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Answer:
The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not.
This means that [tex]n = 100, \pi = \frac{20}{100} = 0.2[/tex]
68% confidence level
So [tex]\alpha = 0.32[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.32}{2} = 0.84[/tex], so [tex]Z = 0.995[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 0.995\sqrt{\frac{0.2*0.8}{100}} = 0.16[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 + 0.995\sqrt{\frac{0.2*0.8}{100}} = 0.24[/tex]
The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).
The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).
What is the p-value?
The p-value is also known by the probability value. It is the probability of getting a result that is either the same or more than the actual value.
In a given data with a number n of people surveyed with a probability of success [tex]\pi[/tex] and a confidence level [tex]1-\alpha[/tex], we have the confidence interval of proportions.
[tex]\pi[/tex] ± [tex]\rm z\sqrt{\pi (1-\pi )} /n[/tex]
where z has a p-value of 1-[tex]\alpha /2[/tex]
A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not.
n = 100, [tex]\pi[/tex] = 20/100 = 0.2
So, [tex]\alpha[/tex] = 0.32 where z has a p-value 1-[tex]0.32 /2 = 0.84[/tex]
z = 0.995
The lower limit of this interval
[tex]\pi -\rm z\sqrt{\pi (1-\pi )} /n\\= 0.2 - 0.995 \sqrt{0.2 \times 0.8} /100\\= 0.16[/tex]
The upper limit of this interval
[tex]\pi -\rm z\sqrt{\pi (1-\pi )} /n\\= 0.2 +0.995 \sqrt{0.2 \times 0.8} /100\\= 0.24[/tex]
Hence, The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).
Learn more about p-value ;
https://brainly.com/question/15407907
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