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Figure the standard deviation of the distribution of means for a population with

ID: 3022800 • Letter: F

Question

Figure the standard deviation of the distribution of means for a population witha, standard deviation of 20 and sample sizes of (a) 10. (b) 11, (c) 100. and (d) 10] ADVANCED TOPIC: Figure the 95% confidence interval (that is, the lower and upper confidence limits) for each part of problem 13. Assume that in each case the researcher's sample has a mean of 80 and the population of individuals is known to follow a normal curve. ADVANCED TOPIC. Figure the 99% confidence interval (that is, the loner - and upper confidence limits) for each part of problem 14. Assume that in each case the researcher's sample has a mean of 50 and that the population of individuals is known to follow' a normal curve.

Explanation / Answer

16.

a)

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 =    50          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    20          
n = sample size =    10          
              
Thus,              
Margin of Error E =    16.29097493          
Lower bound =    33.70902507          
Upper bound =    66.29097493          
              
Thus, the confidence interval is              
              
(   33.70902507   ,   66.29097493   ) [ANSWER]

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

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 =    0.005          
X = sample mean =    50          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    20          
n = sample size =    11          
              
Thus,              
Margin of Error E =    15.53283513          
Lower bound =    34.46716487          
Upper bound =    65.53283513          
              
Thus, the confidence interval is              
              
(   34.46716487   ,   65.53283513   ) [ANSWER]

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

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 =    0.005          
X = sample mean =    50          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    20          
n = sample size =    100          
              
Thus,              
Margin of Error E =    5.151658607          
Lower bound =    44.84834139          
Upper bound =    55.15165861          
              
Thus, the confidence interval is              
              
(   44.84834139   ,   55.15165861   ) [ANSWER]

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

d)

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 =    50          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    20          
n = sample size =    101          
              
Thus,              
Margin of Error E =    5.126091905          
Lower bound =    44.87390809          
Upper bound =    55.12609191          
              
Thus, the confidence interval is              
              
(   44.87390809   ,   55.12609191   ) [ANSWER]

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