#Testing the relationship between Race and Health Insurance in the United States.
The assignment involves testing the relationship between Race and Health Insurance in the United States.
First, let us load all the packages we may need
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## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
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## filter, lag## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, union## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.0 ──## ✓ ggplot2 3.3.3 ✓ purrr 0.3.4
## ✓ tibble 3.1.0 ✓ stringr 1.4.0
## ✓ tidyr 1.1.3 ✓ forcats 0.5.1
## ✓ readr 1.4.0## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()We then load the data
## Sex Age Married Income HoursWk Race USCitizen HealthInsurance Language
## 1 0 38 1 64 40 white 1 1 1
## 2 1 18 0 0 0 black 1 1 1
## 3 0 21 0 4 20 white 1 1 1
## 4 1 17 0 0 0 other 1 1 0
## 5 0 55 1 34 40 other 0 0 0
## 6 1 51 0 30 40 black 1 1 1## Sex Age Married Income HoursWk Race USCitizen HealthInsurance Language
## 1995 1 28 0 70 40 white 1 1 1
## 1996 1 20 0 30 43 white 1 1 1
## 1997 0 28 0 100 45 white 1 1 1
## 1998 1 56 1 125 40 asian 0 1 0
## 1999 1 49 1 50 40 black 1 1 0
## 2000 0 64 1 0 40 black 1 1 0## 'data.frame': 2000 obs. of 9 variables:
## $ Sex : int 0 1 0 1 0 1 1 0 1 0 ...
## $ Age : int 38 18 21 17 55 51 28 46 17 80 ...
## $ Married : int 1 0 0 0 1 0 0 0 0 0 ...
## $ Income : num 64 0 4 0 34 30 13.7 114 0 0 ...
## $ HoursWk : int 40 0 20 0 40 40 40 60 0 0 ...
## $ Race : chr "white" "black" "white" "other" ...
## $ USCitizen : int 1 1 1 1 0 1 1 1 1 1 ...
## $ HealthInsurance: int 1 1 1 1 0 1 0 1 1 1 ...
## $ Language : int 1 1 1 0 0 1 0 0 1 0 ...## Length Class Mode
## Sex 2000 -none- numeric
## Age 2000 -none- numeric
## Married 2000 -none- numeric
## Income 2000 -none- numeric
## HoursWk 2000 -none- numeric
## Race 2000 -none- character
## USCitizen 2000 -none- numeric
## HealthInsurance 2000 -none- numeric
## Language 2000 -none- numeric## [1] "Sex" "Age" "Married" "Income"
## [5] "HoursWk" "Race" "USCitizen" "HealthInsurance"
## [9] "Language"
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## asian black other white
## 0 11 21 30 113
## 1 118 178 122 1407## `summarise()` has grouped output by 'Race'. You can override using the `.groups` argument.| Race | HealthInsurance | Count | Proportions |
|---|---|---|---|
| asian | 0 | 11 | 8.53 |
| asian | 1 | 118 | 91.47 |
| black | 0 | 21 | 10.55 |
| black | 1 | 178 | 89.45 |
| other | 0 | 30 | 19.74 |
| other | 1 | 122 | 80.26 |
| white | 0 | 113 | 7.43 |
| white | 1 | 1407 | 92.57 |
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## Pearson's Chi-squared test
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## data: ins_data
## X-squared = 27.094, df = 3, p-value = 5.627e-06Since the p-value is less than 0.05, we reject the null hypothesis that the race of the citizens is not associated with the health insurance status.
## [1] 2000 10## Warning: glm.fit: algorithm did not converge##
## Call:
## glm(formula = HealthInsurance ~ ., family = binomial(), data = data)
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## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.409e-06 2.409e-06 2.409e-06 2.409e-06 2.409e-06
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## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.657e+01 4.955e+04 -0.001 1.000
## Sex -1.290e-09 1.636e+04 0.000 1.000
## Age 8.309e-11 4.594e+02 0.000 1.000
## Married -3.609e-10 1.708e+04 0.000 1.000
## Income -6.482e-12 1.954e+02 0.000 1.000
## HoursWk -2.820e-11 4.783e+02 0.000 1.000
## Raceblack -1.164e-09 4.390e+04 0.000 1.000
## Raceother -1.179e-09 4.357e+04 0.000 1.000
## Racewhite -4.487e-10 3.637e+04 0.000 1.000
## USCitizen 5.698e-11 3.589e+04 0.000 1.000
## Language -1.009e-09 2.450e+04 0.000 1.000
## HealthInsurance_cat1 5.313e+01 2.959e+04 0.002 0.999
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## (Dispersion parameter for binomial family taken to be 1)
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## Null deviance: 1.1869e+03 on 1999 degrees of freedom
## Residual deviance: 1.1603e-08 on 1988 degrees of freedom
## AIC: 24
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## Number of Fisher Scoring iterations: 25