12. A publishing company has just published a new college textbook. Before the c

Question

12. A publishing company has just published a new college textbook. Before the company decides the price at which to sell this textbook, it wants to know the average price of all such textbooks in the market. The research department at the company took a sample of 36 comparable textbooks and collected information on their prices. This information produced a mean price of $70.50 for this sample. It is known that the standard deviation of the prices of all such textbooks is $4.50. Construct a 90% confidence interval for the mean price of all such college textbooks 13. Twenty-five randomly selected adults who buy books for general reading were asked how much they usually spend on books per year. The sample produced a mean of $1450 and a standard deviation of $300 for such annual expenses. Assume that such expenses for all adults who buy books for general reading have an approximate normal distribution. Determine a 99% confidence interval for the corresponding population mean. 14. A machine at Katz Steel Corporation makes 3-inch-long nails. The probability distribution of the lengths of these nails is normal with a mean of 3 inches and a standard deviation of .1 inch. The quality control inspector takes a sample of 25 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than 2.95 inches or greater than 3.05 inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 25 nails the inspector will conclude that the machine needs an adjustment?

Explanation / Answer

12.
TRADITIONAL METHOD
given that,
standard deviation, =4.5
sample mean, x =70.5
population size (n)=36
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 4.5/ sqrt ( 36) )
= 0.75
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.75
= 1.234
III.
CI = x ± margin of error
confidence interval = [ 70.5 ± 1.234 ]
= [ 69.266,71.734 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =4.5
sample mean, x =70.5
population size (n)=36
level of significance, = 0.1
from standard normal table, two tailed z /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 = [ 70.5 ± Z a/2 ( 4.5/ Sqrt ( 36) ) ]
= [ 70.5 - 1.645 * (0.75) , 70.5 + 1.645 * (0.75) ]
= [ 69.266,71.734 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [69.266 , 71.734 ] 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
[ANSWERS]
best point of estimate = mean = 70.5
standard error =0.75
z table value = 1.645
margin of error = 1.234
confidence interval = [ 69.266 , 71.734 ]

13.
TRADITIONAL METHOD
given that,
sample mean, x =1450
standard deviation, s =300
sample size, n =25
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 300/ sqrt ( 25) )
= 60
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 2.797
margin of error = 2.797 * 60
= 167.82
III.
CI = x ± margin of error
confidence interval = [ 1450 ± 167.82 ]
= [ 1282.18 , 1617.82 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1450
standard deviation, s =300
sample size, n =25
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 2.797
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 1450 ± t a/2 ( 300/ Sqrt ( 25) ]
= [ 1450-(2.797 * 60) , 1450+(2.797 * 60) ]
= [ 1282.18 , 1617.82 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 1282.18 , 1617.82 ] 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

14.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 3
standard Deviation ( sd )= 0.1/ Sqrt ( 25 ) =0.02
sample size (n) = 25
probability that a sample of 25 nails the inspector conclude that the machine needs an adhustment
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 2.95) = (2.95-3)/0.1/ Sqrt ( 25 )
= -0.05/0.02= -2.5
= P ( Z <-2.5) From Standard NOrmal Table
= 0.0062
P(X > 3.05) = (3.05-3)/0.1/ Sqrt ( 25 )
= 0.05/0.02 = 2.5
= P ( Z >2.5) From Standard Normal Table
= 0.0062
P( X < 2.95 OR X > 3.05) = 0.0062+0.0062 = 0.0124

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