STATISTICS 2
LECTURE 1: RECAP, TYPE I AND TYPE II ERRORS
GAUSS CURVE
Normal curve
All measurements and mean
- Between -1ơ & +1ơ 68% of all measurements
- Between -2ơ & +2ơ 95% of all measurements
- Between -3ơ & +3ơ 99,7% of all measurements
- Chance α = 0,05
- N < 30 = student-t
Normal Sample (steekproef) (<30 samples)
z t
Stdev ơ s
mean µ x(met streepje erboven)
EXERCISE CONFICENCE TEST
In researching the amount of active ingredient of a drug, a mean of a series of
measurements of 5 of 49,6 mg active ingredient was found, with a standard deviation ơ =
0,60 mg. the drug should contain 50 mg of active ingredient.
- Is there a significant difference between the found mean and the true value of 50
mg?
n=5
x(met streepje) = 49,6 mg
ơ = 0,60 mg
it should contain 50 mg
95% BI so z value is (=norm.s.inv) is 1,96
H0: x = µ
H1: x ≠ µ or x > µ or x < µ
Method 1: is µ in 95% CI of x
49,6 ± (1,96 * 0,60) / WORTEL 5
BI(95%) = 49,6 ± 0,52
µ = 50 is in this interval so NO significant difference
Method 2: one sample z-test: is z-calc larger than z-critical (z-table)
- with a 95% CI a z-value of -1,96 and +1,96
- Z calc is closer to 0 than z critical.
- NO significant difference between µ and x.
Method 3: p-value
, z calc is 1,491; what area (p value) belongs to this?
- Excel: =NORM.S.VERD(1,491;WAAR)
P = 0,0681 of 6,81%.
1 sited: cut-off value at 95% confidence: p=5%
2 sited: cut-off value at 95% confidence: p=2,5%
- P calc is larger than p cut-off value: NO significant difference between µ and x.
P-value
A change on the right hand sight of the actual value.
EXERCISE
- Is a found mean of 52 (n=6) and a standard deviation of 0,62 mg significantly
larger than 50 mg?
< 30 samples dus t.
s = 0,62 mg.
x = 52.
=NORM.INV(0,95;50;0,62) = 51,02.
t =T.INV.2T(0,05;n-1) = 2,57
52 – (2,57*0,62)/WORTEL(6) = 51,3.
t berekend > t kritisch (table) there’s a difference.
POWER CALCULATIONS
Type I error (α) false positive (male is pregnant)
Type II error (β) false negative (female is pregnant but test says not pregnant)
Type I error
A point of view is
population H0.
- Normal distribution: not
pregnant.
- If lager than x critical
(z-value, 1-sited) then
PRENGANT (but not
necessarily true
(=small change): false
positive.
- The conclusion: the H0 is NOT true, while in fact it IS true.
- Chance is α (unreliability) usually 5%.
Type II error
Point of view population Ha.
LECTURE 1: RECAP, TYPE I AND TYPE II ERRORS
GAUSS CURVE
Normal curve
All measurements and mean
- Between -1ơ & +1ơ 68% of all measurements
- Between -2ơ & +2ơ 95% of all measurements
- Between -3ơ & +3ơ 99,7% of all measurements
- Chance α = 0,05
- N < 30 = student-t
Normal Sample (steekproef) (<30 samples)
z t
Stdev ơ s
mean µ x(met streepje erboven)
EXERCISE CONFICENCE TEST
In researching the amount of active ingredient of a drug, a mean of a series of
measurements of 5 of 49,6 mg active ingredient was found, with a standard deviation ơ =
0,60 mg. the drug should contain 50 mg of active ingredient.
- Is there a significant difference between the found mean and the true value of 50
mg?
n=5
x(met streepje) = 49,6 mg
ơ = 0,60 mg
it should contain 50 mg
95% BI so z value is (=norm.s.inv) is 1,96
H0: x = µ
H1: x ≠ µ or x > µ or x < µ
Method 1: is µ in 95% CI of x
49,6 ± (1,96 * 0,60) / WORTEL 5
BI(95%) = 49,6 ± 0,52
µ = 50 is in this interval so NO significant difference
Method 2: one sample z-test: is z-calc larger than z-critical (z-table)
- with a 95% CI a z-value of -1,96 and +1,96
- Z calc is closer to 0 than z critical.
- NO significant difference between µ and x.
Method 3: p-value
, z calc is 1,491; what area (p value) belongs to this?
- Excel: =NORM.S.VERD(1,491;WAAR)
P = 0,0681 of 6,81%.
1 sited: cut-off value at 95% confidence: p=5%
2 sited: cut-off value at 95% confidence: p=2,5%
- P calc is larger than p cut-off value: NO significant difference between µ and x.
P-value
A change on the right hand sight of the actual value.
EXERCISE
- Is a found mean of 52 (n=6) and a standard deviation of 0,62 mg significantly
larger than 50 mg?
< 30 samples dus t.
s = 0,62 mg.
x = 52.
=NORM.INV(0,95;50;0,62) = 51,02.
t =T.INV.2T(0,05;n-1) = 2,57
52 – (2,57*0,62)/WORTEL(6) = 51,3.
t berekend > t kritisch (table) there’s a difference.
POWER CALCULATIONS
Type I error (α) false positive (male is pregnant)
Type II error (β) false negative (female is pregnant but test says not pregnant)
Type I error
A point of view is
population H0.
- Normal distribution: not
pregnant.
- If lager than x critical
(z-value, 1-sited) then
PRENGANT (but not
necessarily true
(=small change): false
positive.
- The conclusion: the H0 is NOT true, while in fact it IS true.
- Chance is α (unreliability) usually 5%.
Type II error
Point of view population Ha.
