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BASIC STATISTICAL INFERENCE

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BASIC STATISTICAL INFERENCE

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BASIC STATISTICAL INFERENCE

PARAMETERIC TESTS

(QUANTITATIVE DATA)

A. COMPARE BETWEEN TWO MEANS OF POPULATIONS

z-distribution

t-distribution

&

B. COMPARE BETWEEN TWO VARIANCES OF POPULATIONS

f-distribution (fisher’s distribution)

BASIC STATISTICAL INFERENCE

TEST THE NULL HYPOTHESIS

We shall consider here three forms for the alternative hypothesis:

TEST THE ALTERNATIVE HYPOTHESIS

usually represents some postulated value which needs confirmation.

One tailed test

0.05 significant level

Not significant

0.95

Distribution showing 0.05 significant level in one-tailed test

Insignificant difference

P > 0.05

Not significant

?

0.05 significant level

P < 0.05

P < 0.001

P < 0.01

0.95

Distribution showing 0.05 significant level in one-tailed test

Two tailed test

0.05 significant level

0.05 significant level

Not significant

0.95

Distribution showing 0.05 significant level in two-tailed test

Calculated t

Frequency

0.05

0.01

0.001

𝝈

𝝈

Tabulated t

Accept H0

Reject H0

Reject H0

P > 0.05

HYPOTHESIS TESTS ON THE MEAN (LARGE SAMPLES >30)

Mean sample

A given fixed value to be tested

Calculated z

Sample size (>30)

Population standard deviation

HYPOTHESIS TESTS ON THE MEAN (SMALL SAMPLES <30)

Mean sample

A given fixed value to be tested

Calculated z

Sample size (<30)

Sample standard deviation

One sample t-distribution

- To decide if a sample mean is different from a hypothesized population mean.
- You have calculated mean value and standard deviation for the group assuming you have measurement data. where the standard score (t) is:

Degree of freedom (n-1)

t-distribution

- The percentiles values of the t-distribution (tp) are tabulated for a range of values of d.f. and several values of p are represented in a Table .

One sample t-DISTRIBUTION

Example

The mean concentration of cadmium in water sample was 4 ppm for sample size 7 and a standard deviation=0.9 ppm. The allowable limit for this metal is 2 ppm. Test whether or notthe cadmium level in water sample at the allowable limit.

Solution

T cal(2.447) > t tab (2.447)

Reject the null hypothesis

Reject the null hypothesis

T cal(2.447) > t tab (3.707)

Decision:

Thus the cadmium level in water is not at the allowable limit.

Two sample INDEPENDENT t-DISTRIBUTION

Example:

In an New Zealand, Does the average mass of male turtles in location A was significantly higher than Location B?

=

d.f.= n1 + n2 - 2

= 25 + 26 - 2 = 49

Tabulated t at df 59 = 1.671

Thus,tobserved (3.35) > ttabulated (1.67) at α= 0.05

The mass of male turtles in location A is significantly higher than those of location B (reject H0) P<0.05

d.f.= n1 + n2 - 2 = 11+11 -2= 20

=70.45

=60.18

tcalculated(2.209) > ttabulated(2.086) at d.f. 20

Two sample DEPENDENT t-DISTRIBUTION

TESTING THE DIFFERENCE BETWEEN

TWO MEANS OF DEPENDENT SAMPLES

d.f. = n - 1

?

d.f. = 10 – 1= 9

ttabulated at d.f. 10 = 1.833

Fisher’s F-distribution

That is, you will test the null hypothesis H0: σ12 = σ22against an appropriate alternate hypothesis Ha: σ12≠ σ22.

You calculate the F-value as the ratio of the two variances:

- where s12 ≥ s22, so that F ≥ 1.
- The degrees of freedom for the numerator and denominator are n1-1 and n2-1, respectively.
- Compare Fcalc.to a tabulated value Ftab.to see if you should accept or reject the null hypothesis.

Example:

Assume we want to see if a Method 1for measuring the arsenic concentration in soil is significantly more precise than Method 2. Each method was tested ten times, with yielding the following values:

Solution:

So we want to test the null hypothesis H0: σ22 = σ12

against the alternate hypothesis HA: σ22 > σ12

d.f.= 10 – 1 = 9

- The tabulated value for d.f.= 9 in each case, at 1-tailed, 95% confidence level is F9,9 = 3.179.
- In this case, Fcalc < F9,9 tabulated, so we Accept H0that the two standard deviations are equal,so P > 0.05

- We use a 1-tailed test in this case because the only information we are interested in is whether Method 1 is more precise than Method 2