Answer:
CI = (0.674, 0.748)
Step-by-step explanation:
The confidence interval of a proportion is:
CI = p ± SE × CV,
where p is the proportion, SE is the standard error, and CV is the critical value (either a t-score or a z-score).
We already know the proportion:
p = 293/412
p = 0.711
But we need to find the standard error and the critical value.
The standard error is:
SE = √(p (1 − p) / n)
SE = √(0.711 × (1 − 0.711) / 412)
SE = 0.0223
To find the critical value, we must first find the alpha level and the degrees of freedom.
The alpha level for a 90% confidence interval is:
α = (1 − 0.90) / 2 = 0.05
The degrees of freedom is one less than the sample size:
df = 412 − 1 = 411
Since df > 30, we can approximate this with a normal distribution.
If we look up the alpha level in a z score table or with a calculator, we find the z-score is 1.645. That's our critical value. CV = 1.645.
Now we can find the confidence interval:
CI = 0.711 ± 0.0223 * 1.645
CI = 0.711 ± 0.0367
CI = (0.674, 0.748)
So we are 90% confident that the proportion of adults connected to the internet from home is between 0.674 and 0.748.
