• Null hypothesis (𝐻0 ): The job satisfaction distribution for computer programmers
is the same as for information systems managers.
• Alternative hypothesis (𝐻𝑎 ): The job satisfaction distribution for computer
programmers is different from that for information systems managers.
Step 2: Determine the expected frequencies (𝐸𝑖 )
The expected frequencies are calculated by multiplying the total sample size (𝑛 = 500 )
by the population proportions from the ComputerWorld survey.
• Very Satisfied: 𝐸1 = 500 × 0.28 = 140
• Somewhat Satisfied: 𝐸2 = 500 × 0.46 = 230
• Neither: 𝐸3 = 500 × 0.12 = 60
• Somewhat Dissatisfied: 𝐸4 = 500 × 0.10 = 50
• Very Dissatisfied: 𝐸5 = 500 × 0.04 = 20
Note: The sum of the expected frequencies is 140 + 230 + 60 + 50 + 20 = 500 , which is
equal to the sample size.
Step 3: Calculate the chi-square test statistic (𝜒 2 )