1. A random sample of 55 trading days is taken and the volume of a particular st
ID: 3309794 • Letter: 1
Question
1. A random sample of 55 trading days is taken and the volume of a particular stock is recorded. A 95% confidence interval for m is calculated to be (35.01, 44.53). Which of the following is a correct interpretation of this interval?
A. We are 95% confident that the distribution of sample mean trading volumes varies between 35.01 and 44.53 shares.
B. The probability that the population mean trading volume is between 35.01 and 44.53 shares is 95%.
C. We are 95% confident that the population mean trading volume is between 35.01 and 44.53 shares.
D. The probability that a sample of size 55 results in a population mean trading volume between 35.01 and 44.53 is 0.95.
2. Compute E, the margin of error for the Z interval, for a 95% confidence interval for m if a random sample of size n = 40 with a sample mean 25 is drawn from a population with s = 2
A. 0.52 B. 1.645 C. 0.62 D. cannot be computed because the Z interval does not apply
Explanation / Answer
1.
c.
We are 95% confident that the population mean trading volume is between 35.01 and 44.53 shares.
interpretations:
1) we are 95% sure that the interval [ 35.01,44.53 ] 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
2.
TRADITIONAL METHOD
given that,
standard deviation, =2
sample mean, x =25
population size (n)=40
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2/ sqrt ( 40) )
= 0.316
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.316
= 0.62
III.
CI = x ± margin of error
confidence interval = [ 25 ± 0.62 ]
= [ 24.38,25.62 ]
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DIRECT METHOD
given that,
standard deviation, =2
sample mean, x =25
population size (n)=40
level of significance, = 0.05
from standard normal table, two tailed z /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 = [ 25 ± Z a/2 ( 2/ Sqrt ( 40) ) ]
= [ 25 - 1.96 * (0.316) , 25 + 1.96 * (0.316) ]
= [ 24.38,25.62 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [24.38 , 25.62 ] 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
[ANSWERS]
best point of estimate = mean = 25
standard error =0.316
z table value = 1.96
margin of error = 0.62
confidence interval = [ 24.38 , 25.62 ]
option :C
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