Question

The Pique Technique was developed because a target for solicitation is more likely to comply
if mindless refusal is disrupted by a strange or unusual request. Researchers had young people
ask 144 targets for money. The targets were asked either for a standard amount – a quarter
– or an unusual amount – 17 cents. 43.1% of those asked for 17 cents responded, compared
to 30.6% of those asked for a quarter. Each group contained 72 targets. Test whether this
increase is statistically significant at = 0.05. Do this by hand, and then verify your results
using prop.test in R.

Answer 1

Let p1 and p2 be the proportion of respondents among those asked for 17 cents and a quarter respectively.

Null hypothesis H0: p1 = p2

Alternative hypothesis H0: p1 > p2

Sample proportions are,

= 0.431

= 0.306

Pooled proportion, p = (n1 + n2 ) / (n1 + n2)

Here sample size is same for both groups. n1 = n2 = 144

p = (n1 + n2 ) / (n1 + n2) = (144 * 0.431 + 144 * 0.306) / (144 + 144) = 0.3685

Standard error of difference in proportions, SE =

= 0.05685114

Test statistic, z = ( - ) / SE = (0.431 - 0.306) / 0.05685114 = 2.20

P-value = P(z > 2.20) = 0.0139

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that proportion of respondents among those asked for 17 cents is significantly greater than the proportion of respondents asked for a quarter.

Running prop.test in R (without continuity correction), we get the same p-value and hence the conclusion is same as above.

> prop.test(x = c(144*0.431, 144*0.306), n = c(144,144), alternative = c("greater"), correct = FALSE)

   2-sample test for equality of proportions without continuity correction

data: c(144 * 0.431, 144 * 0.306) out of c(144, 144)
X-squared = 4.8344, df = 1, p-value = 0.01395
alternative hypothesis: greater
95 percent confidence interval:
0.03227637 1.00000000
sample estimates:
prop 1 prop 2
0.431 0.306