Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
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
[tex]p_M = \frac{388}{670} = 0.5791[/tex]
[tex]s_M = \sqrt{\frac{0.5791*0.4209}{670}} = 0.0191[/tex]
Female undergraduates:
Of 617, 323 were employed. So
[tex]p_F = \frac{323}{617} = 0.5235[/tex]
[tex]s_F = \sqrt{\frac{0.5235*0.4765}{617}} = 0.0201[/tex]
Distribution of the difference:
[tex]p = p_M - p_F = 0.5791 - 0.5235 = 0.0556[/tex]
[tex]s = sqrt{s_M^2+s_F^2} = \sqrt{0.0201^2 + 0.0191^2} = 0.0277[/tex]
Confidence interval:
The confidence interval is given by:
[tex]p \pm zs[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower bound of the interval is:
[tex]p - zs = 0.0556 - 1.645*0.0277 = 0.01[/tex]
The upper bound of the interval is:
[tex]p + zs = 0.0556 + 1.645*0.0277 = 0.1012[/tex]
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).
