Suppose that we will take a random sample of size n from a p

Suppose that we will take a random sample of size n from a population having mean u and standard deviation o. For each of the following situations, find the mean, variance, and standard deviation of the sampling distribution of the sample mean x.

a. u = 10   o = 2 , n = 25

b. u = 500 , o = .5 , n = 100

c. u = 3 , o = .1 , n = 4

d. u = 100 , o = 1, n= 1,600

&

Find an interval that contains (approximately or exactly) 99.73 percent of all the possible sample means, In which cases must we assume that the population is normally distributed? Why?

Solution

Note that the mean, variance, and standard deviation of the sampling distribution of the sample mean x is given by

u(X) = u
variance(X) = sigma^2 / n
sigma(X) = sigma / sqrt(n)

Thus:

a: u(X) = 10, variance(X) = 0.16, sigma(X) = 0.4
b: u(X) = 500, variance(X) = 0.0025, sigma(X) = 0.05
c: u(X) = 10, variance(X) = 0.0025, sigma(X) = 0.05
d: u(X) = 10, variance(X) = 0.000625, sigma(X) = 0.025

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A. As n = 25 < 30, we assume t distribution only.

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    10          
t(alpha/2) = critical z for the confidence interval =    3.344721741          
s = sample standard deviation =    2          
n = sample size =    25          
              
Thus,              
              
Lower bound =    8.662111303          
Upper bound =    11.3378887          
              
Thus, the confidence interval is              
              
(   8.662111303   ,   11.3378887   )

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B.

As n = 100 > 30, then we can assume z distribution.

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    500          
z(alpha/2) = critical z for the confidence interval =    2.999976993          
s = sample standard deviation =    0.5          
n = sample size =    100          
              
Thus,              
              
Lower bound =    499.8500012          
Upper bound =    500.1499988          
              
Thus, the confidence interval is              
              
(   499.8500012   ,   500.1499988   )

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C. As n = 4 < 30, we only use t distribution.

USING T DISTRIBUTION              
              
Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    3          
t(alpha/2) = critical z for the confidence interval =    9.218701822          
s = sample standard deviation =    0.1          
n = sample size =    4          
              
Thus,              
              
Lower bound =    2.539064909          
Upper bound =    3.460935091          
              
Thus, the confidence interval is              
              
(   2.539064909   ,   3.460935091   )

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D. As n = 1600 > 30, then we can assume z distribution.

USING Z DISTRIBUTION              
              
Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    100          
z(alpha/2) = critical z for the confidence interval =    2.999976993          
s = sample standard deviation =    1          
n = sample size =    1600          
              
Thus,              
              
Lower bound =    99.92500058          
Upper bound =    100.0749994          
              
Thus, the confidence interval is              
              
(   99.92500058   ,   100.0749994   )