Categorial Data Analysis: seat belts and child injuries

[raspberryfields]

This is a supplemental exercise in my textbook:
"Seat Belt Test Study"
792 injured children studied

Hispanic Seatbelt Worn: 31
Hispanic Not worn: 283
Total: 314

Non-hispanic Seatbelt Worn: 148
Non-hispanic Not worn: 330
Total: 478

e) construct a 99% confidence interval for the difference between proportion A and B. A= of the injured hispanic children, what proportion were not wearing seat belts? B= of the injured non-hispanic children, what proportion were not wearing seat belts?
ANSWER: 0.211+/- 0.070

I know that 0.211 is Proportion A-B = 0.901-0.69 = 0.211 what i do not know how to get the 0.070 part. I think that it involves 2.575 x square root of pq/n but i am not sure what the proportions are and how to obtain the answer.

Thanks in advance!

 

[chivox]

Hi. the z value just comes from looking it up in a table. You can find a complete z-value table on the National Institute of Standard and Technology site (a US government site, trust with caution

http://www.itl.nist.gov/div898/handbook/eda/section3/eda3671.htm

If you want the 99% confidence interval, you have to use z from the table at plus-or-minus one-half of (1.00 - 0.99), which is 0.995 and 0.005. That value, according to the site is +/- 2.576.

The variable p represents the proportion of interest, and q is defined as (1 - p), since the probability of either event happening is 1 in total. However, the formula you gave above is appropriate for finding the confidence intervals on a proportion. What you have here is a completely different problem, since it asks you to find the confidence interval on the "difference" between the two proportions.

As far as estimating the confidence interval for the "difference" between two independent proportions, the interval is

\(\displaystyle (p_1 - p_2) \pm z \cdot \sqrt{SE_1^2 + SE_2^2}\)

where SE[sub:2sk8niik]1[/sub:2sk8niik] and SE[sub:2sk8niik]2[/sub:2sk8niik] are the standard errors of the two proportions, and z is the table value for one-half the confidence interval, since it is a plus-or-minus interval.

The standard error of a proportion can be computed as

\(\displaystyle SE = \frac{\sqrt{(p)(1 - p)}}{\sqrt{n}}\)

where p is the proportion of interest (such as 283/314 in the case of A in your problem) and n is the total number in that study (314 in your A case).