Question

The most common injury that occurs among mountain climbers is trauma to the lower extremity (leg). Consequently, rescuers must be proficient in immobilizing and splinting of fractures. The researchers examined the likelihood of needing certain types of splints. A Mountain Rescue study reported that there were 5 shaft splints needed among 336 live casualties. The researchers will use this study to estimate the proportion of all mountain casualties that require a femoral shaft splint. Complete parts a and b below. a. Is the sample large enough to apply the large-sample estimation method of this section? Show why or why not. O A. It is inappropriate because the number of failures in the sample is less than 15. B. It is inappropriate because the number of successes in the sample is less than 15. O C. It is inappropriate because the number of failures in the sample is greater than 15. OD. It is inappropriate because the number of successes in the sample is greater than 15. b. Use Wilson's adjustment to find a 90% confidence interval for the true proportion of all mountain casualties that require a femoral shaft splint. Interpret the result. The 90% confidence interval is ( 11 ). (Round to three decimal places as needed.)

Answer 1

Answer:

a)

Given,

Sample size n = 336

x = 5

Here we can say that the sample size is large enough, if np^ >= 15 & nq^ = n(1-p^) >= 15

np^ = 1 < 15

nq^ = n - x

= 336 - 5

= 331 >= 15

Here both are not satisfied,

so it is inappropriate because the number of successes in the sample is less than 15.

b)

Consider,

Point estimate = (x + 2) / (n + 4)

= (5 + 2) / (336 + 4)

p^ = 0.0206

Now for 90% confidence interval

Interval = p^ +/- z*sqrt((p^(1-p^)/(n+4))

Here for 90% confidence interval, z value is 1.645

= 0.0206 +/- 1.645*sqrt(0.0206(1-0.0206)/(336+4))

= 0.0206 +/- 0.0127

= (0.0206 - 0.0127 , 0.0206 + 0.0127)

= (0.0079 , 0.0333)

Interval = (0.008 , 0.033)