The t test can be used for three purposes: 1) To compare a sample mean to a population mean, 2) To compare means between two independent samples, and 3) To compare values within one sample measured on two different occasions. The document then provides examples and step-by-step solutions for each of these three applications of the t test, demonstrating how to calculate the t statistic and compare it to critical values to determine if any differences are statistically significant.
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Chapter 9: Inferences from Two Samples
9.2: Two Means, Independent Samples
OBJECTIVES:
Run the test of hypothesis for mean difference using paired samples. Construct a confidence interval for the difference in population means using paired samples.
Observation of interest will be the difference in the readings
before and after intervention called paired difference observation.
Paired t test:
A paired t-test is used to compare two means where you have two samples in which observations in one sample can be paired with observations in the other sample.
Examples of where this might occur are:
Before-and-after observations on the same subjects (e.g. students’ test
results before and after a particular module or course).
A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a sphygmomanometer and a dynamap).
When there is a relationship between the groups, such as identical twins.
This test is concerned with the pair-wise differences
between sets of data.
This means that each data point in one group has a related data point in the other group (groups always have equal numbers).
ASSUMPTIONS:
The sample or samples are randomly selected
The sample data are dependent
The distribution of differences is approximately normally
distributed.
Note: The under root is onto the entire numerator and denominator, so you should take the root after solving it entirely
where “t” has (n-1) degrees of freedom and “n” is
the total number of pairs.
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 9: Inferences from Two Samples
9.2: Two Means, Independent Samples
OBJECTIVES:
Run the test of hypothesis for mean difference using paired samples. Construct a confidence interval for the difference in population means using paired samples.
Observation of interest will be the difference in the readings
before and after intervention called paired difference observation.
Paired t test:
A paired t-test is used to compare two means where you have two samples in which observations in one sample can be paired with observations in the other sample.
Examples of where this might occur are:
Before-and-after observations on the same subjects (e.g. students’ test
results before and after a particular module or course).
A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a sphygmomanometer and a dynamap).
When there is a relationship between the groups, such as identical twins.
This test is concerned with the pair-wise differences
between sets of data.
This means that each data point in one group has a related data point in the other group (groups always have equal numbers).
ASSUMPTIONS:
The sample or samples are randomly selected
The sample data are dependent
The distribution of differences is approximately normally
distributed.
Note: The under root is onto the entire numerator and denominator, so you should take the root after solving it entirely
where “t” has (n-1) degrees of freedom and “n” is
the total number of pairs.
Test of significance (t-test, proportion test, chi-square test)Ramnath Takiar
The presentation discusses the concept of test of significance including the test of significance examples of t-test, proportion test and chi-square test.
2. Applications
To compare the mean of a sample with
population mean.
To compare the mean of one sample
with the mean of another independent
sample.
To compare between the values
(readings) of one sample but in 2
occasions. 2
3. 1.Sample mean and population
mean
The general steps of testing hypothesis must be
followed.
Ho: Sample mean=Population mean.
Degrees of freedom = n - 1
−
X −µ
t =
SE
3
4. Example
The following data represents hemoglobin values in
gm/dl for 10 patients:
10.5 9 6.5 8 11
7 7.5 8.5 9.5 12
Is the mean value for patients significantly differ
from the mean value of general population
(12 gm/dl) . Evaluate the role of chance.
4
5. Solution
Mention all steps of testing hypothesis.
8.95 − 12
t= = −5.352
1.80201
10
Then compare with tabulated value, for 9 df, and 5% level of
significance. It is = 2.262
The calculated value>tabulated value.
Reject Ho and conclude that there is a statistically significant difference
between the mean of sample and population mean, and this
difference is unlikely due to chance.
5
7. 2.Two independent samples
The following data represents weight in Kg for 10
males and 12 females.
Males:
80 75 95 55 60
70 75 72 80 65
Females:
60 70 50 85 45 60
80 65 70 62 77 82
7
8. 2.Two independent samples, cont.
Is there a statistically significant difference between
the mean weight of males and females. Let alpha =
0.01
To solve it follow the steps and use this equation.
−
−
X1 − X 2
t=
2 2
(n1 − 1) S1 + (n 2 − 1) S 2 1 1
( + )
n1 + n 2 − 2 n1 n 2
8
9. Results
Mean1=72.7 Mean2=67.17
Variance1=128.46 Variance2=157.787
Df = n1+n2-2=20
t = 1.074
The tabulated t, 2 sides, for alpha 0.01 is 2.845
Then accept Ho and conclude that there is no
significant difference between the 2 means. This
difference may be due to chance.
P>0.01
9
10. 3.One sample in two occasions
Mention steps of testing hypothesis.
The df here = n – 1.
−
d
t =
sd
n
(∑ d ) 2
∑ d2 −
n
sd =
n −1
10
11. Example: Blood pressure of 8
patients, before & after treatment
BP before BP after d d2
180 140 40 1600
200 145 55 3025
230 150 80 6400
240 155 85 7225
170 120 50 2500
190 130 60 3600
200 140 60 3600
165 130 35 1225
Mean d=465/8=58.125 ∑d=465 ∑d2=2917511
12. Results and conclusion
t=9.387
Tabulated t (df7), with level of significance
0.05, two tails, = 2.36
We reject Ho and conclude that there is
significant difference between BP readings
before and after treatment.
P<0.05.
12