BASIC STATISTICAL INFERENCE
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BASIC STATISTICAL INFERENCE
A. COMPARE BETWEEN TWO MEANS OF POPULATIONSB. COMPARE BETWEEN TWO VARIANCES OF POPULATIONS
PARAMETERIC TESTS(QUANTITATIVE DATA)
t-distribution z-distribution
f-distribution (fisher’s distribution)
&
BASIC STATISTICAL INFERENCE
𝝁 ,
TEST THE NULL HYPOTHESIS 𝑯 𝟎 :𝝁=𝝁𝟎
We shall consider here three forms for the alternative hypothesis:
𝑯𝟏 :𝝁>𝝁𝟎 , 𝑯 𝟏 :𝝁<𝝁𝟎 , 𝑯𝟏 :𝝁≠𝝁𝟎 ,
TEST THE ALTERNATIVE HYPOTHESIS 𝑯𝟎 :𝝈=𝝈𝟎
𝑯𝟏 :𝝈>𝝈𝟎𝑯 𝟏:𝝈<𝝈𝟎 𝑯𝟏 :𝝈≠𝝈𝟎
𝝈 ,
Not significant
Distribution showing 0.05 significant level in one-tailed test
0.05 significant level
0.95
One tailed test
P < 0.05 P < 0.01 P < 0.001P > 0.05 Insignificant difference
(𝟏−𝜶 )
Not significant
Distribution showing 0.05 significant level in one-tailed test
0.05 significant level
0.95(𝟏−𝜶 ) 𝜶=𝟎 .𝟎𝟓
𝜶=𝟎 .𝟎𝟓
?
𝑯𝟏 :𝝁>𝝁𝟎 ,
𝑯𝟏 :𝝁<𝝁𝟎 ,
𝑯𝟏 :𝝈>𝝈𝟎
𝑯𝟏 :𝝈<𝝈𝟎
𝑯𝟏 :𝝁≠𝝁𝟎
Distribution showing 0.05 significant level in two-tailed test
0.05 significant level 0.05 significant level
0.95
Two tailed test
Not significant
(𝟏−𝜶 ) 𝜶𝟐=
𝟎 .𝟎𝟓𝟐 =𝟎 .𝟎𝟐𝟓𝜶
𝟐=𝟎 .𝟎𝟓𝟐 =𝟎 .𝟎𝟐𝟓
𝑯𝟏 :𝝈≠𝝈𝟎
Freq
uenc
y
−𝟐𝝈𝝈
0.050.01
Calculated t
𝞵𝝈
0.001
Tabulated t
𝑷>𝟎 .𝟎𝟓
𝑷<𝟎 .𝟎𝟓𝑷<𝟎 .𝟎𝟏𝑷<𝟎 .𝟎𝟎𝟏
Accept H0 Reject H0Reject H0
P > 0.05
𝟏−𝛂
𝑷<𝟎 .𝟎𝟓𝑷<𝟎 .𝟎𝟏
𝑷<𝟎 .𝟎𝟎𝟎𝟏𝑷<𝟎 .𝟎𝟎𝟏
𝒁 𝒔=𝑿−𝝁𝟎
𝝈 /√𝒏
Calculated z
Mean sample
A given fixed value to be tested
Population standard deviation Sample size (>30)
HYPOTHESIS TESTS ON THE MEAN (LARGE SAMPLES >30)
𝒕 𝒔=𝑿−𝝁𝟎
𝑺 /√𝒏
Calculated z
Mean sample
A given fixed value to be tested
Sample standard deviation Sample size (<30)
HYPOTHESIS TESTS ON THE MEAN (SMALL SAMPLES <30)
o To decide if a sample mean is different from a hypothesized population mean.o You have calculated mean value and standard deviation for the group assuming
you have measurement data. where the standard score (t) is:
𝐭𝐜𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞𝐝=𝐗−𝛍❑
𝑺 .𝑬 .
One sample t-distribution
𝐭 𝐭𝐚𝐛𝐮𝐥𝐚𝐭𝐞𝐝 :(𝐝 .𝐟 . ,𝐩𝐯𝐚𝐥𝐮𝐞)Degree of freedom (n-1)
t-distribution
o 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 .
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 not the cadmium level in water sample at the allowable limit.
Example
Solution
T cal (2.447) > t tab (2.447) Reject the null hypothesis
T cal (2.447) > t tab (3.707) Reject the null hypothesis
𝑷<𝟎 .𝟎𝟏Decision:
Thus the cadmium level in water is not at the allowable limit.
One sample t-DISTRIBUTION
Example: In an New Zealand, Does the average mass of male turtles in location A was significantly higher than Location B?
Location A Location B
n 25 2638 35
S 4 3
=
𝑯 𝒂 :𝝁 𝑨>𝝁𝑩
𝑺𝑷𝟐=
(𝒏𝟏−𝟏)𝑺𝟏𝟐+ (𝒏𝟐−𝟏 )𝑺𝟐
𝟐
𝒏𝟏+𝒏𝟐−𝟐
d.f. = n1 + n2 - 2
¿(𝟐𝟓−𝟏 )∗𝟏𝟔+(𝟐𝟔−𝟏 )∗𝟗
𝟐𝟓+𝟐𝟔−𝟐 =𝟔𝟗𝟗𝟓𝟗 =𝟏𝟏 .𝟖𝟓
¿(𝟑𝟖−𝟑𝟓 )
√𝟏𝟏 .𝟖𝟓 ( 𝟏𝟐𝟓 + 𝟏𝟐𝟔 )
=¿ 𝟑√𝟎 .𝟖𝟎
=𝟑 .𝟑𝟓
= 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
Two sample INDEPENDENT t-DISTRIBUTION
Control (X1) Pb (X2)
79 63
83 71
68 46
59 57
81 53
76 46
80 57
74 76
58 52
49 68
68 73
𝐒𝐒𝟏=∑𝐗𝟏𝟐−
(𝐗𝟏 )𝟐
𝐧𝟏
(X1)2
62416889452434816561577664005476336424014624
(X2)2
39695041211632492809211632495776270446245329
775 662 55837 40982
𝐒𝐒𝟏=𝟓𝟓𝟖𝟑𝟕−(𝟕𝟕𝟓 )𝟐
𝟏𝟏 =𝟏𝟐𝟑𝟒 .𝟕𝟑
𝐒𝐒𝟐=𝟒𝟎𝟗𝟖𝟐− (𝟔𝟔𝟐 )𝟐
𝟏𝟏 =𝟏𝟏𝟒𝟏 .𝟔𝟒
=70.45 =60.18
𝐭= 𝐗𝟏−𝐗𝟐
√( 𝐒𝐒𝟏+𝐒𝐒𝟐
(𝐧𝟏+𝐧𝟐 )−𝟐 )( 𝟏𝐧𝟏+ 𝟏𝐧𝟐 )
tcalculated (2.209) > ttabulated (2.086) at d.f. 20
d.f. = n1 + n2 - 2 = 11+11 -2= 20
𝒕= ∑ 𝑫
√𝒏∑ 𝑫𝟐− (𝑫 )𝟐
𝒏−𝟏𝑫=𝑿𝟐−𝑿𝟏
d.f. = n - 1Before (X1) After (X2)
14 0
6 0
4 3
15 20
3 0
3 0
6 1
5 1
6 1
3 0
D
-14-6-15-3-3-5-4-5-3
D2
196361
2599
2516259
𝒕= −𝟑𝟗𝟏𝟒 .𝟖𝟕=−𝟐 .𝟔𝟐𝟑d.f. = 10 – 1= 9ttabulated at d.f. 10 = 1.833 ?
-39 351
TESTING THE DIFFERENCE BETWEEN TWO MEANS OF DEPENDENT SAMPLES
Two sample DEPENDENT t-DISTRIBUTION
1) That is, you will test the null hypothesis H0: σ12 = σ2
2 against an appropriate alternate hypothesis Ha: σ1
2 ≠ σ22 .
2) You calculate the F-value as the ratio of the two variances:
𝑭 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅=𝒔𝟏𝟐
𝒔𝟐𝟐 where s1
2 ≥ 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.
Fisher’s F-distribution
Example:Assume we want to see if a Method 1 for measuring the arsenic
concentration in soil is significantly more precise than Method 2. Each method was tested ten times, with yielding the following values:
Methods Mean (ppm) S.D. (ppm)
Method 1 6.7 0.8Method 2 8.2 1.2
So we want to test the null hypothesis H0: σ22 = σ1
2 against the alternate hypothesis HA: σ2
2 > σ12
Solution:
∵𝑭 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅=𝒔𝟏𝟐
𝒔𝟐𝟐¿𝟏 .𝟐❑
𝟐
𝟎 .𝟖❑𝟐¿𝟐 .𝟐𝟓
o The tabulated value for d.f.= 9 in each case, at 1-tailed, 95% confidence level is F9,9 = 3.179.
o In this case, Fcalc < F9,9 tabulated, so we Accept H0 that the two standard deviations are equal, so P > 0.05
d.f.= 10 – 1 = 9
o 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
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