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1. STATISTICS and
PROBABILITY
Estimation of
Parameters Part I

2. Session Objectives:
At the end of the session, the participants
should be able to
1.illustrate point and interval estimations
2.distinguish between point and interval
estimations
3.identify point estimator for the population
mean
4.compute for the point estimate of the
population mean

3. Session Objectives:
5. identify the appropriate form of the
confidence interval estimator for the population
mean when: (a) the population variance is
known, (b) the population variance is unknown,
and (c) the CLT is to be used
6. illustrate the t-distribution
7. construct a t-distribution
8. identify regions under the t-distribution
corresponding to different t-values
9. identify percentiles using the t-value

4. Session Objectives:
10. compute for the confidence interval
estimate based on the appropriate form of
the estimator for the population mean
11. solve problems involving confidence
interval estimation of the population mean
12. draw conclusion about the population
mean based on its confidence interval
estimate

5. Point and Interval Estimation

6. A point estimator is needed to obtain a
point estimate. It is assumed that the
population is normally distributed when
the sample mean is used as a point
estimate of the population mean. The
point estimator of a sample mean x is the
statistic X.
Example 1: The sample mean x is 45.12 and
the population mean ?????? is 46.51. Hence, the
point estimate is the single value 45.12 and
the point of estimator is the statistic X.

7. A good point estimate is one that is
unbiased. If random sampling was
done in the collection of a set of
data, and a sample mean is
computed out of these data to
approximate the population mean,
then the point estimate x is a good
point estimate.

8. Example 2: A teacher wanted to
determine the average height of Grade
9 students in their school. What he did
was to go to one of the eight sections in
Grade 9 and then took their heights. He
computed for the mean height of the
students and got 165 cm.
a.What is the point estimate?
b.Is it a good point estimate? Why?

10. Sample Mean x
(6, 2) 4
(6, 8) 7
(6, 9) 7.5
(6, 3) 4.5
Each mean x of each sample is a
point estimate of the population
mean ??????.
Consider some samples
of size 3 drawn from the
same population
without replacement
Sample Mean x
(6, 2, 8) 5.3
(6, 2, 9) 5.7
(6, 2, 3) 3.7
(2, 8, 9) 6.3
Each mean x of each sample is a
point estimate of the population
mean ??????.

11. It is better to approximate the population
parameter by determining a range of
values within which the population mean
is more likely to be located. This range
of values is called the interval estimate
or confidence interval. Confidence
intervals use interval estimates to define
a range of values that includes the
parameter being estimated with
specified level of confidence.

13. 95% confidence level means that the
probability of the confidence interval
containing the true population parameter is
95%.
Critical value is the value that indicates the
point beyond which lies the rejection region.
This region does not contain the true
population parameter.

14. Interval Estimate of Population Mean
with Known Variance

16. Example: The mean score of a random
sample of 49 Grade 11 students who
took the periodic test is calculated to be
78. The population variance is known
to be 0.16.
a.Find the margin of error.
b.Find the 95% confidence interval for
estimating the population parameter.

17. Interpretation:
The researcher is 95% confident
that the sample mean x = 78
differs from the population mean
?????? by no more than 0.11. Also,
The researcher is 95% confident
that the population mean ?????? is
between 77.89 and 78.11 when
the mean of the sample is 78.

18. Example 2: Assuming normality, use the
given confidence level and sample data
below to find the following:
a.Margin of error
b.Confidence interval for estimating the
population parameter
Given data:99% confidence level
n = 50
x = 18,000
?????? = 2,500

20. The t-Distribution
The t-distribution, just like the standard
normal curve, is bell-shaped and unimodal.
It is symmetric about t = 0. However, its
variance is greater than 1. This makes it
wider and flatter in the middle. It has more
area in its tails than that of the standard
normal curve. Its shape depends on the
sample size. As the sample size n becomes
larger, the t-distribution gets closer to the
standard normal curve.

24. To find a value in the Table of t Critical
Values, there is a need to adjust the sample
size n by converting it to degree of freedom
df. In
 statistics, the number of 
degrees of
freedom 
is the number of values in the final
calculation of a
 statistic that are free to vary.
The
formula is df = n -1.
Example 1: A student-researcher wants to determine
whether the mean score in mathematics of the 25 students
in Grade 8 Newton is significantly different from the school
mean of 89. The mean and the standard deviation of the
scores of the students in Newton are 95 and 15
respectively. Assume 95% confidence level.

26. The computed value of t is equal to 2 which
is smaller that the tabular value of 2.064.
The computed t value does not fall in the
critical region. Therefore, the mean score of
Grade 8 Newton in Mathematics is the same
with the mean score of all the students
taking up Grade 8 Mathematics.
2.0642
critical region

27. Example 2: A student suspects that the data she
collected for research study do not represent the
target population. Here are the data she collected:
162734
202930
223037
253042
263235
The population mean is 27. Assuming normality in
the target population, is the student’s suspicion
correct? Use 90% confidence level.

29. -1.7610 1.761
1.151
0.050.05

30. The computed value of t is less than the
tabular value of 1.761. Therefore, the
student is wrong in suspecting that the data
are not representative of the target
population.
Example 3: A sample of size n = 20 is a
simple random sample selected from a
normally distributed population. Find the
value of t such that the shaded area to the
left of t is 0.05.

31. Solution:
Find df = n -1 = 20 -1 = 19
0-t t
0.050.05

32. 0
0.025
0.025

33. Example 5: For a t-distribution with 25
degrees of freedom, find the value of t such
that the area to the right of t is 0.05
0
0.050.05

34. Example 6: For a t-distribution with 14
degrees of freedom, find the value of t such
that the area between –t and t is 0.90.
0-t t
0.90
0.050.05

35. Identifying Percentiles Using the t-
Distribution Table
Example 7: The graph of a distribution with df
= 15 is shown below:
a. If the shaded area on the right is 0.05, what is
the area to the left of it?
0

39. Interval Estimate of Population Mean
with Unknown Variance

41. Example 1: The mean and standard
deviation of the content of sample of 10
similar containers are 10.5 liters and 0.352,
respectively.
a.Find a 95% confidence interval for the
actual mean content.
b.Find the lower and upper confidence
limits.
Assume that the contents are approximately
normally distributed.

44. Example 2: The following are randomly
selected scores in Filipino of 12 Grade 10
students: 75, 65, 76, 80, 85, 77, 81, 83, 80,
70, 71, 69
a. Find the 99% confidence interval for the
mean score of Grade 10 students, assuming
that the students’ score is approximately
normally distributed.
b. Find the lower and upper confidence
limits.

45. Solution:

46. Lower confidence
interval
Upper confidence
interval