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?

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".

[tex]p_1[/tex] represent the real population proportion for 1

[tex]\hat p_1 =\frac{20}{400}=0.05[/tex] represent the estimated proportion for sample 1

[tex]n_1=400[/tex] is the sample size required for 1

[tex]p_2[/tex] represent the real population proportion for 2

[tex]\hat p_2 =\frac{20}{200}=0.1[/tex] represent the estimated proportion for 2

[tex]n_2=200[/tex] is the sample size required for Brand B

[tex]z[/tex] represent the critical value for the margin of error

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

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

[tex](\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}}[/tex]

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

[tex]z_{\alpha/2}=1.96[/tex]

And replacing into the confidence interval formula we got:

[tex](0.05-0.1) - 1.96 \sqrt{\frac{0.05(1-0.05)}{400} +\frac{0.1(1-0.1)}{200}}=-0.0967[/tex]

[tex](0.05-0.1) +1.96 \sqrt{\frac{0.05(1-0.05)}{400} +\frac{0.1(1-0.1)}{200}}=-0.00326[/tex]

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 [tex]-0.0967 \leq p_1 -p_2 \leq -0.00326[/tex]

And the interval not contains the 0.