Biostatistics Question

Question

Prior to the initiation of a rubella immunization program in a metropolitan county, a survey revealed that 55 out of a sample of 100 elementary school children in the county had been immunized against the disease. Are these data compatible with the contention that 50% of the elementary school children had been immunized against rubella?

 

A. State the null and (2-sided) alternative hypotheses to answer this question.

 

B. Calculate an appropriate test statistic.

 

C. Construct a 95% confidence interval for the true proportion of children that had been immunized against rubella.

 

Any help getting started on answering this question would be helpful. I'm really looking for overall help, not just an answer.

 

Thanks!

Answer 1

Hello,

The null hypothesis is a very general statement that there is no difference between some number of groups or that there is no relationship between a set of variables. In your case what you would say is that the mean of the sample equals 50 (or is not different than 50): M = 50.

You're dealing with percentages calculated from binary data - in other words, each child could come form 1 of 2 possible groups: immunized or not-immunized. Problems like these are typically dealt with using a binomial test. The formula for a binomial test is:

z = (X - pn)/sqrt(npq)

In this formula X is the number of "success" you observed (in your case, the number of immunized children). n is the number of children total. p is the probability of a child being immunized according to the null hypothesis. In your case, the null hypothesis says that 50% should be immunized so p = .5. q is always 1 - p (in your case, also .5).
Putting it all together, we get: (55 - .5*100)/sqrt(100*.5*.5). When we do the math, this results in a value of 1. So z = 1. To determine whether this value is significant, you have to check it against the critical value that you can find in a normal distribution table (your teacher should have given you one or it should be in your textbook). Whether or not it is significant will depend on what alpha level you are using - by most typical standards (when alpha = .05), this value is not considered significant. This means that we have failed to reject the null hypothesis.

 

The confidence interval is very much like the margin of error in a political poll. It tells us where the true mean is likely to fall. To calculate this, we need 3 things: the mean (which is 55%), the standard error and the critical value.

The standard error for this type of data can be computed using the formula: sqrt((pq)/n).

In this example, we get sqrt((.5*.5)/100)) = sqrt(.25/100) = sqrt(.0025) = 0.05.

Next we need to multiply the standard error by the critical value. Since you want a 95% confidence interval, the critical value is 1.96 (this can be found in the normal distribution table). 1.96*.05 = 0.098.

Last, in order the create the confidence interval, we add and subtract this number from the mean:

.55 + .098 = .648 (or 64.8%)

.55 - .098 = .452 (or 45.2%)

What this means is that the study found that 55% of kids were immunized, but because of sampling error, the true, population percentage could reasonably be anywhere between 45.2% and 64.8%. Notice how this interval includes 50%. This is why the test earlier was non-significant. This interval says that the true value very well could be 50%.

 

Let me know if any of this didn't make sense.