Question

A certain change in a process for manufacturing component parts is being considered. Samples are taken under both the existing and the new process so as to determine if the new process results in an improvement. If 75 of 1500 items from the existing process are found to be defective and 80 of 2000 items from the new process are found to be defective, find a 90% confidence interval for the true difference in the proportion of defectives between the existing and the new process.

Answer 1

Answer:

And the 90% confidence interval for the difference would be given by:(-0.022;0.0017).  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for brand A  

$\hat p_A =\frac{80}{2000}=0.04$ represent the estimated proportion for the new process

$n_A=2000$ is the sample size required for Brand A

$p_B$ represent the real population proportion for brand b  

$\hat p_B =\frac{75}{1500}=0.05$ represent the estimated proportion for the before process

$n_B=1500$ is the sample size required for before process

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.64$  

And replacing into the confidence interval formula we got:  

 $(0.04 -0.05) - 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=-0.022$  

$(0.04 -0.05) + 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=0.0017$  

And the 90% confidence interval would be given (-0.022;0.0017).