Question

A manufacturer receives parts from two suppliers. A simple random sample of 400 parts from Supplier 1 finds 20 defective. A simple random sample of 200 parts from Supplier 2 finds 20 defective. Let p1 and p2 be the proportion of all parts from Supplier 1 and Supplier 2, respectively, that are defective. Would a 95% confidence interval for p1 – p2 contain the value zero?

Answer 1

Answer:

We are confident at 95% that the difference between the two proportions is between $-0.0967 \leq p_1 -p_2 \leq -0.00326$

And the interval not contains the 0.  

Step-by-step explanation:

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".  

$p_1$ represent the real population proportion for 1

$\hat p_1 =\frac{20}{400}=0.05$ represent the estimated proportion for sample 1

$n_1=400$ is the sample size required for 1

$p_2$ represent the real population proportion for 2  

$\hat p_2 =\frac{20}{200}=0.1$ represent the estimated proportion for  2

$n_2=200$ is the sample size required for Brand B

$z$ 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  

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.05-0.1) - 1.96 \sqrt{\frac{0.05(1-0.05)}{400} +\frac{0.1(1-0.1)}{200}}=-0.0967$  

$(0.05-0.1) +1.96 \sqrt{\frac{0.05(1-0.05)}{400} +\frac{0.1(1-0.1)}{200}}=-0.00326$  

And the 95% confidence interval would be given (-0.0967;-0.00326).  

We are confident at 95% that the difference between the two proportions is between $-0.0967 \leq p_1 -p_2 \leq -0.00326$

And the interval not contains the 0.