1. science
2. mathematics
3. statistics

(Population Mean – Large Sample)
INTERVAL ESTIMATION
GIVEN:
A population with population mean µ and population standard deviation σ is
given. The population size is large and the population mean µ is unknown.
AIM:
To take a random sample of large sample size n (n ≥ 30) , with replacement or
without replacement, from the population and then find an interval which may
contain µ .
METHOD
1st step.
Select a random sample of large sample size n (n ≥ 30) from the population.
2nd step.
Compute the sample mean x and the sample standard deviation s.
3rd step.
Choose a number α which is between 0 and 1. Usually α = 0.05 so that
1 − α = 95 % or α = 0.10 so that 1 − α = 90 % .
σ
4th step.
Compute x ± z α / 2
5th step.
The interval with endpoints x ± z α / 2
n
.
interval for the population mean µ .
σ
n
is called a 100(1 − α)% confidence
If 1 − α = 95 % , the interval is called a 95% confidence interval for µ .
If 1 − α = 90 % , the interval is called a 90% confidence interval for µ .
REMARK
σ
σ ⎞
⎛
As there are many samples, there are many intervals ⎜ x − z α / 2
, x + zα / 2
⎟ . It can be
n
n⎠
⎝
shown that 100(1 − α)% of these intervals contain µ . Therefore the interval with endpoints
σ
x ± zα / 2
obtained from a sample is called a 100(1 − α)% confidence interval for the
n
population mean µ .
QUESTION What is z α / 2 ?
ANSWER
If 1 − α = 95 % so that α = 0.05 , use the standard normal distribution table to
get z α / 2 = z 0.025 = 1.96 .
If 1 − α = 90% so that α = 0.10 , use the standard normal distribution table to
get z α / 2 = z 0.5 = 1.645 .
1
QUESTION What should I do if σ is unknown?
ANSWER
When the sample size is large (n ≥ 30) , the population standard deviation σ
can be approximated by the sample deviation s. Hence:
σ
x ± zα / 2
≈ x ± zα / 2
n
s
n
EXAMPLE 1
200 tablets are selected from a chemist’s stock. The mean aspirin content of the 200 tablets is
80 mg with a standard deviation of 1.8 mg. Find a 95% confidence interval for the mean
aspirin content of all tablets in stock.
SOLUTION
Let µ be the mean aspirin content of all tablets in stock.
The endpoints of a 95% confidence interval for µ are given by:
x ± zα / 2
σ
n
≈ x ± zα / 2
s
n
1 .8
= 80 ± z 0.025
200
= 79.75 and 80.25
A 95% confidence interval for µ is approximately:
79.75 < µ < 80.25
EXAMPLE 2
The mean and the standard deviation for the quality grade-point averages of a random sample
of 36 college seniors are calculated to be 2.6 and 0.3 respectively. Find a 95% confidence
interval for the mean of the entire senior class.
SOLUTION
Let µ and σ be the mean and the standard deviation of the grade-point averages of the entire
senior class respectively.
The endpoints of a 95% confidence interval for µ are given by:
x ± zα / 2
σ
n
≈ x ± zα / 2
s
= 2.6 ± z 0.025
2
n
0 .3
36
= 2.6 ± 1.96
0 .3
36
= 2.50 and 2.70
A 95% confidence interval for µ is approximately
2.50 < µ < 2.70 .
3
SUPPLEMATARY READING
DERIVATION OF THE CONFIDENCE INTERVAL
(a)
In the above figure, a population of large population size is given. Since the sample
X −µ
size is large, the variable
has a standard normal distribution.
σ
n
(b)
In the above figure, α = 1 − 95% = 0.05 and z α / 2 = z 0.025 = 1.96 .
(c)
For 1 − α = 95% of the random samples drawn from the population, we have
− zα / 2 <
X −µ
< zα / 2 .
σ
n
Hence
− zα / 2 <
µ−X
< zα / 2
σ
n
4
or
σ
X − zα / 2
(d)
n
σ
< µ < X + zα / 2
n
.
Size of error
Refer to ( c ).
For
1 − α = 95% of the random samples drawn from the population, we have
X −µ
< zα / 2 .
σ
− zα / 2 <
n
Hence
− zα / 2
σ
n
< X − µ < zα / 2
or
X − µ < zα / 2
zα / 2
σ
n
σ
n
σ
n
,
.
is called the size of error.
Understanding the size of error
For 100(1 − α)% of the random samples, µ is in a neighbourhood of X and the
σ
difference between µ and X is less than z α / 2
.
n
END
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EXAMPLE 3
A research worker wants to determine the average time it takes a mechanic to rotate the tires
of a car, and she wants to be able to assert with 95% confidence that the mean of her sample
is off by at most 0.50 minute. If she can presume from past experience that the true standard
deviation is 1.6 minutes, how large a sample will she have to take?
SOLUTION
µ = true average time in minutes
σ = true standard deviation in minutes
n = sample size
We have:
⎛
⎞
⎜
⎟
X −µ
P ⎜ −1.96 <
< 1.96 ⎟ = 0.95
σ
⎜
⎟
⎜
⎟
n
⎝
⎠
1.96σ ⎞
⎛ 1.96σ
< X −µ <
P⎜ −
⎟ = 0.95
n
n ⎠
⎝
Hence:
1.96σ
n
= 0.50
⎛ 1.96σ ⎞
⎛ 1.96 × 1.6 ⎞
n=⎜
⎟ ≈⎜
⎟ = 39.3
⎝ 0.50 ⎠
⎝ 0.50 ⎠
2
2
Rounded up to the nearest integer, n = 40.
EXAMPLE 4
A random sample of 50 mathematics marks out of a total of 200 showed a mean of 75 and a
standard deviation of 10. Find a 95% confidence interval for the mean of the 200 marks.
SOLUTION
The population size is small and the sampling is without replacement. Thus:
⎞
⎛
⎟
⎜
X −µ
⎟
⎜
P⎜ − 1.96 <
< 1.96 ⎟ = 0.95
σ N−n
⎟⎟
⎜⎜
n N −1
⎠
⎝
6
( N = 200, n = 50)
⎞
⎛
⎟
⎜
µ−X
⎟
⎜
P⎜ − 1.96 <
< 1.96 ⎟ = 0.95
σ N−n
⎟⎟
⎜⎜
N
1
−
n
⎠
⎝
⎛
σ
P⎜⎜ X − 1.96
n
⎝
N−n
σ
< µ < X + 1.96
N −1
n
N−n ⎞
⎟ = 0.95
N − 1 ⎟⎠
The endpoints of a confidence interval for the mean (µ) of all 200 marks are:
x ± 1.96
σ
n
N−n
s N−n
≈ x ± 1.96
N −1
n N −1
10 200 − 50
= 75 ± 1.96
50 200 − 1
= 72.6 and 77.4
A 95% confidence interval for µ is
72.6 < µ < 77.4 .
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HOW TO MAKE A GUESS – SOME OBSERVATIONS
We try to make a guess about a population mean µ given that the population standard
deviation is σ = 0.974 . Suppose we want to test the three possibilities: µ < 15 , µ = 15 and
µ > 15 .
X − 15
.
We select many random samples of size n = 50 and calculate values of the statistic
σ
n
The following diagrams are helpful in making a decision.
The above diagrams show that an observation of the selected statistic may give a hint of the
possible value of µ .
8
Large-Sample TESTS CONCERNING THE MEAN OF A POPULATION WITH
KNOWN VARIANCE
AIM
To set up two hypotheses about the population mean µ . The first is called the
null hypothesis and is denoted by H 0 . The second is called the alternative
hypothesis and is denoted by H1 . The null hypothesis and the alternative
hypothesis are given in the table below:
Suppose you want to test one
of the following hypotheses
against the other.
µ = µ0
µ ≠ µ0
H0 : µ = µ0
H1 : µ ≠ µ 0
METHOD
µ ≥ µ0
µ < µ0
H0 : µ = µ0
µ = µ0
µ < µ0
H1 : µ < µ 0
µ ≤ µ0
µ > µ0
H0 : µ = µ0
µ = µ0
µ > µ0
H1 : µ > µ 0
To compute the value of a statistic and then determine whether the null
hypothesis or the alternative hypothesis should be accepted. The decision may
be correct or wrong (refer to the table below).
IF:
H 0 is true
H 0 is true
H 1 is true
H 1 is true
Decision
Reject H 0
Accept H 0
Remark
Type I error
Decision is correct
Accept H 0
Type II error
Reject H 0
(i.e. accept H1 )
(i.e. accept H1 )
Decision is correct
The number α = P(H 0 is rejected | H 0 is true) is called the level of
significance. To make a decision, refer to the table below.
Suppose you want to test one of
the following hypotheses against
the other.
µ = µ0
µ ≠ µ0
Test statistic: Z =
n
H0 : µ = µ0
Reject H 0 if Z < −z α / 2 or
H1 : µ ≠ µ 0
Z > zα / 2 .
Reject H 0 if Z < −z α .
µ ≥ µ0
µ < µ0
H0 : µ = µ0
µ = µ0
µ < µ0
H1 : µ < µ 0
µ ≤ µ0
µ > µ0
H0 : µ = µ0
µ = µ0
µ > µ0
H1 : µ > µ 0
9
X − µ0
σ
Reject H 0 if Z > z α .
EXAMPLE 5
A manufacturer of sports equipment has developed a new synthetic fishing line. It is claimed
that the mean breaking strength is 15 pounds. Would you agree with this claim if a random
sample of 50 lines had a mean breaking strength of 14.8 pounds with a standard deviation of
0.5 pound? Use a 0.05 level of significance.
SOLUTION
Aim: Let µ be the mean breaking strength. We want to test whether µ = 15 pounds is true
or not. We proceed as follows:
1.
Null hypothesis
H o : µ = 15 pounds
Alternative hypothesis
H1 : µ ≠ 15 pounds
2.
α = 0.05
3.
Critical region:
Reject H 0 if
X − 15
X − 15
< −z α / 2 = −1.96 or
> 1.96 .
σ
σ
50
4.
Calculations:
X − 15 14.8 − 15
≈
= −2.828
0 .5
σ
50
5.
Decision:
50
50
Since − 2.828 < −1.96 , reject H o and conclude that the average
strength is less than 15 pounds.
Understanding the level of significance and type I error
We say that a type I error has been committed if we reject the null hypothesis when it is true.
For this example, we have:
⎛
⎜
X − 15
P(the null hypothesis is rejected | µ = 15 ) = P⎜
< −z α / 2 or
⎜ σ
⎜
⎝ 50
= 0.05
⎞
⎟
X − 15
> zα / 2 ⎟
⎟
σ
⎟
50
⎠
Type II error
We say that a type II error has been committed if we accept the null hypothesis when it is
false.
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COMPUTER SIMULATION
We can use computer simulation to understand the method used to solve the above example.
We assume a population with σ = 0.794 .
Case 1:
µ = 15
The above figure shows that we usually accept the null hypothesis and the probability of
committing a type I error is 0.05.
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Case 2:
µ < 15
If µ = 14.5 < 15 , the above figure shows that we usually reject the null hypothesis and
conclude that µ < 15 .
Case 3:
µ > 15
If µ = 15.5 > 15 , the above figure shows that we usually conclude that µ > 15 .
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EXAMPLE 6
A manufacturer of sports equipment has developed a new synthetic fishing line. It is claimed
that the mean breaking strength is less than 15 pounds. Would you agree with this claim if a
random sample of 50 lines was tested and found to have a mean breaking strength of 14.88
pounds with a standard deviation of 0.5 pound? Use a 0.05 level of significance.
SOLUTION
Aim: Let µ be the mean breaking strength. We want to test whether µ < 15 pounds is true
or not. We proceed as follows:
1.
Null hypothesis
H 0 : µ = 15 pounds
Alternative hypothesis
H1 : µ < 15 pounds
4.
α = 0.05
5.
Critical region:
Reject H 0 if
X − 15
< − z α = −1.645 .
σ
50
4.
Calculations:
X − 15 14.88 − 15
≈
= −1.697
σ
0 .5
50
5.
Decision:
50
Since − 1.697 < −1.645 , reject H o and conclude that the
mean breaking strength is less than 15 pounds.
Understanding the level of significance and the type I error
We say that a type I error has been committed if we reject the null hypothesis when it is true.
For this example, we have:
⎛
⎞
⎜
⎟
X − 15
⎜
P(the null hypothesis is rejected | µ = 15 ) = P
< −z α ⎟
⎜ σ
⎟
⎜
⎟
⎝ 50
⎠
=α
= 0.05
Type II error
We say that a type II error has been committed if we accept the null hypothesis when it is
false.
13
COMPUTER SIMULATION
We can use computer simulation to understand the method used to solve the above example.
We assume a population with σ = 0.794 .
Case 1:
µ = 15
If µ = 15 , the above figure shows that we usually accept the null hypothesis and the
probability of committing a type I error is 0.05.
14
Case 2:
µ > 15
If µ = 15.5 > 15 , the above figure shows that we usually do not reject the null hypothesis and
conclude that the population mean is greater than or equal to 15.
Case 3:
µ < 15
If µ = 14.5 < 15 , the above figure shows that we usually conclude that µ < 15 .
15
EXAMPLE 7
A manufacturer of sports equipment has developed a new synthetic fishing line. It is claimed
that the mean breaking strength is more than 15 pounds. Would you agree with this claim if a
random sample of 50 lines had a mean breaking strength of 15.1 pounds with a standard
deviation of 0.5 pound? Use a 0.05 level of significance.
SOLUTION
Aim: Let µ be the mean breaking strength. We want to test whether µ > 15 pounds is true
or not. We proceed as follows:
1.
Null hypothesis
H 0 : µ = 15 pounds
Alternative hypothesis
H1 : µ > 15 pounds
6.
α = 0.05
7.
Critical region:
Reject H 0 if
X − 15
> z α = 1.645 .
σ
50
4.
Calculations:
X − 15 15.1 − 15
≈
= 1.4142
σ
0 .5
50
5.
Decision:
50
Since 1.4142 < 1.645 , do not reject H o and conclude that the
mean breaking strength is not more than 15 pounds.
Understanding the level of significance and the type I error
We say that a type I error has been committed if we reject the null hypothesis when it is true.
For this example, we have:
⎛
⎞
⎜
⎟
X − 15
⎜
P(the null hypothesis is rejected | µ = 15 ) = P
> zα ⎟
⎜ σ
⎟
⎜
⎟
⎝ 50
⎠
=α
= 0.05
Type II error
We say that a type II error has been committed if we accept the null hypothesis when it is
false.
16
COMPUTER SIMULATION
We can use computer simulation to understand the method used to solve the above example.
We assume a population with σ = 0.794 .
Case 1:
µ = 15
If µ = 15 , the above figure shows that we usually accept the null hypothesis and the
probability of committing a type I error is 0.05.
17
Case 2:
µ < 15
If µ = 14.5 < 15 , the above figure shows that we usually do not reject the null hypothesis and
conclude that µ ≤ 15 .
Case 3: µ > 15
If µ = 15.5 > 15 , the above figure shows that we usually conclude that µ > 15 .
18
EXERCISE
1.
A random sample of 100 automobile owners shows that an automobile is driven on
the average 14500 miles per year, in the state of Virginia, with a standard deviation of
2400 miles.
(b)
Construct a 99% confidence interval for the average number of miles an
automobile is driven annually in Virginia.
What can we assert with 99% confidence about the possible size of our error if
we estimate the average number of miles driven by car owners in Virginia as
14500 miles per year?
Ans.
(a) (13882,15118)
(a)
2.
An efficiency expert wishes to determine the average time that it takes to drill 3 holes
in a certain metal clamp. How large a sample will he need to be 95% confident that
his sample will be within 15 seconds of the true mean? Assume that it is known from
previous studies that the true standard deviation is 40 seconds.
Ans.
3.
664
It is claimed that an automobile is driven on the average less than 12000 miles per
year. To test this claim, a random sample of 1000 automobile owners are asked to
keep a record of the miles they travel. Would you agree with this claim if the random
sample showed an average of 11850 miles and a standard deviation of 2400 miles?
Use a 0.01 level of significance.
Ans.
5.
28
Past records indicate that the lengths of rods produced by a machine have a mean of
500 and a standard deviation of 5. Find the sample size needed if there should be a
99% confidence of the error in the sample estimate not exceeding 0.5.
Ans.
4.
(b) 618
The average is not less than 12000 miles
The average height of the males in the freshman class of a certain college has been
68.5 inches, with a standard deviation of 2.7 inches. Is there reason to believe that
there has been an increase in the average height if a random sample of 50 males in the
present freshman class have an average height of 69.7 inches? Use a 0.02 level of
significance.
Ans.
The average height is greater than 68.5 inches
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