For anybody who has had this question 0 and 5 :(
1 answer:
You might be interested in
Answer:
The 90% confidence interval for the difference of proportions is (0.01775,0.18225).
Step-by-step explanation:
Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
p1 -> 1993
20 out of 100, so:
p2 -> 1997
10 out of 100, so:
Distribution of p1 – p2:
Confidence interval:
In which
z is the z-score that has a p-value of .
90% confidence level
So , z is the value of Z that has a p-value of
, so
.
The lower bound of the interval is:
The upper bound of the interval is:
The 90% confidence interval for the difference of proportions is (0.01775,0.18225).