Question

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 100 employees. Medium-sized companies have 100 to 999 employees, and large companies have 1000 or more employees. The questionnaire sent to 225 employees asked whether or not the employee had health insurance and then asked the enployee to indicate the size of the company Size of Company Small Mediumm Large Health Insurance Yes 35 64 90 Total 50 75 100 No 15 10 a. Conduct a test of independence to determine whether health insurance coverage is independent of the size of the company. What is the p-value? Compute the value of the χ2 test statistic (to 2 decimals) The p-value is 1-Select your answer- Using α .05 level of significance, what is your conclusion? Select your answer 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 Based on the percentages calculated above, what can you conclude?

Answer 1

Answer:

a) $\chi^2 = \frac{(35-42)^2}{42}+\frac{(15-8)^2}{8}+\frac{(64-63)^2}{63}+\frac{(11-12)^2}{12}+\frac{(90-84)^2}{84}+\frac{(10-16)^2}{16}=10.07$  

$p_v = P(\chi^2_{2} >10.069)=0.0065$

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent for this case.

b) Small

$\% Yes =\frac{35}{50} x100=70\%$

$\% No =\frac{15}{50} x100=30\%$

Medium

$\% Yes =\frac{64}{75} x100=85.33\%$

$\% No =\frac{11}{75} x100=14.67\%$

Large

$\% Yes =\frac{84}{100} x100=84\%$

$\% No =\frac{16}{100} x100=16\%$

On this case we see that the percentage of people with no health insurance in small companies (30%) is significantly higher than for the medium (14.67%)  and large companies (16%)

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Size if Company             Yes      No       Total

Small                                 35       15         50

Medium                             64       11          75

Large                                 90       10         100

Total                                  189      36         225

Part a

We need to conduct a chi square test in order to check the following hypothesis:

H0: There variable size of company is independent from the variable Health Insurance

H1: There variable size of company is dependent from the variable Health Insurance

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2=\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{189*50}{225}=42$

$E_{2} =\frac{36*50}{225}=8$

$E_{3} =\frac{189*75}{225}=63$

$E_{4} =\frac{36*75}{225}=12$

$E_{5} =\frac{189*100}{225}=84$

$E_{6} =\frac{36*100}{225}=16$

And the expected values are given by:

Size if Company             Yes      No       Total

Small                                 42       8          50

Medium                             63       12          75

Large                                 84       16         100

Total                                  189      36         225

And now we can calculate the statistic:

$\chi^2 = \frac{(35-42)^2}{42}+\frac{(15-8)^2}{8}+\frac{(64-63)^2}{63}+\frac{(11-12)^2}{12}+\frac{(90-84)^2}{84}+\frac{(10-16)^2}{16}=10.07$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >10.069)=0.0065$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(10.069,2,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent for this case.

Part b

First for this case we have this:

$\% Yes =\frac{189}{225} x100=84\%$

$\% No =\frac{36}{225} x100=16\%$

So we see that the proportion of people with health insurance is clear higher than the proportion without it.

Now if we find the proportion for each type of company we have this:

Small

$\% Yes =\frac{35}{50} x100=70\%$

$\% No =\frac{15}{50} x100=30\%$

Medium

$\% Yes =\frac{64}{75} x100=85.33\%$

$\% No =\frac{11}{75} x100=14.67\%$

Large

$\% Yes =\frac{84}{100} x100=84\%$

$\% No =\frac{16}{100} x100=16\%$

On this case we see that the percentage of people with no health insurance in small companies (30%) is significantly higher than for the medium (14.67%)  and large companies (16%)