Question

A random sample of 10 venture capital investments in the fiber optics business sector yielded the following data, in millions of dollars. Assume that the population standard deviation is $2.02 million. Complete parts (a) through (d) below. (Note: The sum of the data is $57.42 million.) 5.19 2.76 2.31 9.55 10.03 3.42 5.63 5.44 5,48 7.61 a. Find a 99% confidence interval for the mean amount, , of all venture capital investments in the fiber optics business sector. The confidence interval for p is from $ million to $ million. (Round to two decimal places as needed.) b. The 95% confidence interval is from $4.49 million to $6.99 million. Why is the confidence interval in part (a) longer than this confidence interval? Compared to the 95% confidence interval, the value of is in the 99% confidence interval. c. Draw a graph to display both confidence intervals. OA. OB. 96% CI for 99% CI for 99% CI for 95% CI for 7.5 5.5 Oc. OD. 96% CI 95% CI for for 99% CI 99% CI for for Click to select your answer(s).

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.02
sample mean, x =5.742
population size (n)=10
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.02/ sqrt ( 10) )
= 0.64
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.64
= 1.65
III.
CI = x ± margin of error
confidence interval = [ 5.742 ± 1.65 ]
= [ 4.1,7.39 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.02
sample mean, x =5.742
population size (n)=10
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 5.742 ± Z a/2 ( 2.02/ Sqrt ( 10) ) ]
= [ 5.742 - 2.576 * (0.64) , 5.742 + 2.576 * (0.64) ]
= [ 4.1,7.39 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [4.1 , 7.39 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.02
sample mean, x =5.742
population size (n)=10
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.02/ sqrt ( 10) )
= 0.64
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.64
= 1.25
III.
CI = x ± margin of error
confidence interval = [ 5.742 ± 1.25 ]
= [ 4.49,6.99 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.02
sample mean, x =5.742
population size (n)=10
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 5.742 ± Z a/2 ( 2.02/ Sqrt ( 10) ) ]
= [ 5.742 - 1.96 * (0.64) , 5.742 + 1.96 * (0.64) ]
= [ 4.49,6.99 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [4.49 , 6.99 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
99% sure that the interval [4.1 , 7.39 ]
95% sure that the interval [4.49 , 6.99 ]
option:A
diagram shows