Question

A random sample of 121 checking bail bonds around the Court House in Houston showed an average hourly bail bond of $280. The standard deviation of the population is known to be $66.

a. Is it necessary to know anything about the shape of the distribution of the bail bond in order to make         an interval estimate of the mean of all the bail bonds? Explain.

b. Find the standard error of the mean.

c. Give a point estimate of the population mean.

d. Construct a 90% confidence interval estimates for the mean.

Answer 1

a.
yes,
it is necessary to know anything about the shape of the distribution of the bail bond in order to make an interval estimate of the mean of all the bail bonds
because A random sample of 121 checking bail bonds around the Court House in Houston.

TRADITIONAL METHOD
given that,
standard deviation, σ =66
sample mean, x =280
population size (n)=121
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 66/ sqrt ( 121) )
= 6
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 6
= 9.87
III.
CI = x ± margin of error
confidence interval = [ 280 ± 9.87 ]
= [ 270.13,289.87 ]
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DIRECT METHOD
given that,
standard deviation, σ =66
sample mean, x =280
population size (n)=121
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 280 ± Z a/2 ( 66/ Sqrt ( 121) ) ]
= [ 280 - 1.645 * (6) , 280 + 1.645 * (6) ]
= [ 270.13,289.87 ]
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interpretations:
1. we are 90% sure that the interval [270.13 , 289.87 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
b.
standard error = 6
c.
point estimate for mean is 280$
d.
90% sure that the interval [270.13 , 289.87 ]