Respuesta :
Answer:
[tex]SE=\sqrt{\frac{0.505 (1-0.505)}{832}+\frac{0.0311(1-0.0311)}{145}}=0.0226[/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".
1) Data given and notation
[tex]X_{1}=420[/tex] represent the number of cats diagnosing malignant lymphoma from homes where herbicides were used regularly
[tex]X_{2}=17[/tex] represent the number of cats diagnosing malignant lymphoma from homes where NO herbicides were used regularly
[tex]n_{1}=832[/tex] sample 1 selected
[tex]n_{2}=145[/tex] sample 2 selected
[tex]\hat p_{1}=\frac{420}{832}=0.505[/tex] represent the proportion of of cats diagnosing malignant lymphoma from homes where herbicides were used regularly
[tex]\hat p_{2}=\frac{17}{145}=0.0311[/tex] represent the proportion of cats diagnosing malignant lymphoma from homes where NO herbicides were used regularly
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]p_1 -p_2[/tex] parameter of interest
2) Solution to the problem
We are interested on the standard error for the difference of proportions and is given by this formula:
[tex]SE=\sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_{1}}+\frac{\hat p_2 (1-\hat p_2)}{n_{2}}[/tex]
And if we replace the values given we got:
[tex]SE=\sqrt{\frac{0.505 (1-0.505)}{832}+\frac{0.0311(1-0.0311)}{145}}=0.0226[/tex]
