(R) 2-Sample Test for Equality of Proportions

In today’s article we are going to revisit in greater detail, a topic which was reviewed in a prior article.

What the 2-Sample Test for Equality of Proportions seeks to achieve, is an assessment of differentiation as it pertains to one survey group’s response, as measured against another.

To illustrate the application of this methodology, I will utilize a prior example which was previously published to this site (10/15/2017).

Example:

A pollster took a survey of 1300 individuals, the results of such indicated that 600 were in favor of candidate A. A second survey, taken weeks later, showed that 500 individuals out of 1500 voters were now in favor with candidate A. At a 10% significant level, is there evidence that the candidate's popularity has decreased.

# Model Hypothesis #

# H0: p1 - p2 = 0 #

# (The proportions are the same) # 

# Ha: p1 - p2 > 0 #

# (The proportions are NOT the same) #

# Disable Scientific Notation in R Output #

options(scipen = 999)

# Model Application #

prop.test(x = c(600,500), n=c(1300,1500), conf.level = .95, correct = FALSE)

Which produces the output:

2-sample test for equality of proportions without continuity correction

data: c(600, 500) out of c(1300, 1500)
X-squared = 47.991, df = 1, p-value = 0.000000000004281
alternative hypothesis: two.sided
95 percent confidence interval:
0.09210145 0.16430881
sample estimates:
prop 1 prop 2
0.4615385 0.3333333

We are now prepared to state the details of our model’s application, and the subsequent findings and analysis which occurred as a result of such.

Conclusions:

A 2-Sample Test for Equality of Proportions without Continuity Correction was performed to analyze whether the poll survey results for Candidate A., significantly differed from subsequent poll survey results gathered weeks later. A 90% confidence interval was assumed for significance.

There was a significant difference in Candidate A’s favorability score as from the initial poll findings: 46% (600/1300), as compared to Candidate A’s favorability score the subsequent poll findings: 33% (500/1500); χ2 (1, N = 316) = 47.99, p > .10.

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That's all for now.

I'll see you next time, Data Heads.

-RD