Respuesta :
Answer:
a) [tex] ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019[/tex]
b) [tex] ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245[/tex]
c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of [tex]\hat p [/tex] the margin of error change for part a and b.
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
If solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
Part a
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=\pm 1.96[/tex]
If we replace the values into equation (a) for 1983 we got:
[tex] ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019[/tex]
Part b
Since is the same confidence level the z value it's the same.
If we replace the values into equation (a) for 2008 we got:
[tex] ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245[/tex]
Is the margin of error the same in parts (a) and (b)? Why or why not?
On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of [tex]\hat p [/tex] the margin of error change for part a and b.
The margin error is 0.019 and 0.0245 for the respective years 1983 , 2008.
What is Margin of error ?
The margin of error is the range of values below and above the sample statistic in a confidence interval.
In 1983, 87% of 19-year-olds had a driver’s license.
After 25 years the percentage had dropped to 75%.
these results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008.
At a 95% confidence,
Since our interval is at 95% of confidence, our significance level would be given by
∝ =1-0.95 = 0.05
∝/2 = 0.025
And the critical value would be given by:
[tex]\rm z_{\alpha/2}[/tex] = ±1.96
[tex]Margin \;Error = z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]
[tex]Margin \;Error = 1.96\sqrt{\dfrac{0.87(1-0.87)}{1200}}[/tex]
= 0.019
Since is the same confidence level the z value it's the same.
If we replace the values into equation for 2008 we got:
[tex]Margin \;Error = 1.96\sqrt{\dfrac{0.75(1-0.75)}{1200}}[/tex]
=0.0245
Margin Error is not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008.
To know more about Margin of Error
https://brainly.com/question/10501147
#SPJ5
