The document discusses confidence intervals and hypothesis testing for population means, explaining the distinction between point estimates and confidence interval estimates. It includes specific examples involving z-tests and t-tests to determine whether a population mean differs from a specified value based on sample data. Various scenarios of testing, such as with known and unknown population variances, are demonstrated through calculations on sample means and standard deviations.

1. Normal & t-test
Confidence interval
Md. Saiful Islam
Dept. of Pharmaceutical Sciences
North South University

2. Confidence Interval
• Confidence Interval :Whenever we draw
a sample from population, we eventually
want to use the statistics to estimate
population values or population
parameters.
• There are two ways of estimating a
population parameter: a point estimate and
a confidence interval estimate.

3. • A point estimate of the population mean µ is
the sample mean, , computed from a random
sample of the population. A frequently used point
estimate for the population standard deviation σ
is s , the sample standard deviation.
• Confidence interval is the numbers in which
we have a specified degree of assurance that the
value of the parameter was captured. A
confidence interval allows us to estimate the
unknown parameter µ and give a margin of
error indicating how good our estimate is.
• CI for z test :95% CI of µ =
• CI for t test :95% CI of µ =
)/(
2
nzx σα±
)/(
2
nstx α±
x

4. Normal test
When sampling is from a normally distributed
population and the population variance is
known, the test statistic for testing Ho:
EX. If random sample of size 36 is drawn from a
normal population with mean and variance
are respectively 27 and 16 respectively. Can
we conclude the mean age of this population is
different from 30 years?
n
x
z
/σ
µ−
=

5. Hypothesis Testing: A Single
Population Mean
Calculation of test statistic
We have z= (27-30)/ 0.67 = - 4.5
We reject hypothesis. We conclude that
population mean is different from 30 years.

6. Hypothesis Testing: A Single
Population Mean
Testing Ho by means of a confidence
interval. The 95% confidence interval of
population mean is given by
Upper bound=27+ 1.31=28.31
Lower bound= 27-1.31=25.69
The age lies between 28.31 to 25.69 years
)/( nzx σα±

7. Hypothesis Testing: A Single
Population Mean
Sampling from a normally distributed
population: Population Variance is unknown
The test statistic for testing Ho: Population
mean= µ = µo is
Statistic is
Example: Will we be able to conclude that the
mean BMI for the population is 35 .
ns
x
t
/
µ−
=

8. Hypothesis Testing: A Single
Population Mean
Can we reject the hypothesis that
population mean is equal to 35.
Body Mass Index ( BMI) measurements
for 14 male subjects are given below.
Subject 1 2 3 4 5 6 7 8 9 10 11 12 13 14
BMI 23 25 21 37 39 21 23 24 32 57 23 26 32 45

9. Hypothesis Testing: A Single
Population Mean
Data: The data consist of BMI measurements on
14 subjects as given above.
Assumptions: The 14 subjects constitute a simple
random sample from a population of similar
subjects. We assume that BMI measurements
in this population are approximately normally
distributed.
Hypotheses: Population mean 35
Population is not equal to 35
Test statistic is with d.f is n-1,
t= (30.5-35)/2.8434
ns
x
t
/
µ−
=
ns
x
t
/
µ−
=

10. Hypothesis Testing: A Single
Population Mean
The calculated value of t = -1.58
With 13 degree of the value of t is –2.16
Since computed value of t is less than the table
value. We accept the hypothesis. Based on the
data the mean population from which the
sample drawn may be 35.
CI for t statistic:
Upper bound of CI=30.5+2.16*2.84=36.6.
Lower bound of CI=30.5-2.16*2.84=24.36
ns
x
t
/
µ−
=
)/( ntx σα±

11. Hypothesis Testing: A Single
Population Mean
Hypothesis Testing: population standard deviation is
not known
If the population standard deviation is not
known , the usual practice is to use the sample
standard deviation as an estimate. The test
statistic for testing Ho= µ = µo,
then , which when Ho is true , is
distributed approximately as the standard
normal distribution if n is large.
ns
x
t
/
µ−
=
ns
x
t
/
µ−
=

12. Hypothesis Testing: A Single
Population Mean
Ex. A study was conducted to describe the
menopausal status , menopausal symptoms,
energy expenditure, and aerobic fitness of
healthy midlife women and to determine the
relationship among these factors. The mean
score of maximum oxygen uptake for a sample
242 was 33.3 with a standard deviation of
12.14. The researcher wishes to know if, on the
basis of these data, one may conclude that the
mean score for a population of such women is
greater than 30
ns
x
t
/
µ−
=

13. Hypothesis Testing: A Single
Population Mean
Data Maximum oxygen uptake for 242
women with mean = 33.3 and s = 12.14
Assumptions: The data constitute a simple
random from a population of healthy
midlife women similar to those in the
sample.
Hypotheses Ho: µ grater than equal to 30
Ha: µ is greater than 30
ns
x
t
/
µ−
=

14. Hypothesis Testing: A Single
Population Mean
Data Maximum oxygen uptake for 242
women with mean = 33.3 and s = 12.14
Assumptions: The data constitute a simple
random from a population of healthy
midlife women similar to those in the
sample.
Hypotheses Ho: µ grater than equal to 30
Ha: µ is greater than 30
ns
x
t
/
µ−
=

15. Hypothesis Testing: A Single
Population Mean
Given the above test statistic
We have z= ( 33.3-30)/0.7804= 3.3/0.7804
= 4.23
We reject the hypothesis since computed
value is greater than table value. We
conclude that the mean score for the
sampled population is greater than 30.
ns
x
t
/
µ−
=