The concentration of lead in blood was measured in a sample of 21 children from

Question

The concentration of lead in blood was measured in a sample of 21 children from a large school in Flint, MI. The sample mean was calculated to be 46.12 ng/mL and the standard deviation was 5.8 ng/mL.

Of the choices below, what is the maximum level of confidence (i.e., 95% confidence is greater than 90% confidence) that you can say that the mean concentration in these students’ blood is less than the CDC’s recommended maximum blood concentration of 50 ng/mL?

HINT: remember to calculate your values with the correct number of significant figures before answering.

Explanation / Answer

FOR 90% CONFIDENCE:

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    46.12          
t(alpha/2) = critical t for the confidence interval =    1.724718243          
s = sample standard deviation =    5.8          
n = sample size =    21          
df = n - 1 =    20          
Thus,              
Margin of Error E =    2.182913382          
Lower bound =    43.93708662          
Upper bound =    48.30291338          
              
Thus, the confidence interval is              
              
(   43.93708662   ,   48.30291338   ) [LESS THAN 50!]

*********************

FOR 95% CONFIDENCE:

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    46.12          
t(alpha/2) = critical t for the confidence interval =    2.085963447          
s = sample standard deviation =    5.8          
n = sample size =    21          
df = n - 1 =    20          
Thus,              
Margin of Error E =    2.640128347          
Lower bound =    43.47987165          
Upper bound =    48.76012835          
              
Thus, the confidence interval is              
              
(   43.47987165   ,   48.76012835   ) [LESS THAN 50]

***********************

for 99% confidence:

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    46.12          
t(alpha/2) = critical t for the confidence interval =    2.84533971          
s = sample standard deviation =    5.8          
n = sample size =    21          
df = n - 1 =    20          
Thus,              
Margin of Error E =    3.601243365          
Lower bound =    42.51875663          
Upper bound =    49.72124337          
              
Thus, the confidence interval is              
              
(   42.51875663   ,   49.72124337   ) [STILL LESS THAN 50!]

****************************

Hence, we can say it is less than 50 at A MAXIMUM OF 99% CONFIDENCE. [ANSWER]

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