A simple random sample of size n is drawn. The sample mean, x is found to be 17.

Question

A simple random sample of size n is drawn. The sample mean, x is found to be 17.8,and the sample standard deviation,s,is found to be 4,7 a.construct a 95%confidence interval about mu if the sample size, n, is 34. b.construct a 95% confidence interval about mu if the sample size,n,is 51. How does increasing the sample size affect the margin of error, E? construct a 99% confidence interval about mu if the sample size,n,is 34. compare the results to those obtained in part How does the level of affect the size of the margin of erres, E?

Explanation / Answer

a)

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 =    17.8          
t(alpha/2) = critical t for the confidence interval =    2.034515297          
s = sample standard deviation =    4.7          
n = sample size =    34          
df = n - 1 =    33          
Thus,              
Margin of Error E =    1.639907526          
Lower bound =    16.16009247          
Upper bound =    19.43990753          
              
Thus, the confidence interval is              
              
(   16.16009247   ,   19.43990753   ) [ANSWER]

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b)

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 =    17.8          
t(alpha/2) = critical t for the confidence interval =    2.008559112          
s = sample standard deviation =    4.7          
n = sample size =    51          
df = n - 1 =    50          
Thus,              
Margin of Error E =    1.321896301          
Lower bound =    16.4781037          
Upper bound =    19.1218963          
              
Thus, the confidence interval is              
              
(   16.4781037   ,   19.1218963   ) [ANSWER]

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OPTION B: The margin of error decreases.

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c)

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 =    17.8          
t(alpha/2) = critical t for the confidence interval =    2.733276642          
s = sample standard deviation =    4.7          
n = sample size =    34          
df = n - 1 =    33          
Thus,              
Margin of Error E =    2.203139462          
Lower bound =    15.59686054          
Upper bound =    20.00313946          
              
Thus, the confidence interval is              
              
(   15.59686054   ,   20.00313946   ) [ANSWER]

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Comparing parts a and c, we see a larger margin of error, so

OPTION A: The margin of error increases. [ANSWER]

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