benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sample data below show the number of companies providing health insurance for small, medium, and large companies. For the purposes of this study, small companies are companies that have fewer than employees. Medium-sized companies have to employees, and large companies have or more employees. The questionnaire sent to employees asked whether or not the employee had health insurance and then asked the enployee to indicate the size of the company. Health Insurance Size of Company Yes No Total Small 31 19
Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sample data below show the number of companies providing health insurance for small, medium, and large companies. For the purposes of this study, small companies are companies that have fewer than employees. Medium-sized companies have to employees, and large companies have or more employees. The questionnaire sent to employees asked whether or not the employee had health insurance and then asked the enployee to indicate the size of the company.
Health Insurance | |||||
Size of Company | Yes | No | Total | ||
Small | 31 | 19 | 50 | ||
Medium | 70 | 5 | 75 | ||
Large | 88 | 12 | 100 |
a. Conduct a test of independence to determine whether health insurance coverage is independent of the size of the company. What is the -value?
Compute the value of the x^2 test statistic (to 2 decimals).
b. A newspaper article indicated employees of small companies are more likely to lack health insurance coverage. Calculate the percentages of employees without health insurance based on company size (to the nearest whole number).
Small | |
Medium | |
Large |
NOTE: As the significance level, i.e, is not mentioned, I am taking it as 5%, i.e, 0.05.
a).
The null hypothesis (H0) states that health insurance coverage is independent of the size of the company.
The alternative hypothesis (Ha) states that health insurance coverage is dependent of the size of the company.
First we will depict the information in a proper manner. The given frequencies are the observed frequencies. It is shown as:
Observed Frequencies (O) | |||
Yes | No | Row Total | |
Small | 31 | 19 | 31+19=50 |
Medium | 70 | 5 | 70+5=75 |
Large | 88 | 12 | 88+12=100 |
Column Total | 31+70+88=189 | 19+5+12=36 | (189+36) or (50+75+100)=225 |
Now, we have to calculate the expected frequencies. The formula for expected frequencies is:
The following table shows the calculations of the expected freqncuies:
Expected Frequencies (E) | ||
Yes | No | |
Small | 50*189/225=42 | 50*36/225=8 |
Medium | 75* 189/225=63 | 75*36/225=12 |
Large | 100*189/225=84 | 100*36/225=16 |
The following table shows the required calculations for the test statistic.
O | E | (O-E)2 | (O-E)2/E |
31 | 42 | 121 | 2.881 |
19 | 8 | 121 | 15.125 |
70 | 63 | 49 | 0.7778 |
5 | 12 | 49 | 4.0833 |
88 | 84 | 16 | 0.1905 |
12 | 16 | 16 | 1 |
The formula and calculations for the test statistic is as follows;
Thus, the test statistic=24.06
The degree of freedom for the test is calculated as:
Df=(Number of rows-1)*(Number of column-1)
=(3-1)*(2-1)
=2
Now, observe and df=2 in the table to obtain the p-value.
The p-value obtained is less than 0.00001
The decision rule states that if the p-value is less than the significance level, we Reject the null hypothesis.
In this case, 0.00001<0.05, Hence We Reject the null hypothesis.
Therefore, It is concluded that health insurance coverage is dependent of the size of the company.
Step by step
Solved in 2 steps