Inference for Difference of Two Proportions

Dr. Mine Dogucu

Confidence Interval

Example

Do those who take college level science courses and those who don’t have different rates of belief in life after death? Below are the responses from General Social Survey in 2018.

Belief in Life After Death
Yes No
College Science Course
 Yes 375 75
No 485 115

Understanding the Question

Response: Belief in Life After Death (categorical)
Explanatory: College Science Course

What did we observe?

Belief in Life After Death Among College Science Course Takers

\(p_{science} = \frac{375}{375+75} = 0.8333333\)

\(n_{science} = 450\)

Belief in Life After Death Among Non - College Science Course Takers

\(p_{noscience} = \frac{485}{485+115} = 0.8083333\)

\(n_{noscience} = 600\)

What did we observe?

It seems like there are more after life believers among college science course takers (~83%) when compared to those who did not take college science course (~80.83%). But now that we have taken statistics course we cannot only rely on comparison of sample statistics. We know we have to think about population parameters.

CLT for two proportions

If conditions are met then according to CLT \((p_1 - p_2) \sim \text{approximately } N(\pi_1 - \pi_2, {\frac{\pi_1(1-\pi_1)}{n_1} + \frac{\pi_2(1-\pi_2)}{n_2}})\)

Recall that the standard deviation of the the sampling distribution is the standard error.

Standard error for difference of two proportions

\(\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\)

Conditions

Independence: Within each group data have to be independent from each other. The two groups have to be independent from one another.

GSS utilizes some form of random sampling so we would expect independence within each group. People either have taken a college level science class or they have not taken so we can assume that the groups are independent from one another.

There needs to be at least 10 successes and 10 failures in each group.

We have seen that all the values in the contingency table were greater than 10.

Confidence Interval Construction

CI = \(\text{point estimate} \pm \text { critical value} \times \text{standard error}\)

CI for difference of two proportions = \(p_1 - p_2 \pm \text { critical value} \times \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\)

Steps

  1. Calculate point estimate.
  2. Calculate critical value.
  3. Calculate standard error.
  4. Construct the confidence interval.

CI for difference of two proportions = \(p_1 - p_2 \pm \text { critical value} \times \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\)

p1 <- 0.83
p2 <- 0.8083
n1 <- 450
n2 <- 600

point_estimate <- p1 - p2

cv <- qnorm(0.975)
cv
[1] 1.959964

CI for difference of two proportions = \(p_1 - p_2 \pm \text { critical value} \times \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\)

se <- sqrt((p1*(1-p1)/n1)+(p2*(1-p2)/n2))

point_estimate - cv*se
[1] -0.02516763
point_estimate + cv*se
[1] 0.06856763

95%CI for difference of two proportions is (-0.025,0.069)

Hypothesis Testing

Steps

  1. Set hypotheses
  2. Identify Sampling Distribution of \(H_0\)
  3. Calculate p-value
  4. Make a Decision and a Conclusion.

Example

Is there a relationship between taking a college level science class and belief in life after death?

Step 1. Set hypotheses

\[H_0: \pi_1 = \pi_2\]

\[H_A: \pi_1 \neq \pi_2\]

Step 2. Identify Sampling Distribution of \(H_0\)

If conditions are met then according to CLT \((p_1 - p_2) \sim \text{approximately } N(\pi_1 - \pi_2, {\frac{\pi_1(1-\pi_1)}{n_1} + \frac{\pi_2(1-\pi_2)}{n_2}})\)

Assuming that the null is true then \[\pi_1 = \pi_2\] so we cannot use different \(p_1\) and \(p_2\) in place of \(\pi_1\) and \(\pi_2\).

Pooled estimate of a proportion when \(H_0: \pi_1 = \pi_2\)

\(p_{pooled} = \frac{\text{number of total successes}}{\text{number of total cases}} = \frac{p_1n_1+p_2n_2}{n_1+n_2}\)

\(SE = \sqrt{\frac{p_{pooled}(1-p_{pooled})}{n_1}+\frac{p_{pooled}(1-p_{pooled})}{n_2}}\)

We also use the pooled proportion when checking conditions for success-failure counts.

Calculating pooled proportion

\(p_{pooled} = \frac{\text{number of total successes}}{\text{number of total cases}} = \frac{p_1n_1+p_2n_2}{n_1+n_2}\)

p1 <- 0.83
p2 <- 0.8083
n1 <- 450
n2 <- 600

p_pooled <- (p1*n1+p2*n2)/(n1+n2)

p_pooled
[1] 0.8176

Calculating standard error

\(SE = \sqrt{\frac{p_{pooled}(1-p_{pooled})}{n_1}+\frac{p_{pooled}(1-p_{pooled})}{n_2}}\)

sqrt((p_pooled*(1-p_pooled)/n1) + (p_pooled*(1-p_pooled)/n2))
[1] 0.02408217

Step 3. Calculate p-value

p1 - p2
[1] 0.0217

How likely are we to observe a difference of proportions in samples that is at least as extreme as (0.0217)?

If the null hypothesis is true then

se <- sqrt((p_pooled*(1-p_pooled)/n1) + (p_pooled*(1-p_pooled)/n2))

pnorm(0.0217, mean = 0, sd = se, lower.tail = FALSE) #P(p1-p2>0.0217)
[1] 0.1837725
pnorm(-0.0217, mean = 0, sd = se) #P(p1-p2<-0.0217)
[1] 0.1837725
pnorm(-0.0217, mean = 0, sd = se)*2 #p-value
[1] 0.367545

Step 4. Make a Decision and a Conclusion.

If the null hypothesis were true ( \(\pi_1 - \pi_2 = 0\) ) then the probability of observing a difference of proportions in the sample that is at least extreme as 0.0217 would be 0.37. In other words, p-value = 0.37 which is not less than 0.05. This implies that the observed value ( \(p_1 - p_2 = 0.0217\) ) is not significantly different than 0. We fail to reject the null and conclude that we have not found any evidence against the null.

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