Computing the test statistic without the mean and standard deviation?

Question

In the past when elementary school children in a large city were screened, 18% were found to have auditory defects.  In a recent study of 200 children in the city who had been screened within the past month, 51 were found to have such defects.  Do these data suggest a significant increase in the proportion of children having auditory defects?  Test at . Round your answers to three decimal places.

(a). Without rounding any interim calculations, compute the test statistic: 



(b). Using your answer from part (a), compute the p-value:

Answer 1

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.