I really don't get these answers but if you could help me that'll mean a lot thank you for answering my question if you did
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Answer:
The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).
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.
Male undergraduates:
670, 388 were employed. So
Female undergraduates:
Of 617, 323 were employed. So
Distribution of the difference:
Confidence interval:
The confidence interval is given by:
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 between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).