Download Chapter 8) 21. What is sampling error? Sample error is the

Chapter 8)
21. What is sampling error?
Sample error is the difference between a sample statistic and its corresponding population
parameter
Could the value of the sampling error be zero?
Yes, when the sample statistic is equal to its corresponding population parameter
If it were zero, what would this mean?
That means that our estimation (statistic value) is equal to the parameter
22. List the reasons for sampling.
1) To contact the whole population would be would be time consuming
2) The cost of studying all the items in a population may be prohibitive
3) The physical impossibility of checking all the items in the population
4) The destructive nature of some tests
5) The sample results are adequate
Give an example of each reason for sampling.
1) If we want to know the proportion of smokers on a population of a certain country we must
select a sample, if we ask every person the time involved is huge
2) If we want to test a certain drug we must select a sample because if we select the hole
population the cost is so expensive
3) If we want to study the bacterial level in a river we select some parts of the river, is impossible
to select all the river.
4) If we have a product that we want to sell we select some of them to prove them, we can´t test
all the units because we want to sell them
5) On some cases there is no a great difference if we consider the population or a sample, for
example if we want to study an index price of a product
34. Information from the American Institute of Insurance indicates the mean amount of life
insurance per household in the United States is $110,000. This distribution follows the normal
distribution with a standard deviation of $40,000.
a) If we select a random sample of 50 households, what is the standard error of the mean?
𝜎𝑋̅ =
𝜎
√𝑛
=
40,000
√50
=5,656.85
b) What is the expected shape of the distribution of the sample mean ?
Sample mean follows a normal distribution, (bell shaped) with mean = 110,000
and sd =
40,000
√50
=5,656.85
c) What is the likehood of selecting a sample with a mean of at least 112,000?
𝑋̅ has a normal distribution with mean = 110,000
and sd =
40,000
𝑋̅−110,000
√50
40,000
√50
, Z=
P(𝑋̅ ≥ 112,000) = 𝑃 (𝑍 ≥
has a standard normal distribution
112,000−110,000
40,000
√50
) = 𝑃(𝑍 ≥ 0.354) = 0.362
d) What is the likehood of selecting a sample with a mean of more than 100,000 but less than
112,000?
𝑋̅ has a normal distribution with mean = 110,000
and sd =
40,000
𝑋̅−110,000
√50
40,000
√50
, Z=
has a standard normal distribution
P(100,000 ≤ 𝑋̅ ≤ 112,000) = 𝑃 (
100,000−110,000
40,000
√50
≤𝑍≤
112,000−110,000
40,000
√50
)=
𝑃(−1.768 ≤ 𝑍 ≤ 0.354) = 0.638
Chapter 9)
32. A state meat inspector in Iowa has been given the assignment of estimating the mean net
weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights
cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01
pounds, with a standard deviation of 0.03 pounds.
a. What is the estimated population mean?
Answer: 3.01
b. Determine a 95 percent confidence interval for the population mean.
CI = 𝑥̅ ±t
𝑠
√𝑛
= 3.01± t
0.03
=3.01±
√36
2.03
0.03
=
√36
(3,3.02)
t=t(/2,n-1) = t (0.025,35)=2.03
Answer: (3,3.02)
34. A recent survey of 50 executives who were laid off from their previous position revealed it
took a mean of 26 weeks for them to find another position. The standard deviation of the sample
was 6.2 weeks. Construct a 95 percent confidence interval for the population mean.
CI = 𝑥̅ ±t
𝑠
√𝑛
= 26± t
6.2
√50
=26± 2.01
6.2
√50
= (24.238,27.762)
t=t(/2,n-1) = t (0.025,49)=2.01
Answer: (24.238,27.762)
Is it reasonable that the population mean is 28 weeks? Justify your answer.
No, because 28 is not in the CI
36. The American Restaurant Association collected information on the number of meals eaten
outside the home per week by young married couples. A survey of 60 couples showed the sample
mean number of meals eaten outside the home was 2.76 meals per week, with a standard
deviation of 0.75 meals per week. Construct a 97 percent confidence interval for the population
mean.
CI = 𝑥̅ ±t
𝑠
√𝑛
= 2.76± t
0.75
=2.76±
√60
t=t(/2,n-1) = t (0.015,59)=2.224
Answer: (24.238,27.762)
2.224
0.75
√60
= (2.751,2.769)