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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".
[tex]p_A[/tex] represent the real population proportion for brand A
[tex]\hat p_A =\frac{80}{2000}=0.04[/tex] represent the estimated proportion for the new process
[tex]n_A=2000[/tex] is the sample size required for Brand A
[tex]p_B[/tex] represent the real population proportion for brand b
[tex]\hat p_B =\frac{75}{1500}=0.05[/tex] represent the estimated proportion for the before process
[tex]n_B=1500[/tex] is the sample size required for before process
[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_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}}[/tex]
For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex](0.04 -0.05) - 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=-0.022[/tex]
[tex](0.04 -0.05) + 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=0.0017[/tex]
And the 90% confidence interval would be given (-0.022;0.0017).
Using the z-distribution, it is found that the 90% confidence interval for the true difference in the proportion of defectives between the existing and the new process is (-0.0017, 0.0217).
Samples:
For the sample of items from the existing process, we have that:
[tex]n_E = 1500, p_E = \frac{75}{1500} = 0.05, s_E = \sqrt{\frac{0.05(0.95)}{1500}} = 0.0056[/tex]
For the sample of items from the new process, we have that:
[tex]n_N = 2000, p_N = \frac{80}{2000} = 0.04, s_N = \sqrt{\frac{0.04(0.96)}{2000}} = 0.0044[/tex]
Distribution of differences:
For the distribution of differences, the mean and the standard error are given by:
[tex]\overline{x} = p_E - p_N = 0.05 - 0.04 = 0.01[/tex]
[tex]s = \sqrt{s_E^2 + s_N^2} = \sqrt{0.0056^2 + 0.0044^2} = 0.0071[/tex]
Interval:
The confidence interval is:
[tex]\overline{x} \pm zs[/tex]
- 90% confidence level, hence using a z-distribution calculator, the critical value is z = 1.645.
Then:
[tex]\overline{x} - zs = 0.01 - 1.645(0.0071) = -0.0017[/tex]
[tex]\overline{x} + zs = 0.01 + 1.645(0.0071) = 0.0217[/tex]
The 90% confidence interval for the true difference in the proportion of defectives between the existing and the new process is (-0.0017, 0.0217).
To learn more about the z-distribution, you can take a look at https://brainly.com/question/25779801
