Respuesta :
Answer:
a) Private colleges
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.749 - 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.724[/tex]
[tex]0.749 + 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.774[/tex]
The 95% confidence interval would be given by (0.724;0.774)
The width of the interval is: 0.774-0.724=0.05
Public colleges
[tex]0.719 - 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.680[/tex]
[tex]0.719 + 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.758[/tex]
The 95% confidence interval would be given by (0.680;0.758)
The width of the interval is: 0.758-0.680=0.078
So then we can conclude that for public colleges we have a wider interval
b) Private colleges
[tex] ME= 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}= 0.025[/tex]
Public colleges
[tex] ME= 1.96\sqrt{\frac{0.719(1-0.719)}{505}}= 0.039[/tex]
We see a larger margin of error for the public colleges
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
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}=-1.96, z_{1-\alpha/2}=1.96[/tex]
Private colleges
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.749 - 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.724[/tex]
[tex]0.749 + 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.774[/tex]
The 95% confidence interval would be given by (0.724;0.774)
The width of the interval is: 0.774-0.724=0.05
Public colleges
[tex]0.719 - 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.680[/tex]
[tex]0.719 + 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.758[/tex]
The 95% confidence interval would be given by (0.680;0.758)
The width of the interval is: 0.758-0.680=0.078
So then we can conclude that for public colleges we have a wider interval
Part b
Private colleges
[tex] ME= 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}= 0.025[/tex]
Public colleges
[tex] ME= 1.96\sqrt{\frac{0.719(1-0.719)}{505}}= 0.039[/tex]
We see a larger margin of error for the public colleges
In 2004 the ACT inc was reported to have a 74% of 1644 random selection of th colleges freshmen that retuned next year. The retention was 71.95 and 505 students enrolled in the public school 74.9% among reenrolled and 1139 students in private.
- As per the confidence interval is "a range of values that’s likely to the include a population value with the certain degree of confidence. It is often expressed as % in the lower and upper level.
- As per the margins of the error is the range of the values below and the above the sample statistic with the same confidence interval.
- The Normal distribution, is symmetric about the mean, showing that data near the mean are more frequent in data.
Learn more about the that 74% of 1644 randomly selected college.
brainly.com/question/14396460.