Hypothesis testing

1. T- Test & Z- Test
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HYPOTHESIS TESTING
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2. STANDARD DEVIATION ?

3. Standard deviation (SD or σ) is a measure that is used to quantify the amount
of variation of a set of data values.
STANDARD DEVIATION
X -observation value
X – Mean
N- no of observations

4. DEGREE OF FREEDOM ?

5.  There are five balloons: one blue, one
red, one yellow, one pink, & one green.
 If 5 students (n=5) are each to select one
balloon only 4 will have a choice of color
(df=4). The last person will get whatever
color is left
 df = N-1
DEGREE OF FREEDOM

6.  It is used in research purpose
 Observations < 30
 Identification of one tiled or two tiled
Less than/ More than – one tiled
Equal or not equal – Two tiled
T - TEST

7. One Sample T-test (Single Mean)
T - TEST
μ –Assumed Value
X – Mean
S- Standard deviation
n- No. of observations

8.  Six students are chosen at random from the class and given a math
proficiency test. The professor wants the class to be able to score
above 70 on the test. The six students get scores of 62, 92, 75, 68, 83,
and 95. Can the professor have 90 percent confidence that the
mean score for the class on the test would be above 70?
SINGLE MEAN T TEST

9. null hypothesis: H 0: μ = 70
alternative hypothesis: H a : μ > 70
 First, compute the sample
mean and standard deviation
SINGLE MEAN T TEST

10. The number of degrees of freedom for the
problem is 6 – 1 = 5.
The value in the t‐table for t (.10,5) is 1.476.
Because the computed t‐value of 1.71 is larger
than the critical value in the table, the null
hypothesis can be rejected, and the professor
has evidence that the class mean on the math
test would be at least 70
SINGLE MEAN T TEST

11. COMBINED MEAN T-TEST

12.  Problem: Sam Sleep researcher 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. He brings sixteen participants into his sleep lab and
randomly assigns them to one of two groups. In one group he has participants sleep for
eight hours and in the other group he has them sleep for four. The next morning he
administers the SCAT (Sam's Cognitive Ability Test) to all participants. (Scores on the
SCAT range from 1-9 with high scores representing better performance).
SCAT 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
COMBINED MEAN T-TEST

13. The number of degrees of freedom for the problem is 14
The value in the t‐table for t (.05,14) is 2.145.
COMBINED MEAN T-TEST

14. Z-TEST

15.  It is used in research purpose
 Observations ≥ 30
 Identification of one tiled or two tiled
Less than/ More than – one tiled
Equal or not equal – Two tiled
Z - TEST

16.  Level of significants
Z - TEST
Los One tiled Two tiled
1% 2.33 2.58
5% 1.65 1.96
10% 1.28 1.65

17. In the population, the average IQ is 100 with a standard deviation of 15.
A team of scientists wants to test a new medication to see if it has either
a positive or negative effect on intelligence, or no effect at all. A sample
of 30 participants who have taken the medication has a mean of 140.
Did the medication affect intelligence, using alpha = 0.05?
SINGLE MEAN Z-TEST

18. Medication significantly affected intelligence, z = 14.60, p < 0.05.
SINGLE MEAN Z-TEST

19. COMBINED MEAN Z-TEST
z

20. COMBINED MEAN Z-TEST
The amount of a certain trace element in blood is known to vary with a
standard deviation of 14.1 ppm (parts per million) for male blood donors
and 9.5 ppm for female donors. Random samples of 75 male and 50
female donors yield concentration means of 28 and 33 ppm, respectively.
What is the likelihood that the population means of concentrations of the
element are the same for men and women?

21. COMBINED MEAN Z-TEST
Null hypothesis: H 0: μ 1 = μ 2
alternative hypothesis: H a : μ 1 ≠ μ 2
Reject null hypothesis

22. THANK YOU