A friend of mine ask me to give sample T test and Z test, however I told her I cannot do it at once since I don't have any book or computer here in the middle of barrio and WIFI seems impossible. But she insist since she need it very bady. So I gave he a simple explanation as follows ; T test for me is comparing the means of two realated group like comparing your grades in mathematics and english, which of them have a significant difference or better grade from the other. And Z test is a test for proportion if you data or sample ar in normal distribution.
Two Sample t-Test
Purpose
To compare responses from two groups. These two groups can come from different experimental treatments, or different natural "populations".Assumptions
- each group is considered to be a sample from a distinct population
- the responses in each group are independent of those in the other group
- the distributions of the variable of interest are normal
How It Works
- The null hypothesis is that the two population means are equal to each other.
To test the null hypothesis, you need to calculate the following values:
(the means of the two samples), s12, s22 (the variances of the two samples), n1, n2 (the sample sizes of the two samples), and k (the degrees of freedom).
- Compute the t-statistic.
- Compare the calculated t-value, with k degrees of freedom, to the critical t value from the t distribution table at the chosen confidence level and decide whether to accept or reject the null hypothesis.
*Reject the null hypothesis when: calculated t-value > critical t-value
- Note: This procedure can be used when the distribution variances from the two populations are not equal, and the sample sizes are not equal.
Example: Miss Brigitte Angayen Young a successful scientist hypothesizes
that people who are allowed to sleep for only four hours will score
significantly lower than people who are allowed to sleep for eight hours
on a cognitive skills test. She brings sixteen participants into her
sleep lab and randomly assigns them to one of two groups. In one group
she has participants sleep for eight hours and in the other group she
has them sleep for four. The next morning she administers the BCAT (Briggite's
Cognitive Ability Test) to all participants. (Scores on the BCAT range
from 1-9 with high scores representing better performance).
| BCAT scores | ||||||||
| 8 hours sleep group (X) | 5 | 7 | 5 | 3 | 5 | 3 | 3 | 9 |
| 4 hours sleep group (Y) | 8 | 1 | 4 | 6 | 6 | 4 | 1 | 2 |
| x | (x-Mx)2 | y | (y - My)2 |
| 5 | 0 | 8 | 16 |
| 7 | 4 | 1 | 9 |
| 5 | 0 | 4 | 0 |
| 3 | 4 | 6 | 4 |
| 5 | 0 | 6 | 4 |
| 3 | 4 | 4 | 0 |
| 3 | 4 | 1 | 9 |
| 9 | 16 | 2 | 4 |
| Sx=40 | S(x-Mx)2=32 | Sy=32 | S(y-My)2=46 |
| Mx=5 |  |
My=4 |  |
*(according to the t significant probability table with df = 14, t must be at least 2.145 to reach p < .05, so this difference is not statistically significant)
t significant probability table is a long table I hope your instructor could relate to that
Interpretation: Briggite's hypothesis was not
confirmed. She did not find a significant difference between those who
slept for four hours versus those who slept for eight hours on cognitive
test performance.
So even Briggite's friend goes out having jam every night they are still smart. Try mo
Z-test
Used
to compare a proportion created by a random sample to a proportion originating from or
thought to represent the value for the entire population.
As an example, to make sure your
random sample of 100 subjects is not biased regarding a person’s sex, you could
compare the proportion of women in the sample to the known proportion of women in the
underlying population as reported in census data or by some other reliable source.
Problem:
Historical data indicates that about 10% of your agency's clients believe they were given
poor service. Now under new management for six months, a random sample of 110 clients
found that 15% believe they were given poor service.
Pu
= .10
Ps
= .15
n
= 110
Assumptions
Independent
random sampling
Nominal
level data
Large
sample size
State
the Hypothesis
Ho:
There is no statistically significant difference between the historical proportion of
clients reporting poor service and the current proportion of clients reporting poor
service. 
If
2-tailed test
Ha:
There is a statistically significant difference between the historical proportion of
clients reporting poor service and the current proportion of clients reporting poor
service. 
If
1-tailed test
Ha:
The proportion of current clients reporting poor service is significantly greater than the
historical proportion of clients reporting poor service.
Set
the Rejection Criteria
Use
z-distribution
table to estimate critical value
If
2-tailed test, Alpha .05, Zcv = 1.96
If
1-tailed test, Alpha .05, Zcv = 1.65
Compute
the Test Statistic
Estimate
Standard Error
p
= population proportion
q
= 1 - p
n
= sample size
Test
Statistic
Decide
Results of Null Hypothesis
If
a 2-tailed test was used
Since
the test statistic of 1.724 did not meet or exceed the critical value of 1.96, there is
insufficient evidence to conclude there is a statistically significant difference between
the historical proportion of clients reporting poor service and the current proportion of
clients reporting poor service.
If
a 1-tailed test was used
Since
the test statistic of 1.724 exceeds the critical value of 1.65, you can conclude the
proportion of current clients reporting poor service is significantly greater than the
historical proportion of clients reporting poor service.
However there are different types of T test the other are the student's T test, Paired T test and ect.
for each significance level, the Z-test has a single critical value
(for example, 1.96 for 5% two tailed) which makes it more convenient
than the Student's t-test
which has separate critical values for each sample size. Therefore,
many statistical tests can be conveniently performed as approximate
Z-tests if the sample size is large or the population variance known.
If the population variance is unknown (and therefore has to be
estimated from the sample itself) and the sample size is not large, the
Student t-test may be more appropriate. I found this in wikipedia .
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