Kristine K. answered • 20d

Berkeley Grad, 10 Years Experience, 95th Percentile GRE, SAT and ACT

In this case, we can get the mean and the standard deviation from the percentages.

The children either have a defect or do not, meaning that the variable we are interested in is binary, (0,1). (Called a Bernoulli variable.) The mean of any binary variable is the proportion of 1's in the data. So here, the mean from the first screening is .18 because 18% of the students had defects.

Similarly, the mean from the second screening is .255, because 25.5% of the students had a defect. (We know this because 51/200 = .255.)

Again, because the variable is binary, we know that it's variance follows the formula:

V = p(1 - p)

where p is the mean.

In this example, the first screening group had a variance of .18(1-.18) = 0.15.

In the second group, variance is .255(1-.255) = 0.19.

The numerator of the t statistic is then: .255 - .18

The denominator is the square root of the sum of the variances divided by the number of observations: √(.15/10000 + .19/200) = 0.031.

(They didn't actually say the number of observations in the first study, but did say that it was all children in a large city, so conservatively I estimated 10,000. The larger the number gets, the less the exact value matters.)

The value of the t statistic is then 0.075/.031 = 2.42. You can compare this to a table of t statistics to get the p-value, or use an online calculator.

I hope this helps! If not, feel free to follow up in a comment or message.