Question 9.8: Suppose 93 out of 104 pupils have passed the final examinati......
Suppose 93 out of 104 pupils have passed the final examination at a certain school.
(a) Calculate a 95%confidence interval for the probability of failing the examination both by manual calculations and by using R, and compare the results.
(b) At county level 3.2% of pupils failed the examination. Are the school’s pupils worse than those in the whole county?
If a student fails then x_{i} = 1.We know that \sum{x_{i}} = 11 and n = 104.
(a) Using \bar{x} = p as point estimate, we get \hat{p} =\frac{11}{ 104} ≈ 0.106 = 10.6 %. Using z_{{1−α}/{2}}= z_{0.975} = 1.96, we can calculate the confidence interval as
0.106 ± 1.96 ·\sqrt{\frac{0.106 · 0.894}{104} } = [0.047; 0.165] .
Using R we get:
binom.test(11,104)$conf.int
[1] 0.05399514 0.18137316
This result is different because the above command does not use the normal approximation. In addition, p is rather small which means that care must be exercised when using the results of the confidence interval with normal approximation in this context.
(b) The point estimate of 10.6%is substantially higher than 3.2 %. The lower bound confidence interval is still larger than the proportion of failures at county level. This indicates that the school is doing worse than most other schools in the county.