Assignment
Instructions
You are going to use Excel to do a chi-square analysis, as shown in the videos this week.
The table shows data on vaccine effectiveness. Data from 2789 subjects has been collected on whether they have been vaccinated and the severity of symptoms.
Severe Symptoms Moderate Symptoms Mild Symptoms Row Totals
Not Vaccinated 595 580 575 1750
Vaccinated 374 300 365 1039
Column Totals 969 880 940 2789
Turn in the following on an Excel spreadsheet
State the null and alternative hypotheses
Create a table for the expected frequencies
Find the chi-square test statistic
Find the p-value for the test statistic
At a significance level of 0.05, determine if the null hypothesis is accepted or rejected
State the conclusion
In a Excel document, turn in these questions with your responses numbered and written in detail.
Chi-Square
We have seen many processes for dealing with numerical data, including independent proportions. But what about when we have interaction between variables? Think about how there are differences in how genders or races have differences in how they perceive things. Many things are independent when dealing with proportions, but more frequently you will see dependence, which means there is interaction between variables and not just in the sense of intervention and outcome. It’s important to determine if there is this dependence. The chi-square (chi rhymes with sky) distribution is a skewed distribution which is the distribution that dependent proportions follow.
A team of doctors wants to see if the reasons for heart surgery differ between men and women. The table below lists different causes of surgery by gender.
Congestive Heart Failure Diabetes Three-vessel Disease Row Total
Male 225 75 300 600
Female 250 96 230 676
Column Total 475 171 530 1176
The chi-square test of independence will indicate if these proportions are dependent or independent, however, it will not determine in what way they are dependent. Further testing is done to make that determination, which will not be covered here. With any hypothesis test, we start by deciding on a significance level. Here, the researchers decide on a significance level of 0.01 since there are different preexisting conditions that might be better monitored. Next, we state the null and alternative hypotheses. It’s always assumed in a chi-square test of independence that there is no interaction between the variables.
Ho: The row and column variables are independent (no interaction between preexisting condition and gender)
HA: The row and column variables are dependent (there is an interaction between a preexisting condition and gender)
The chi-square test is tedious to run by hand, so we speed up the process using technology. The test is dependent on having expected frequencies in the cells which are calculated by multiplying row and column totals, then dividing by sample size. This is demonstrated in the cells below, in parentheses.
Congestive Heart Failure Diabetes Three-vessel Disease Row Total
Male 225 (600*475/1176 = 242.3) 75 (600*171/1176 = 87.2) 300 (600*560/1176 = 270.4) 600
Female 250 (576*475/1176 = 232.7) 96 (576*171/1176 = 83.8) 230 (576*530/1176 = 259.6) 576
Column Total 475 171 530 1176
The chi-square test statistic is found by subtracting the observed-expected frequencies, squaring them, dividing them by the expected frequencies, and adding those up. This is where technology will be helpful. This video shows how to perform chi-square tests of independence in Excel.
Perform Chi-Square Test Of Independence In Excel (Including P-Value!) (YouTube, 08:34)
Using Excel, we get a chi-square test statistic of
�
2
=12.655
with degrees of freedom of (number of rows – 1)(number of columns – 1) = (2 – 1)(3 – 1) = 2 degrees of freedom
Using Excel as shown in the video, we get a p-value of 0.00179. Since the significance level is 0.01, and the p-value is smaller than that, the null hypothesis is rejected. The data indicate that these preexisting conditions and gender are dependent. With this information, the team can go to work on carefully screening patients for these conditions and providing the appropriate care.
Continue to the Lecture Videos section.
