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PREPARED BY:
- IS A STATISTICAL PROCEDURE USED
TO TEST AN ALTERNATIVE
HYPOTHESIS AGAINST A NULL
HYPOTHESIS.
TO DETERMINE WHETHER TWO
SAMPLES’ MEAN ARE DIFFERENT WHEN
VARIANCES/STANDARD DEVIATIONS ARE
KNOWN AND SAMPLE IS LARGE ( n ≥ 30 ).
A Z-VALUE( A WHOLE NUMBER/
DECIMAL NUMBER ) WHICH WILL BE
COMPARED TO THE VALUE OF ALPHA(
LEVEL OF SIGNIFICANCE) TO MAKE A
DECISION/ CONCLUSION.
FOR THE FORMULA TO BE
USED :
𝜇 is called mu, 𝜎 is called sigma
This is the formula for
Standard Deviation:
A population is a collection of persons, objects or items of
interest.
A sample is a portion of the whole and, if properly taken, is
representative of the whole.
The population mean is represented by the Greek
letter mu (μ). It is given by the formula
𝜇 =
𝑥
𝑁
The capital Greek letter sigma (𝚺) is commonly used in
mathematics to represent a summation of all the numbers in
a grouping. N is the number of terms in the population.
The sample mean is represented by x bar . It is given by the
formula
𝑋=
𝑥
𝑛
1.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?
𝑋 = 140
𝜇 = 100
𝛿 = 15
n = 30
Ho : 𝜇 = 140
Ha : 𝜇 ≠ 140
=
√
140 − 100
15
30
= 14.60
Base on the result of the
computation the z-stat is
14.60. The decision is to
reject the Null Hypothesis
and accept the
Alternative Hypothesis .
Therefore the medication
significantly affected the
intelligence.
2. How do UMD students measure up on the older version of
the verbal GRE? We know that the population average on
the old version of the GRE (from ETS) was 554 with a
standard deviation of 99. Our sample of 90 UMD students
had an average of 568. Is the 14 point difference in
averages enough to say that UMD students perform better
than the general population?
Given in problem: μM = μ = 554, σ = 99
M = 568, N = 90
Remember that if we use distribution of means, we are using
a sample and need to use standard error.
436.10
90
99

N
M


Given in problem: μM = μ = 554, σ = 99 M = 568, N = 90
Consult z table for z = 1.34
436.10
90
99

N
M


  34.1
436.10
)554568(





M
MM
z


The formula Z-TEST is a
STATISTICAL PROCEDURE used to
test null hypothesis against alternative
hypothesis where the formula is
Where the data needed for the formula
is the mean population( 𝑋), sample
mean(𝜇),standard deviation(𝛿) and the
no.of respondents(n).
n
s
x
t 0

is a statistical test which is widely used
to compare the mean of two groups of
samples. It is therefore to evaluate
whether the means of the two sets of data
are statistically significantly different from
each other.
to be specific, for use with a single
group or sample of data.
when the population
variance is unknown.
A T-Value ( A WHOLE NUMBER/ DECIMAL NUMBER )
have to read in t test table the critical value
of t distribution corresponding to the
significance level alpha of your choice
(5%). To make Decision / Conclusion.
FOR THE FORMULA TO
BE USED :
a set of values with size n, with
mean 𝑋 and with standard
deviation S. The comparison of
the observed mean 𝑋, of the
population to a theoretical value
μ.
A research group claims that
heavy traffic causes
employees to be late by three
days.
EMPLOYEE LATE
1 5
2 2
3 4
4 4
5 5
6 3
7 5
8 3
9 4
10 5
11 3
12 5
13 4
14 5
15 2
X-bar = 3.9333
S = 1.1547
n = 15
Ho : LATE = 3 DAYS
Ha : LATE ≠ 3 DAYS
n
s
x
t 0

=
3.9333 ─ 3
1.1547
√15
= 3.1303
Base on the result of the
computation the t-stat is
3.1303. The decision is to
reject the Null Hypothesis
and accept the
Alternative Hypothesis .
Therefore the number of
times that the employees
are late is not equal to
three.
The formula T-TEST is a
STATISTICAL PROCEDURE
compare the mean of two groups
of samples
Where the data needed for the
formula is the mean population( 𝑋),
sample mean(𝜇),standard
deviation(𝑆) and the no.of
respondents(n).
n
s
x
t 0

As to use this formula make sure
that the research is all about the
population and sample mean.
The difference between the two
mean is the basis.
Also by using the formula it will
determine if a certain hypothesis
need to reject or accept.

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Math Investigation Project

  • 2.
  • 3. - IS A STATISTICAL PROCEDURE USED TO TEST AN ALTERNATIVE HYPOTHESIS AGAINST A NULL HYPOTHESIS.
  • 4. TO DETERMINE WHETHER TWO SAMPLES’ MEAN ARE DIFFERENT WHEN VARIANCES/STANDARD DEVIATIONS ARE KNOWN AND SAMPLE IS LARGE ( n ≥ 30 ).
  • 5. A Z-VALUE( A WHOLE NUMBER/ DECIMAL NUMBER ) WHICH WILL BE COMPARED TO THE VALUE OF ALPHA( LEVEL OF SIGNIFICANCE) TO MAKE A DECISION/ CONCLUSION.
  • 6. FOR THE FORMULA TO BE USED : 𝜇 is called mu, 𝜎 is called sigma
  • 7. This is the formula for Standard Deviation: A population is a collection of persons, objects or items of interest. A sample is a portion of the whole and, if properly taken, is representative of the whole. The population mean is represented by the Greek letter mu (μ). It is given by the formula 𝜇 = 𝑥 𝑁 The capital Greek letter sigma (𝚺) is commonly used in mathematics to represent a summation of all the numbers in a grouping. N is the number of terms in the population. The sample mean is represented by x bar . It is given by the formula 𝑋= 𝑥 𝑛
  • 8. 1.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?
  • 9. 𝑋 = 140 𝜇 = 100 𝛿 = 15 n = 30 Ho : 𝜇 = 140 Ha : 𝜇 ≠ 140
  • 10. = √ 140 − 100 15 30 = 14.60 Base on the result of the computation the z-stat is 14.60. The decision is to reject the Null Hypothesis and accept the Alternative Hypothesis . Therefore the medication significantly affected the intelligence.
  • 11. 2. How do UMD students measure up on the older version of the verbal GRE? We know that the population average on the old version of the GRE (from ETS) was 554 with a standard deviation of 99. Our sample of 90 UMD students had an average of 568. Is the 14 point difference in averages enough to say that UMD students perform better than the general population? Given in problem: μM = μ = 554, σ = 99 M = 568, N = 90 Remember that if we use distribution of means, we are using a sample and need to use standard error. 436.10 90 99  N M  
  • 12. Given in problem: μM = μ = 554, σ = 99 M = 568, N = 90 Consult z table for z = 1.34 436.10 90 99  N M     34.1 436.10 )554568(      M MM z  
  • 13. The formula Z-TEST is a STATISTICAL PROCEDURE used to test null hypothesis against alternative hypothesis where the formula is Where the data needed for the formula is the mean population( 𝑋), sample mean(𝜇),standard deviation(𝛿) and the no.of respondents(n).
  • 15. is a statistical test which is widely used to compare the mean of two groups of samples. It is therefore to evaluate whether the means of the two sets of data are statistically significantly different from each other.
  • 16. to be specific, for use with a single group or sample of data. when the population variance is unknown.
  • 17. A T-Value ( A WHOLE NUMBER/ DECIMAL NUMBER ) have to read in t test table the critical value of t distribution corresponding to the significance level alpha of your choice (5%). To make Decision / Conclusion.
  • 18. FOR THE FORMULA TO BE USED : a set of values with size n, with mean 𝑋 and with standard deviation S. The comparison of the observed mean 𝑋, of the population to a theoretical value μ.
  • 19. A research group claims that heavy traffic causes employees to be late by three days.
  • 20. EMPLOYEE LATE 1 5 2 2 3 4 4 4 5 5 6 3 7 5 8 3 9 4 10 5 11 3 12 5 13 4 14 5 15 2 X-bar = 3.9333 S = 1.1547 n = 15 Ho : LATE = 3 DAYS Ha : LATE ≠ 3 DAYS
  • 21. n s x t 0  = 3.9333 ─ 3 1.1547 √15 = 3.1303 Base on the result of the computation the t-stat is 3.1303. The decision is to reject the Null Hypothesis and accept the Alternative Hypothesis . Therefore the number of times that the employees are late is not equal to three.
  • 22. The formula T-TEST is a STATISTICAL PROCEDURE compare the mean of two groups of samples Where the data needed for the formula is the mean population( 𝑋), sample mean(𝜇),standard deviation(𝑆) and the no.of respondents(n). n s x t 0 
  • 23. As to use this formula make sure that the research is all about the population and sample mean. The difference between the two mean is the basis. Also by using the formula it will determine if a certain hypothesis need to reject or accept.