Lecture 9 - Hypothesis testing
82
Hypothesis; A statement about one or more population.
The hypothesis usually concern with the parameters of the
population about which statement is made.
The purpose of the hypothesis testing is to help the
clinician, researcher, and administrator in reaching a
decision concerning a population based on results of a
sample that drawn from this population.
The procedures of hypothesis testing:
Understand the nature of the data (to determine the
particular test employed).
State the hypothesis
a) Null hypothesis or tested hypothesis (H
0
): Hypothesis
of no difference, hypothesis of equality.
b) Alternative hypothesis (H
A
): It disagree the null
hypothesis (e.g. there is difference).
Find the tabulated Z, t or X
2
values (type of the test
depends on the type of the data) according to
⍺ (usually
0.05).This will present the "critical values" that separate
the acceptance region from rejection regions.
Find the calculated Z, t or X
2
values (type of the test
depends on the type of the data).
Compare between the tabulated and calculated values, If
the calculated value falls in the acceptance area ➨ we
accept the H
0
-hypothesis, but If the calculated value falls
in the rejection area ➨ we reject the H
0
-hypothesis in
favoring the alternative one (H
A
).
Conclusion. We accept the H
0
-hypothesis ➨We conclude
that there is no difference or association, but if we reject
the H0-hypothesis then we favoring the alternative one
(H
A
) and we conclude the H
A
may be true.
1) Hypothesis testing for single population mean and
known population variance. ''Calculated Z-value= (x-
µ) /(∂/√n)''
A certain breed of rats show as a mean weight gain of 65
gm during the first 3 months of life with a variance of 10
gm
2
. A sample of 16 of these rats were fed a new diet
from birth until the age of 3 months, their mean weight
gain was 60.75 gm. Does this mean that the new diet case
reduction in Wt gain at 0.05 level of significant? Test this
hypothesis.
(H
0
): x = µ (no difference).
(H
A
): x ≠ µ (difference exists)
Tabulated Z:
⍺= 0.05 1- ⍺= 95%
Z = ±1.96 (critical value).
Calculated Z = (x-µ) / (∂/√n)
= 60.57- 65 / √10 /√16 = -5.38
Comparison: since the calculated Z value > tabulated Z
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the difference
between x & µ is statistically significant.
Conclusion: we may conclude that the new diet cause
reduction in WT (-ve value).
2) Hypothesis testing for single population mean and
unknown population variance (Sample size ≤ 30).
''Calculated t-value= (x-µ) /(S/√n)''
Ex: In the previous example, if the population variance
was unknown and the sample SD= 3.84 gm. Does this
mean that the new diet case reduction in Wt gain at 0.05
level of significant? Test this hypothesis.
(H
0
): x = µ (no difference).
(H
A
): x ≠ µ (difference exists)
Tabulated t:
⍺= 0.05 1- ⍺/2= 0.975, df= n-1
t = ±2.1315 (critical value).
Calculated t = (x-µ) / (S/√n)
= 60.57- 65 / 3.84 /√16 = -4.43
Comparison: since the calculated t value > tabulated t
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the difference
between x & µ is statistically significant.
Conclusion: we may conclude that the new diet cause
reduction in Wt (-ve value).
3) Hypothesis testing for the difference between two
population means when population variances are
known.
''Calculated Z-value= (x
1
-x
2
)-(µ
1
-µ
2
) / √
[(∂
2
1
/n
1
) + (∂
2
2
/n
2
)]
Ex: 70 patients suffering from epileptic fit were dividing
into two groups equally. Group 1 placed on treatment
(Tegretol, 200mg x 2) and group 2 placed on placebo. The
mean of the number of seizures experienced during the
period of treatment by the two groups were 15 and 24
consequently, the population variances were 8 and 12, do
these data provide sufficient evidence to indicate that
''Tegretol'' is effective drug in reducing the number of
seizure at 0.05 level of significant
(H
0
): µ
1
=µ
2
(no difference).
(H
A
): µ
1
≠µ
2
(difference exists)
Tabulated Z:
⍺= 0.05 1- ⍺= 95%
Lecture 9 - Hypothesis testing
83
Z = ±1.96 (critical value).
Calculated Z = (x
1
-x
2
)-(µ
1
-µ
2
) / √ [(∂
2
1
/n
1
) + (∂
2
2
/n
2
)]
= (15 – 24) – 0 / √ (8 /35) + (12/35) = -11.9
Comparison: since the calculated Z value > tabulated Z
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the difference
between µ
1
and µ
2
is statistically significant.
Conclusion: we may conclude that ''Tegretol'' is
effective in reducing no. of seizures.
4) Hypothesis testing for the difference between two
population means when population variances are
unknown (Sample size ≤ 30).
''Calculated t-value= (x
1
-x
2
)-(µ
1
-µ
2
) / √ [(S
2
1
/n
1
) +
(S
2
2
/n
2
)].
Ex: Median nerve conducting velocity values were
recorded for 10 subjects with a diagnosis of mercury
poisoning, similar determination also were made for 15
apparently healthy subjects. The results were as follow:
Group
N
Mean(sec\mm)
S(sec\mm)
With
poisoning
10 55
6
Healthy
15 63
5
Do these data provide sufficient evidence to indicate that
nerve conducting velocity was affected by mercury
poisoning?
(H
0
): µ
1
=µ
2
(no difference).
(H
A
): µ
1
≠µ
2
(difference exists)
Tabulated t:
⍺= 0.05 1- ⍺/2 = 0.975 df= 10+15-2= 23
t = ±2.068 (critical value).
Calculated t = (x
1
-x
2
)-(µ
1
-µ
2
) / √ [(S
2
1
/n
1
) + (S
2
2
/n
2
)]
= (55 – 63) – 0 / √ (36 /10) + (25/15) = -3.485
Comparison: since the calculated t value > tabulated t
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the difference
between µ
1
and µ
2
is statistically significant.
Conclusion: we may conclude that mercury poisoning
reduce nerve conduction.
5) Hypothesis testing for single population proportion
(P)
"Calculated Z= (P-P)/√P (1-P)/n"
Ex: Suppose we are interesting in knowing what
proportion of automobile driver regularly wear seat belts.
In survey of 300 adults, 123 said they regularly were seat
belts. Can we conclude from these data that in this sample
the proportion who regularly wears seat belts is not 50%?
(H
0
): P = P (no difference).
(H
A
): P ≠ P (difference exists)
Tabulated Z:
⍺= 0.05 1- ⍺= 95%
Z = ±1.96 (critical value).
Calculated Z = (P-P)/√P (1-P)/n
(0.41- 0.5) /√0.5 (1-0.5)/300 = -3.11
Comparison: since the calculated Z value > tabulated Z
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the difference
between p & p is statistically significant.
Conclusion: we may conclude that in this sample the
proportion who regularly wears seat belts is not 50%
(less, -ve value).
6) Hypothesis testing for the difference between two
populations proportions (P
1
-P
2
). We also use z-test and
the formula is;
Calculated Z = (P
1
-P
2
) - (P
1
-P
2
) /√ [(P
1
(1- P
1
)/ n
1
) + P
2
(1- P
2
)/ n
2
)]
Ex: In a study of DM, we have the following results
obtained from samples of male and female. Male n
1
=150,
no. of DM=21. Female n2=200, no. of DM=48. Can we
conclude from these data that there is a difference in the
proportion of DM between the two samples?
(H
0
): P
1
= P
2
(no difference).
(H
A
): P
1
≠ P
2
(difference exists)
Tabulated Z:
⍺= 0.05 1- ⍺= 95%
Z = ±1.96 (critical value).
Calculated Z = (P
1
-P
2
) - (P
1
-P
2
) /√ [(P
1
(1- P
1
)/ n
1
) + P
2
(1- P
2
)/ n
2
)]
(0.14- 0.24)-0 /√ [0.14 (1-0.14)/150] + [0.24(1-0.24)/200]
= -58.38
Comparison: since the calculated Z value > tabulated Z
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the difference
between P
1
& P
2
is statistically significant.
Conclusion: we may conclude that in this sample the
proportion who regularly wears seat belts is not 50%
(less, -ve value).
Lecture 9 - Hypothesis testing
84
7) Hypothesis testing in pair comparison.
Calculated t= [d- µd] / [Sd/√n]
Ex: A group of 15 boys (12 years old), were measured for
height by 2 nurses, the results were as in the table below.
Do these data justify the conclusion that that there is a
difference in the accuracy of the 2 nurses?
d (mean difference) = ∑d / n = 3.7 / 15 = 0.75 Cm
Sd =√ [n ∑d
2
- (∑d)
2
] / [n(n-1)]
= √ [15(3.05) – (3.7)
2
]\ [15(15-1)] = 0.33
(H
0
): µd = 0 (no difference).
(H
A
):µd ≠ µ (difference exists)
Tabulated t:
⍺= 0.05 1- ⍺/2= 0.975, df= n-1
t = ±2.144 (critical value).
Calculated t = [d- µd] / [Sd/√n]
= 0.75- 0 / 0.33 /√15 = 3.125
Comparison: since the calculated t value > tabulated t
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that µd ≠ µ, the
difference is statistically significant.
Conclusion: we may conclude that there is a difference
in the accuracy of height measurement between the 2
nurses.
8) Hypothesis testing in Chi-Square distribution (X
2
-
test).
Calculated X
2
= ∑ (O-E)
2
/E
Ex: In a study for the association between increasing
diastolic blood pressure and CVA development, 200
individuals were followed for 5 years, the results is shown
in below table. Do these data suggest an association
between increasing diastolic blood pressure and
development of CVA? Test a reasonable hypothesis. (
⍺=
0.05)
Diast. B.P
(mm Hg)
Development of CVA
Yes No
Total
70-79
1
49
50
80-89
4
46
50
90-99
6
44
50
≥ 100
13
37
50
Total
24
176
200
(H
0
): No association between increasing diastolic BP &
CVA.
(H
A
): Association exists.
Tabulated X
2
: (
⍺= 0.05), df =(r-1)(c-1) =3
From X
2
-distribution table ➨The tabulated value is 7.815
Calculated X
2
=
∑ (O-E)
2
/E
From the table above (observed values), we calculate the
expected values for each cell in the table using the
formula: E= [Raw margin X Column margin] / Grand
total.
Expected values (E).
Diast. B.P
(mm Hg)
Development of CVA
Yes No
Total
70-79
6
44
50
80-89
6
44
50
90-99
6
44
50
≥ 100
6
44
50
Total
24
176
200
Calculated X
2
= 14.78
Comparison: since the calculated X
2
value > tabulated X
2
(falls in the rejection area), so we reject (H
0
) in favoring
the alternative one (H
A
) which states that the association
between increasing diastolic blood pressure and
development of CVA is statistically significant.
Conclusion: we may conclude that the increasing in
diastolic blood pressure lead to development of CVA.
No.
Height (cm)
Difference (d)
(N
2
-N
1
)
d
2
Nurse 1
Nurse 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
142.9
150.9
151.9
158.1
151.2
160.2
157.8
150.1
142.1
159.9
141.9
140.8
147.1
143.6
139.9
143
151.5
152.1
158
151.5
160.5
158
150
142.5
160
142
141
148
144
141
0.1
0.6
0.2
-0.1
0.3
0.3
0.2
-0.1
0.4
0.1
0.1
0.2
0.9
0.4
1.1
0.01
0.36
0.04
0.10
0.09
0.09
0.04
0.01
0.16
0.01
0.01
0.04
0.81
0.16
1.21
∑d=3.7
∑d
2
=3.05