A random sample of 25 was drawn from a normal distribution

. A random sample of 25 was drawn from a normal distribution whose standard deviation is 5. The sample mean is 80. Determine the 95% confidence interval estimate of the population mean.

b. Repeat Part a with a sample size of 100.

c. Repeat Part a with a sample size of 400.

d. Describe what happens to the confidence interval estimate when the sample size increases.

Solution

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 =    80          
t(alpha/2) = critical t for the confidence interval =    2.063898562          
s = sample standard deviation =    5          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
Margin of Error E =    2.063898562          
Lower bound =    77.93610144          
Upper bound =    82.06389856          
              
Thus, the confidence interval is              
              
(   77.93610144   ,   82.06389856   ) [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    80          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    5          
n = sample size =    100          
              
Thus,              
Margin of Error E =    0.979981992          
Lower bound =    79.02001801          
Upper bound =    80.97998199          
              
Thus, the confidence interval is              
              
(   79.02001801   ,   80.97998199   ) [ANSWER]

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c)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    80          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    5          
n = sample size =    400          
              
Thus,              
Margin of Error E =    0.489990996          
Lower bound =    79.510009          
Upper bound =    80.489991          
              
Thus, the confidence interval is              
              
(   79.510009   ,   80.489991   ) [ANSWER]

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

d)

As we can see, as sample size increases, the confidence interval becomes narrower. [ANSWER]

. A random sample of 25 was drawn from a normal distribution whose standard deviation is 5. The sample mean is 80. Determine the 95% confidence interval estimat
. A random sample of 25 was drawn from a normal distribution whose standard deviation is 5. The sample mean is 80. Determine the 95% confidence interval estimat

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