Answer :
Answer:
The 90% confidence interval is (0.0131, 0.0845).
Step-by-step explanation:
Before finding 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.
Old process:
125 out of 580, so:
[tex]p_O = \frac{125}{580} = 0.2155[/tex]
[tex]s_O = \sqrt{\frac{0.2155*0.7845}{580}} = 0.0171[/tex]
New process:
130 out of 780. So
[tex]p_N = \frac{130}{780} = 0.1667[/tex]
[tex]s_N = \sqrt{\frac{0.1667*0.8333}{780}} = 0.0133[/tex]
Distribution of the difference:
[tex]p = p_O - p_N = 0.2155 - 0.1667 = 0.0488[/tex]
[tex]s = \sqrt{s_O^2+s_N^2} = \sqrt{0.0171^2 + 0.0133^2} = 0.0217[/tex]
Confidence interval:
[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 = 0.0488 - 1.645*0.0217 = 0.0131[/tex]
The upper bound of the interval is:
[tex]p + zs = 0.0488 + 1.645*0.0217 = 0.0845[/tex]
The 90% confidence interval is (0.0131, 0.0845).