Respuesta :
Answer:
[tex]ME=z_{\alpha/2} SE= 2.58*0.082=0.21156[/tex]
The 99% confidence interval would be given (0.09844;0.5216).
We are confident at 99% that the difference between the two proportions is between [tex]0.09844 \leq p_B -p_A \leq 0.5216[/tex]
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".
[tex]p_A[/tex] represent the real population proportion for plant A
[tex]\hat p_A =0.75[/tex] represent the estimated proportion for plant A
[tex]n_A[/tex] is the sample size required for plant A
[tex]p_B[/tex] represent the real population proportion for plant b
[tex]\hat p_B =0.44[/tex] represent the estimated proportion for plant B
[tex]n_B[/tex] is the sample size required for plant 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_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 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
The standard error is given by:
[tex]SE=z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}=0.082[/tex]
The margin of error is:
[tex]ME=z_{\alpha/2} SE= 2.58*0.082=0.21156[/tex]
And replacing into the confidence interval formula we got:
[tex](0.75-0.44) - 2.58(0.082)=0.09844[/tex]
[tex](0.75 -0.44) + 2.58(0.082)=0.5216[/tex]
And the 99% confidence interval would be given (0.09844;0.5216).
We are confident at 99% that the difference between the two proportions is between [tex]0.09844 \leq p_B -p_A \leq 0.5216[/tex]
