Say we have a population distribution N4 From this populatio

Say we have a population distribution N(,4). From this population, we have a random sample of size 30 with sample mean being 2.19. Construct 90%, 95%, and 99% confidence intervals of . Compare the length of these intervals, tell me your conclusion.

Solution

Note: I assumed the second entry \"4\" is the variance, so I used standard deviation = 2.

a)

90% confidence:

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 =    0.05          
X = sample mean =    2.19          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    2          
n = sample size =    30          
              
Thus,              
Margin of Error E =    0.600615624          
Lower bound =    1.589384376          
Upper bound =    2.790615624          
              
Thus, the confidence interval is              
              
(   1.589384376   ,   2.790615624   ) [ANSWER]

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b)

95% confidence:

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 =    2.19          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2          
n = sample size =    30          
              
Thus,              
Margin of Error E =    0.715677657          
Lower bound =    1.474322343          
Upper bound =    2.905677657          
              
Thus, the confidence interval is              
              
(   1.474322343   ,   2.905677657   ) [ANSWER]

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c)

99% confidence:

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 =    0.005          
X = sample mean =    2.19          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    2          
n = sample size =    30          
              
Thus,              
Margin of Error E =    0.940559876          
Lower bound =    1.249440124          
Upper bound =    3.130559876          
              
Thus, the confidence interval is              
              
(   1.249440124   ,   3.130559876   ) [ANSWER]

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As we can see, the larger the confidence level, the wider the confidence interval.

Thus, larger confidence levels yield wider confidence intervals.