During a blood-donor program conducted during finals week for college students,

Question

During a blood-donor program conducted during finals week for college students, a blood-pressure reading is taken first, revealing that out of 200 donors, 29 have hypertension. All answers to three places after the decimal. A 95% confidence interval for the true proportion of college students with hypertension during finals week is ( Incorrect: Your answer is incorrect. , ). We can be 80% confident that the true proportion of college students with hypertension during finals week is with a margin of error of . Unless our sample (of 200 donors) is among the most unusual 10% of samples, the true proportion of college students with hypertension during finals week is between and . The probability, at 60% confidence, that a given college donor will have hypertension during finals week is , with a margin of error of . Assuming our sample of donors is among the most typical half of such samples, the true proportion of college students with hypertension during finals week is between and . We are 99% confident that the true proportion of college students with hypertension during finals week is , with a margin of error of . Assuming our sample of donors is among the most typical 99.9% of such samples, the true proportion of college students with hypertension during finals week is between and . Covering the worst-case scenario, how many donors must we examine in order to be 95% confident that we have the margin of error as small as 0.01? Using a prior estimate of 15% of college-age students having hypertension, how many donors must we examine in order to be 99% confident that we have the margin of error as small as 0.01?

Explanation / Answer

During a blood-donor program conducted during finals week for college students, a blood-pressure reading is taken first, revealing that out of 200 donors, 29 have hypertension. All answers to three places after the decimal. A 95% confidence interval for the true proportion of college students with hypertension during finals week is ( Incorrect: Your answer is incorrect. , ).

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.145          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024897289          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.04879779          
lower bound = p^ - z(alpha/2) * sp =   0.09620221          
upper bound = p^ + z(alpha/2) * sp =    0.19379779          
              
Thus, the confidence interval is              
              
(   0.09620221   ,   0.19379779   ) [ANSWER]

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We can be 80% confident that the true proportion of college students with hypertension during finals week is with a margin of error of .

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.145          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024897289          
              
Now, for the critical z,              
alpha/2 =   0.1          
Thus, z(alpha/2) =    1.281551566          
Thus,              
Margin of error = z(alpha/2)*sp =    0.03190716   [ANSWER]

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Unless our sample (of 200 donors) is among the most unusual 10% of samples, the true proportion of college students with hypertension during finals week is between and .

This means 1 - 0.10 = 0.90 or 90% confidence.

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.145          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024897289          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.040952396          
lower bound = p^ - z(alpha/2) * sp =   0.104047604          
upper bound = p^ + z(alpha/2) * sp =    0.185952396          
              
Thus, the confidence interval is              
              
(   0.104047604   ,   0.185952396   ) [ANSWER]

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The probability, at 60% confidence, that a given college donor will have hypertension during finals week is , with a margin of error of .

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.145          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024897289          
              
Now, for the critical z,              
alpha/2 =   0.2          
Thus, z(alpha/2) =    0.841621234          
Thus,              
Margin of error = z(alpha/2)*sp =    0.020954087          
lower bound = p^ - z(alpha/2) * sp =   0.124045913          
upper bound = p^ + z(alpha/2) * sp =    0.165954087          
              
Thus, the confidence interval is              
              
(   0.124045913   ,   0.165954087   ) [ANSWER]

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