Inferential Statistics-Part-I mtech.pptx

Inferential Statistics
Why do you need inferential statistics?
 Making an inference about the population from a sample.
 Allow us to use the information learned from descriptive
statistics.
 It extends beyond the immediate data.
 Used to infer from sample data what is the population might
think.
 Used to make judgements about the probability that an observed
difference between groups is a dependable one or one that might
have happened by chance in the study.

Statistical inference is the procedure by which we reach a conclusion about a population
on the basis of the information contained in a sample drawn from that population.
Statistical Inference (Two General Areas)
 Estimation
 Hypothesis Testing
Estimation process use sample data to calculate some statistics that serves as an
approximation of the corresponding parameter of the population from which the
sample was drawn.
Types of estimate
Point estimate: A point estimate is a single numerical value used to estimate the
corresponding population parameter.
Interval estimate: An interval estimate consists of two numerical values defining a range
of values that, with a specified degree of confidence, most likely includes the parameter
being estimated.

A single computed value is referred to as an estimate. The rule that tells us
how to compute this value, or estimate, is referred to as an estimator.
Estimators are usually presented as formulas. For example,
is an estimator of the population mean, µ. The single numerical value that
results from evaluating this formula is called an estimate of the parameter
µ.
In many cases, a parameter may be estimated by more than one estimator.
One of the desired property of a good estimator is unbiasedness.
An estimator, say, T, of the parameter θ is said to be an unbiased estimator
of θ if E(T) = θ.

The sampled population is the population from which one actually
draws a sample.
The target population is the population about which one wishes to
make an inference.
Only when the target population and the sampled population are the
same it is possible for one to use statistical inference procedures to
reach conclusions about the target population. If the sampled
population and the target population are different, the researcher
can reach conclusions about the target population only on the basis
of nonstatistical considerations like extrapolating findings to the
target population.

CONFIDENCE INTERVAL FOR A POPULATION MEAN
Suppose a researchers wishes to estimate the mean of
some normally distributed population.
Draw a random sample of size n from the normally
distributed population and compute , which is used as a
point estimate of µ. Random sampling inherently
involves chance, cannot be expected to be equal to µ.
It would be much more meaningful, therefore, to
estimate µ by an interval that somehow communicates
information regarding the probable magnitude of µ.

CONFIDENCE INTERVAL FOR A POPULATION MEAN
If sampling is from a normally distributed
population, the sampling distribution of the sample
mean will be normally distributed with a mean
equal to the population mean µ, and a variance is
equal to . From our knowledge of normal
distributions, approximately 95% of the possible
values of constituting the distribution are within
two standard deviations of the mean. The two
points that are two standard deviations from the
mean will contain approximately 95%.

Example:
Suppose a researcher, interested in obtaining an estimate of the average level of some
enzyme in a certain human population, takes a sample of 10 individuals, determines the
level of the enzyme in each, and computes a sample mean of . Suppose further it is
known that the variable of interest is approximately normally distributed with a variance of
45. We wish to estimate µ.

An approximate 95% confidence interval for µ is given by
Interval Estimate Components:
The interval estimate contains in its centre the point estimate of µ. The 2 we recognize as a
value from the standard normal distribution that tells us within how many standard errors
lie approximately 95% of the possible values of . This value of z is referred to as the
reliability coefficient. The last component, , is the standard error, or standard deviation of
the sampling distribution of . In general, then, a interval estimate may be expressed as
follows:
Sampling from a normal distribution with known variance, an interval estimate for µ is
where is the value of z to the left of which lies and to the right of which lies of the area
under its curve.

Interpreting Confidence Intervals:
How do we interpret the interval given by Expression ?
In the present example the reliability coefficient is equal to 2. We say
that in repeated sampling approximately 95% of the intervals
constructed by the above expression will include the population
mean. This interpretation is based on the probability of occurrence of
different values of . We may generalize this interpretation by
designating the total area under the curve of that is outside the
interval as and the area within the interval as 1- .
𝛂 𝛂

Probabilistic Interpretation: In repeated sampling, from a normally distributed
population with a known standard deviation, 100(1- ) percent of all intervals of
𝛂
the form will in the long run include the population mean µ.
The quantity (1- ), in this case 0.95, is called the
𝛂 confidence coefficient (or
confidence level), and the interval is called a confidence interval for µ. When
(1- )=0.95,
𝛂 the interval is called the 95% confidence interval for µ.
Practical Interpretation: When sampling is from a normally distributed
population with known standard deviation, we are 100(1- ) percent confident
𝛂
that the single computed interval, , contains the population mean.
Precision: The quantity obtained by multiplying the reliability factor by the
standard error of the mean is called the precision of the estimate. This quantity
is also called the margin of error.

Sampling from Nonnormal Populations:
As noted, it will not always be possible or prudent to assume that the
population of interest is normally distributed. Thanks to the central limit
theorem, this will not deter us if we are able to select a large enough sample.
We have learned that for large samples, the sampling distribution of is
approximately normally distributed regardless of how the parent population is
distributed.
Example: Punctuality of patients in keeping appointments is of interest to a
research team. In a study of patient flow through the offices of general
practitioners, it was found that a sample of 35 patients was 17.2 minutes late
for appointments, on average. Previous research had shown the standard
deviation to be about 8 minutes. The population distribution was felt to be
nonnormal. What is the 90% confidence interval for µ, the true mean amount
of time late for appointments?

Since the sample size is fairly large (greater than 30), and since the population
standard deviation is known, we draw on the central limit theorem and
assume the sampling distribution of to be approximately normally
distributed. From the normal distribution Table, we find the reliability
coefficient corresponding to a confidence coefficient of 0.90 to be about
1.645. The standard error is
Therefore, 90% percent confidence interval for µ is
Frequently, when the sample is large enough for the application of the central
limit theorem, the population variance is unknown. In that case, we use the
sample variance as a replacement for the unknown population variance in the
formula for constructing a confidence interval for population mean.

t-distribution
One does not have knowledge of the population mean and
variance, and cannot use the statistic to construct Confidence
intervals of the mean. The statistic,
is normally distributed when the population is normally
distributed and is at least approximately normally distributed
when n is large, regardless of the functional form of the
population, we cannot make use of this fact because is
𝜎
unknown.

t-distribution
Most logical solution, we use the sample Standard Deviation (SD),
as an approximation of . This is justifiable when n ≥ 30 and also the use of
𝜎
normal distribution theory to construct confidence interval for
When we have a small sample an alternative procedure for constructing a
confidence interval is the use of Student’s t distribution or simply t
distribution.
William S Gosset was a statistician employed by the Guinness brewing
company which had stipulated that he cannot publish his own name.
(GOSSET -pseudonym of student) student’s t distribution.

t-distribution
Properties of t distribution:
1. It has a mean = 0
2. symmetrical about the mean.
3. In general, it has a variance greater than 1, but the
variance approaches 1 as the sample size becomes
large. For degree of freedom (df) > 2, the variance of t
distribution is df/(df-2), since here df = n - 1 for n > 3
we may write the variance of the t distribution
= (n-1)/(n-3)
4. Variable t ranges from -∞ to +∞
5. Compared to the normal distribution the t distribution
is less peaked in the centre and has higher tails.
6. The t distribution approaches the normal distribution
as n - 1 approaches infinity

t-distribution
Confidence interval using t
Source of reliability coefficient is different
When sampling is from a normal distribution whose standard
deviation , is unknown the 100(1-𝛼) percent confidence interval for
the population mean µ is given by
Applicable: strictly valid when the sample be drawn from a normal distribution

t-distribution: Applicability
When to use t distribution:
The t distribution can be used with any statistic having a bell-shaped distribution
(approximately normal).
The sampling distribution of a statistic should be bell-shaped. If any of the following
conditions apply:
The population distribution is normal.
The population distribution is symmetric, unimodal, and without outliers, and the
sample size is at least 30.
The population distribution is moderately skewed, unimodal, without outliers and the
sample size is at least 40.
The sample size is greater than 40, without outliers
The t distribution should not be used with small samples from populations that are
not approximately normal.

t-distribution: Example
A researcher conducted a study to evaluate the effect of on job body
mechanism instruction on the work performance of newly employed young
workers. He used two randomly selected groups of subjects, an experimental
group and a control group. The experimental group received one hour of back-
school training provided by an occupational therapist. The control group did
not receive this training. A criterion-referenced body mechanics evaluation
checklist was used to evaluate each worker’s lifting, lowering, pulling and
transferring of objects in the work environment. A correctly performed task
received a score of 1. The 15 control subjects made a mean score of 11.53 on
the evaluation with a standard deviation of 3.681. We assume that these 15
controls behave as a random sample from a population of similar subjects. We
wish to use these sample data to estimate the mean score for the population.

Standard normal and t-distribution Tables

t-distribution: Example
Ans: We may use the sample mean 11.53 as a point estimate of the population mean but since
the population standard deviation is unknown, we must assume that the population of values
to be at least approximately normally distributed before constructing a confidence interval for
µ. Let us assume that such an assumption is reasonable and that a 95% confidence interval is
desired. We have our estimator and our standard error is We need now to find the
x̄
reliability coefficient, the value of t associated with a confidence coefficient of 0.95 and n-1
=14 degree of freedom. Since a 95% confidence interval leaves 0.05 of the area under the
curve of t to be equally divided between the two tails, we need the value of t to the right of
which lies 0.025 of the area.
From T distribution table at 14 degrees of freedom: t0.975 = 2.1448
Hence the 95% confidence interval
11.53 ± 2.1448 × (0.9504) = 11.53 ± 2.04 = [9.49, 13.57]
This interval may be interpreted from both the probabilistic and practical points of view. We
are 95% of confidence that the true population mean µ is somewhere between 9.49 and 13.57
because, in repeated sampling, 95% of intervals constructed in like manner will include µ.

Deciding between z and t
When we construct a confidence interval for a population mean, we must decide whether to use
a value of z or a value of t as the reliability factor. To make an appropriate choice we must
consider sample size, whether the sampled population is normally distributed, and whether the
population variance is known.

Confidence interval for the difference between two population means
Confidence interval for the difference between population means provides information
that is helpful and deciding whether or not it is likely that the two population means are
equal. When the constructed interval does not include zero, we say that the interval
provides evidence that the two-population means are not equal. When the interval
includes zero, we say that the population mean may be equal.
Sampling from Normal Populations with known variences
Population variance are known the 100(1-)% confidence interval for µ1 - µ2 is given
Sampling from Non-normal Populations
The construction of a confidence interval for the difference between two population
means when sampling is from non-normal populations proceeds in the same manner as
sampling from normal populations if the sample sizes n1 and n2 are large. Again, this is a
result of the central limit theorem. If the population variances are unknown, we use the
sample variances to estimate them.

Example: Despite common knowledge of the adverse effects of
doing so, many women continue to smoke while pregnant. A
researcher examined the effectiveness of a smoking cessation
program for pregnant women. The mean number of cigarettes
smoked daily at the close of the program by the 328 women who
completed the program was 4.3 with a standard deviation of 5.22.
Among 64 women who did not complete the program, the mean
number of cigarettes smoked per day at the close of the program
was 13 with a standard deviation of 8.97. We wish to construct a 99
percent confidence interval for the difference between the means of
the populations from which the samples may be presumed to have
been selected.

No information is given regarding the shape of the distribution of cigarettes smoked per day.
Since our sample sizes are large, however, the central limit theorem assures us that the sampling
distribution of the difference between sample means will be approximately normally distributed
even if the distribution of the variable in the populations is not normally distributed. We may
use this fact as justification for using the z statistic as the reliability factor in the construction of
our confidence interval. Also, since the population standard deviations are not given, we will use
the sample standard deviations to estimate them. The point estimate for the difference between
population means is the difference between sample means, 4.3 – 13.0 = - 8.7. From normal
distribution Table, we find the reliability factor to be 2.58. The estimated standard error is
Our 99 percent confidence interval for the difference between population means is
- 8.7 ± 2.58 (1.1577) = [- 11.7; - 5.7]
We are 99 percent confident that the mean number of cigarettes smoked per day for women
who complete the program is between 5.7 and 11.7 lower than the mean for women who do
not complete the program.

t distribution and the difference between means
when population variances are unknown, and we wish to estimate
the difference between two population means with a confidence
interval we can use the t distribution as a source of reliability factor
if certain assumptions are met.
We must know or willing to assume, that the two sampled
populations are normally distributed.
Regarding unknown population variances, two situations may occur:
Situation–I: population variances are equal,
Situation–II: population variances are not equal.

Situation–I: population variances are equal
If the assumption of equal population variance is justified, the two samples may be
considered as the estimates of the same quantity, the common variance. Obtain a
pooled estimate of the common variance. Pooled variance is the weighted average of
the sample variances. Sample variances are weighted by their degree of freedom
100(1 - α) percent confidence interval for µ1 - µ2
degree of freedom used in determining the value of t in n1 + n2 – 2

Example: population variances are equal
The purpose of a researcher study was to determine the effect of long term
exercise intervention on corporate executives enrolled in a supervised
fitness program. Data were collected on 13 subjects (the exercise group)
who voluntarily entered a supervised exercise program and remained active
for an average of 13 years and 17 subjects (the secondary group) who
elected not to join the fitness program. Among the data collected on the
subjects was the maximum number of sit-ups completed in 30 seconds. The
exercise group has a mean and standard deviation for this variable of 21.0
and 4.9 respectively. The mean and standard deviation for the sedentary
group were 12.1 and 5.6 respectively. We assume that the two populations
of overall muscle condition measures are approximately normally
distributed and that the two population variances are equal. We wish to
construct a 95% confidence interval for the difference between the means
of the populations represented by these two samples.

Example: population variances are equal
Pooled estimate of the common population variance
from t distribution table (13+17-2) = 28 degree of freedom and desired 0.95 confidence
interval, reliability factor 2.0484
Confidence interval = (21.0 - 12.1) 2.0484 8.9 4.0085 =
We are 95% confident that the difference between the population mean is somewhere
between 4.9 and 12.9. We can say this because we know that if we were to repeat the
study many, many times and compute confidence intervals in the same way, about 95% of
the intervals would include the difference between the population mean.
Since the interval does not include zero, we conclude that the population means are not
equal.
We can interpret this interval that the difference between the two population means is
estimated to be 8.9 and we are 95% confident that the true value lies between 4.9 and
12.9.

Population variance is not equal
When one is unable to conclude that the variances of two populations of
interest are equal even though the two populations may be assumed to
be normally distributed, it is not proper to use t distribution.
Solutions has been proposed by many researchers. But the problem
resolves around the fact that the quality
does not follow t-distribution with n1 + n2 - 2 degree of freedom when
the population variances are not equal.

Population variance is not equal
The solution proposed by Cochran consists of completing the reliability factor by the
following formula:
Where for n1 - 1 degrees of freedom, and for n2 - 1 degrees of freedom.
An approximate 100(1- ) percent confidence interval for µ
𝛼 1 - µ2 is given by

Example: population variances are not equal
The purpose of a research study was to determine the effect of long-term
exercise intervention on corporate executives enrolled in a supervised
fitness program. Data were collected on 13 subjects (the exercise group)
who voluntarily entered a supervised exercise program and remained
active for an average of 13 years and 17 subjects (the secondary group)
who elected not to join the fitness program. Among the data collected on
the subjects were the maximum number of sit-ups completed in 30
seconds. The exercise group has a mean and standard deviation for this
variable of 21.0 and 4.9 respectively. The mean and standard deviation
for the sedentary group were 12.1 and 5.6 respectively. We assume that
the two populations of overall muscle condition measures are
approximately normally distributed and that the two population
variances are not equal. We wish to construct a 95% confidence interval
for the difference between the means of the populations represented by
these two samples.

Example: population variances are not equal
We will use Cochran reliability factor t’. From t distribution Table with 12 degrees of freedom
and
. Similarly, with 16 degrees of freedom and . We now compute
we now construct the 95 percent confidence interval for the difference between the two
population means.
Since the interval does not include zero, we conclude that the population means are not equal.
We can interpret this interval that the difference between the two population means is
estimated to be 8.9 and we are 95% confident that the true value lies between 7.9348 and
16.8348.

Determination of sample size for estimating mean
Planning of any survey experiment - How large a sample to take?
Larger than needed - wasteful of resource
Smaller than needed - lead to a result of no practical use
objective:- Interval estimation should have
1) a narrow interval
2) high reliability.
Total width of the interval is twice the magnitude of the quality :
Increasing reliability means a larger reliability coefficient---> increase interval.

Fixed reliability coefficient and reduce standard error
standard error = , is fixed the only way is to increase n --> take a
larger sample
How large?--> depends on the desired degree of reliability and the
desired interval width
sampling with replacement from an infinite or sufficiently large
population formula
sampling from small finite population without replacement formula

Sample size estimation formulas require the and population variance
is unknown.
The most frequently used source for estimation of are:
A pilot or preliminary sample may drawn from the population and
computed sample variance(S2
) may be used to estimate . Observations
used in the pilot sample may be computed on a part of the final
sample:
n(the computed sample size) - n1(the pilot sample size)
= n2(the number of observation needed to satisfy the total sample size
requirement)
Estimates of Sigma square may be available from previous or similar
studies.

Inference about population variance
Confidence interval for the variance of a normally
distributed population
Distribution normal? - used sample variance as an
approximate estimator of population variance
Wonder about the quality? - check whether the sample
variance is an unbiased estimator of population variance.
To be unbiased - average value of the sample variance over
all possible sample must be equal to the population variance.
E() =

Draw all possible samples of size two from the
population consisting of the values 6, 8, 10, 12 and
14.
If we compute the sample variance
For each of the possible samples, we obtain the
sample variances as shown in the table
Sampling with replacement
E(s2
), the expected value of the mean of the sample
variance, (0 + 2 + ….. + 2 + 0)/25 = 8
Hence E(s2
) =
S E C O N D D R A W
6 8 10 12 14
F
I
R
S
T
D
R
A
w
6
8
10
12
14
0
2
8
18
32
2
0
2
8
18
8
2
0
2
8
18
8
2
0
2
32
18
8
2
0

Sampling without replacement: the expected value of s2
,
(0 + 2 + ….. + 2 + 0)/10 = 10
Then E(s2
) = where sampling is with replacement. Results justify the use of for computing
the sample variance.
E(s2
) when sampling is without replacement.
Interval estimation of a population variance
Success depends on our ability to find an approximate sampling distribution
Confidence interval for are usually based on the sampling distribution of
If sample of size n are drawn from a normally distributed population, this quality [ ] has a
distribution known as Chi square distribution with n - 1 degrees of freedom.
To obtain a 100(1-α)% confidence interval for we select values of from the table in such a
way that α/2 is to the left of smaller value of and α/2 is to be the right of the larger values
of .

The 100(1-α) % confidence interval for is
Confidence interval for , population standard deviation is
Method is widely used but have some draw back. Normality of the population is crucial.
Estimator is not in the centre of the confidence interval because distribution is not
symmetric.

A random sample of 20 nominally measured 2 mm diameter steel ball bearings is
taken and the diameters are measured precisely. The measurements, in mm, are
as follows:
2.02 1.94 2.09 1.95 1.98 2.00 2.03 2.04 2.08 2.07
1.99 1.96 1.99 1.95 1.99 1.99 2.03 2.05 2.01 2.03
Assuming that the diameters are normally distributed with unknown mean  and
unknown variance 2
,
a) Find a 2-sided 95% confidence interval for the variance 2
b) Find a 2-sided confidence interval for the standard deviation 

From the data we calculate and , and .
Hence,
There are 19 degrees of freedom and the critical values of the Chi-square
distribution and
The confidence interval for 2
is
The confidence interval for  is

Inferential Statistics-Part-I mtech.pptx

Inference about population variances
Confidence interval for the ratio of the variances of the two normally distributed populations:
One way of comparing two variances is to compute their ratio
To use t distribution for constructing a confidence interval for the difference between population
means requires that the population variances be equal. If the two variances are equal, the ratio
will be one. If the confidence interval for the ratio of two populations variances includes 1, we
conclude that the two populations variances, may in fact, be equal.
This is a form of inference, and we must rely on some sampling distribution. This time the
distribution of is utilised provided the following assumptions are met.
Assumption:
1. and are computed from independent samples of size n1 and n2
2. Sample have been drawn from two normally distributed populations
3. let
If the assumption are met, follows a F distribution

F-distribution: If U and V are independent chi-square random variables with r1 and r2
degrees of freedom, respectively, then: follows an F distribution with r1 numerator degrees
of freedom and r2 denominator degrees of freedom.
This F distribution depends on two-degrees-of-freedom values, one corresponding to the
value of n1 – 1 used in computing, and the other corresponding to the value of n2 – 1 used in
computing . They are usually referred to as the numerator degree of freedom and
denominator degree of freedom.
A confidence interval for at 100(1-α) % confidence is constructed by
the values from F table to the left lies α/2 of the area
the value from F table to the right lies α/2 of the area

The values of at the intersection of the column headed df1 and the row labelled df2. If we
have extensive table of F distribution, finding would be no trouble. To include every
possible percentile of F would make a very lengthy table. A relationship exists to compute

Problem#
The variability in the thickness of oxide layers in
semiconductor wafers is a critical characteristic, where
low variability is desirable. A company is investigating two
different ways to mix gases so as to reduce the variability
of the oxide thickness. We produce 16 wafers with each
gas mixture and our results indicate that the standard
deviation is s1 = 1.96Å and s2 = 2.13Å for the two
mixtures. What is the 95% confidence interval for the ratio
between the two variances?

Given information:
Sample size of population 1: n1 = 16;
sample standard deviation for sample from population 1: s1 = 1.96;
Sample size of population 2: n2 = 16;
Sample standard deviation for sample from population 2: s2 = 2.13.
Since we are looking for a 95% confidence interval, we need two
f values:

Inferential Statistics-Part-I mtech.pptx

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