Suppose that we will take a random sample of size n from a population having mea
ID: 3437441 • Letter: S
Question
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?
Explanation / Answer
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
******************
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 )
***********************
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 )
******************
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 )
***************************
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 )
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