(Statistics) How to Find 95% Confidence Interval from a distribution table?

Question

This table indicates the break down of what operating system people are using in America and Australia.

Assuming that the relevant proportion is the same in both Australia and America, use this data to find an approximate 95% confidence interval for the percentage of people in these countries who use an alternative operating system. Interpret this interval

I found the percentages by for America and Australia (142/246 and 84/154 respectively) but i'm not sure what to do after this. Do I have to find the standard deviation and sub into the 95% confidence interval formula?

TIA

Answer 1

I assume you want two confidence intervals, one for America and one for Australia.

For America: The traditional 95% CI for the binomial proportion $p$ is based on $\hat p = 142/246$ and $n = 246.$ The interval is $$\hat p \pm 1.96 \sqrt{\frac{\hat p(1-\hat p)}{n}},$$ which computes to $(0.516, 0.639)$.

A somewhat more accurate 95% CI is the 'Agresti` or 'Plus Four' CI, based on $\tilde n = n + 4$ and $\tilde p = (142 + 2)/\tilde n$. The interval is $$\tilde p \pm 1.96 \sqrt{\frac{\tilde p(1-\tilde p)}{\tilde n}},$$ which computes to $(0.515, 0.637)$.

Computations in R statistical software are shown below:

pm = c(-1,1)
n = 246;  p.h = 142/246
p.h + pm*1.96*sqrt(p.h*(1-p.h)/n)
## 0.5155032 0.6389683
n.t = 250;  p.t = 144/250
p.t + pm*1.96*sqrt(p.t*(1-p.t)/n.t)
## 0.5147395 0.6372605

In some countries the 'Wilson' interval is used. For 95% CIs it gives results very similar to Agresti's. See Wikipedia, or perhaps your text.

For Australia: CIs can be computed similarly.

Comparison: If you want to compare the two proportions to see if they differ between countries, there are several ways to do this. Here is output from Minitab statistical software which includes an approximated test and CI based on normal approximation and 'Fisher's exact test' based on a hypergeometric distribution.

There is no evidence of a different in rates between the two countries: (a) the CI includes 0; (b) the P-values of the tests are far above 5%.

MTB > PTwo 246 142 154 84;
SUBC>   Confidence 95.0;
SUBC>   Test 0.0;
SUBC>   Alternative 0.  # Two-sided test

Test and CI for Two Proportions 

Sample    X    N  Sample p
1       142  246  0.577236
2        84  154  0.545455

Difference = p (1) - p (2)
Estimate for difference:  0.0317812
95% CI for difference:  (-0.0681956, 0.131758)
Test for difference = 0 (vs ≠ 0):  Z = 0.62  P-Value = 0.533

Fisher’s exact test: P-Value = 0.536