Respuesta :
Answer:
[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]
And for this case the confidence interval is given by: (0.275, 0.305)
We can estimate the proportion difference as:
[tex]\hat p_D = \frac{0.275+0.305}{2}=0.29[/tex]
And the margin of error would be:
[tex] ME=0.305-0.29=0.015[/tex]
So then for this case the possibl two options are:
We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%.
We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.
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 of Latinos who view global warming as a serious problem
[tex]\hat p_A=0.75 [/tex] represent the estimated proportion Latinos who view global warming as a serious problem
[tex]n_A[/tex] is the sample size required of Latinos who view global warming as a serious problem
[tex]p_B[/tex] represent the real population proportion of white who view global warming as a serious problem
[tex]\hat p_B =0.46[/tex] represent the estimated proportion of whitewho view global warming as a serious problem
[tex]n_B[/tex] is the sample size required of white
[tex]z[/tex] represent the critical value for the margin of error
Solution to the problem
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]
And for this case the confidence interval is given by: (0.275, 0.305)
We can estimate the proportion difference as:
[tex]\hat p_D = \frac{0.275+0.305}{2}=0.29[/tex]
And the margin of error would be:
[tex] ME=0.305-0.29=0.015[/tex]
So then for this case the possibl two options are:
We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%.
We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.
Using confidence interval concepts, the correct option is:
We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.
x% confidence interval:
A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.
In this problem:
The 95% confidence interval based on this sample difference of the proportions of Latinos and Whites is (0.275, 0.305).
- The sample proportion is the mean of these bounds, hence [tex]\overline{p} = \frac{0.275 + 0.305}{2} = 0.29[/tex].
- The margin of error is half the difference, hence: [tex]M = \frac{0.305 - 0.275}{2} = 0.015[/tex]
Hence, the correct option is:
We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.
A similar problem is given at https://brainly.com/question/24204959