Добавил:
Sekretar
kiopkiopkiop18@yandex.ru
t.me/Prokururor I Вовсе не секретарь, но почту проверяю
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_155_библиотеки_им_акад_М_И_Перельмана
.pdf
11.2 Comparing the Mean of a Population with a Fixed Value 229
The Variance of Population is Unknown
Example 11.2 The recommended average level of energy intake for women aged
20–30 is 7725 kcal/day. In a random sample of 11 healthy women in the same
age group, the average energy intake of 10 days was measured and the values of
5260–5470–5640–6180–6390–6515–6805–7515–8230–8770–7515 were recorded.
Assuming that the distribution of the above population is normal, is their daily energy
intake equal to the recommended level?
Solution: We can extract and calculate the following measures:
μ
0
= 7725, n = 11, x = 6753.6,
S = 1142.1 and Se =
S
√
n
=
1142.1
√
11
= 344.4
Since the variance of the population is not given, instead of the z distribution, the
t distribution with degree of freedom df = n−1 = 11–1 = 10 should be used. Its
critical value from the t table is equal to
t
(
0.975,10
)
= 2.228 and the 95% confidence
interval is obtained as follows:
95% CI = x ± t
(
0.975,10
)
× Se = 6753.6 ± 2.228 × 344.4
=
(
5986 to 7521
)
The 95% confidence interval does not include the recommended value. Therefore,
the average energy intake of women in t his population cannot be considered equal
to the recommended level and is less than that.
Using the Test:
The null and alternative hypotheses are:
H
0
: μ = 7725
H
A
: μ /= 7725
.
Sample size is 11 and significant level is 0.05.
Test statistic is t =
(
x−μ
0
)
S
√
n
.
Critical value is t
(
0.975,10
)
= 2.228. If the test statistic is greater than the critical
value of 2.228, then we reject the null hypothesis (
H
0
).
According to sample data,
t =
(
x − μ
0
)
σ
√
n
=
(
6753.6 − 7725
)
344.4
√
11
=−2.82.
Since
|
−2.82
|
> 2.228
, then we reject the null hypothesis (H
0
).
Since the difference between the mean of energy intake level of women aged 20–
30 and 7725 is significant, we cannot assume that the mean of energy intake level of
women aged 20–30 is 7725 (Fig.
11.1).
To execute this test by SPSS, first we go to Analyze\Compare Means\One-Sample
T-Test to open another window (Fig.
11.2).
https://t.me/med1917
230 11 Hypothesis Testing
Fig. 11.1 Steps to run t-test in SPSS
Fig. 11.2 Data entry in
SPSS
In this window, move the variable name to be tested to the “Test Variable (s)” box
and type the hypothetical constant value in the “Test Value” box. After pressing the
OK button, the test will run.
Example 11.3 Run the hypothesis test for the previous example (energy intake of
women 20–30 years old) with SPSS.
Solution: The data should be entered in a single column in SPSS as follows.
Then, as shown in Fig. 11.2, select the SPSS path and enter the interested variable
in the related box and press the OK button to have the following output.
One-sample statistics
N Mean Std. deviation Std. error mean
Energy intake 11 6753.64 1142.123 344.363
https://t.me/med1917
11.2 Comparing the Mean of a Population with a Fixed Value 231
One-sample test
Test value = 0
t df Sig. (two-tailed) Mean
difference
95% CI of the difference
Lower Upper
Energy
intake
19.612 10 0.000 6753.636 5986.35 7520.93
As can be seen, the answers are equal to manual calculations. Here the p-value
is given in the fourth column. p = 0.000 is smaller than the significance level (α);
therefore, the null hypothesis is rejected.
Example 11.4 A pharmaceutical company has stated that the percentage of alcohol
in its cough syrup is 5%. The factory quality control unit, in order to investigate
this claim, purchased 10 bottles of that syrup from different pharmacies, and after
chemical analysis, the percentages of alcohol (4.81, 4.85, 4.70, 4.86, 4.86, 5.16, 4.80,
4.72, 5.04, 5.20) is obtained: Assuming the distribution of the percentage of alcohol
in the bottles is distributed normally, do the sample observations confirm the factory
claim?
Solution: Given information on the exercise is:
n = 10 and x = 4.90,
Note that in this example the average percentage of alcohol (not the proportion
of alcoholic syrups) is considered. On the other hand, since the variance of the
population is unknown, we have to find its variance from the sample data and use
the t distribution with degree of freedom df = n−1 = 10–1 = 9 instead of z.
S
2
=
(
5.20 − 4.90
)
2
+
(
5.04 − 4.90
)
2
+ ··· +
(
4.81 − 4.90
)
2
+
n − 1
=
0.2734
9
= 0.0304
The standard deviation is S =
√
0.0304 = 0.174 and the standard error of
estimation is
Se =
S
√
n
=
0.174
√
10
= 0.055.
Its critical value from the table is t
(
0.975,10
)
= 2.262. Then, the 95% confidence
interval is obtained as follows:
95% CI = x ± t
(
0.975,9
)
× Se = 4.90 ± 2.262 × 0.055 = 4.90 ± 0.124
=
(
4.776 to 5.024
)
The 95% confidence interval includes the factory’s claimed percentage. Therefore,
the average alcohol of these syrups can be considered equal to the claimed amount
of 5%.
https://t.me/med1917
232 11 Hypothesis Testing
Using the Test:
The null and alternative hypotheses are:
H
0
: μ = 5
H
A
: μ /= 5
.
Sample size is 10 and significant level is 0.05.
Test statistic is t =
(
x−μ
0
)
S
√
n
.
Critical value is t
(
0.975,9
)
= 2.262. If the test statistic is greater than the critical
value of 2.262, then we reject the null hypothesis (
H
0
).
According to sample data,
t =
(
x − μ
0
)
σ
√
n
=
(
4.9 − 5
)
0.174
√
10
=−1.818.
Since
|
−1.818
|
< 2.262
, then we don’t reject the null hypothesis (H
0
).
The mean percentage of alcohol in syrups produced by this factory can be
considered to be 5%. Because, the difference between the mean of the sample with
the claimed percentage of 5 is not statistically significant.
To execute the significant test in SPSS, first go to Analyze\Compare Means\One-
Sample T Test, to open the following window.
In this window, select the name of the variable (Syroup) from the list on the left
and enter it in the right box titled Test Variable (s). In the Test Value box, enter the
value 0.05, and then confirm the program by pressing OK to run and have the output
as below.
One-sample statistics
N Mean Std. deviation Std. error mean
Syroup 10 4.9000 0.17429 0.05512
https://t.me/med1917
11.3 Comparing the Mean of Two Populations 233
One-sample test
Test value = 0
t df Sig. (two-tailed) Mean
difference
95% CI of the
difference
Lower Upper
Syroup 88.903 9 0.000 4.90000 4.7753 5.0247
11.3 Comparing the Mean of Two Populations
The Variance of the Two Populations is Known.
Example 1.5 In a study, the age of 180 patients with coronavirus was recorded. The
mean age of 49 deaths was 68.5 years with a standard deviation of 15.07 years and
the mean age of 131 recovered patients was 55.69 years with a standard deviation
of 18.08 years. Can the average age of the deaths be equal to the average age of the
recovered ones? Similar studies show that the standard deviation of the age of the
dead and the recovered is 19 and 14 years, respectively.
Solution: The given information in this example is:
n
1
= 49, x
1
= 68.50, S
1
= 15.07,σ
2
1
=
(
19
)
2
= 361
n
2
= 131, x
2
= 55.69, S
2
= 18.08,σ
2
2
=
(
14
)
2
= 196
In order to obtain the pooled standard error of the two groups, the known standard
deviations of the groups, which are given in the example, must be used, and the
samples standard deviation cannot be applied:
Se =
σ
2
1
n
1
+
σ
2
2
n
2
=
361
49
+
196
131
=
√
8.863 = 2.98
The samples are large enough and the 95% confidence interval of the age mean
difference between the two groups of recovers and deaths can be estimated using the
normal approximation (
z
0.975
= 1.96).
CI95% =
(
x
1
− x
2
)
± z
0.975
× Se =
(
68.50 − 55.69
)
± 1.96 × 2.98
= 12.81 ± 5.84 =
(
6.97 to 18.64
)
The estimated interval does not include the value zero. As a result, based on this
example, the null hypothesis is rejected. That is, the mean age in the two groups of
recovers and deaths cannot be equated.
https://t.me/med1917
234 11 Hypothesis Testing
Using the Test:
The null and alternative hypotheses are:
H
0
: μ
1
= μ
2
or μ
1
− μ
2
= 0
H
A
: μ
1
/= μ
2
or μ
1
− μ
2
/= 0
.
Sample size in first and second groups is 49 and 131, respectively, and default
significant level is 0.05.
Test statistic is
z =
(
x
1
− x
2
)
σ
2
1
n
1
+
σ
2
2
n
2
Critical value is z
0.975
= 1.96. If the test statistic is greater than the critical value
of 1.96, then we reject the null hypothesis (
H
0
).
The test statistic according to sample data in two groups is:
z =
(
68.5 − 55.69
)
361
49
+
196
131
=
12.81
2.98
= 4.30
Since 4.30 > 1.96, then we reject the null hypothesis (H
0
).
The mean age of the two groups cannot be considered equal. That is, the age
mean difference between the two groups recovered and the deaths was significant
with zero.
Example 11.6
In a malaria-screening program, 150 children aged 1–4 years were
tested for the presence of the P. falciparum parasite in their blood. The parasite was
present in 70 children with a mean hemoglobin level of 10.6 per 100 ml. In 80
children without parasites, the mean hemoglobin level was 11.5 g per 100 ml. Can
the mean hemoglobin level among children infected with the parasite be equal to
that of children without the parasite? Similar studies show a standard deviation of
hemoglobin levels among children in both groups which is 1.4 g.
Solution: The following information is provided in the example:
σ
1
= σ
2
= 1.4, x
1
= 10.6, x
2
= 11.5, n
1
= 70 and n
2
= 80
Because the standard error of the two groups is equal, the pooled standard error
can be obtained by the following formula:
Se = σ ×
1
n
1
+
1
n
2
= 1.4 ×
1
70
+
1
80
= 0.229
The samples are large enough and the 95% confidence interval of the difference
in hemoglobin level between the two groups can be estimated using the normal
approximation (
z
0.975
= 1.96).
https://t.me/med1917
11.3 Comparing the Mean of Two Populations 235
CI95% =
(
x
1
− x
2
)
± z
0.975
× Se =
(
11.50 − 10.6
)
± 1.96 × 0.229
= 0.90 ± 0.45 =
(
0.45 to 1.35
)
The estimated interval does not include the value zero. As a result, based on this
example, the null hypothesis is rejected. That is, the mean hemoglobin level in the
two groups cannot be equated.
Using the Test:
The null and alternative hypotheses are:
H
0
: μ
1
= μ
2
H
A
: μ
1
/= μ
2
or
H
0
: μ
1
− μ
2
= 0
H
A
: μ
1
− μ
2
/= 0
Sample size in first and second groups is 70 and 80, respectively, and default
significant level is 0.05.
Test statistic is:
z =
(
x
1
− x
2
)
σ ×
1
n
1
+
1
n
2
Critical value is z
0.975
= 1.96. If the test statistic is greater than the critical value
of 1.96, then we reject the null hypothesis (
H
0
).
The test statistic according to sample data in two groups is:
z =
(
11.5 − 10.6
)
1.4 ×
1
70
+
1
80
=
0.90
0.229
= 3.93
Since 3.93 > 1.96, then we reject the null hypothesis (H
0
).
The mean hemoglobin level of the two groups cannot be considered equal. That
is, the mean difference of hemoglobin levels between two groups with zero was
significant.
The Variance of the Two Populations is Unknown
Example 11.7 A study was performed to compare the birth weight of infants whose
mothers smoked with those whose mothers did not smoke. Sample mean and standard
deviation of weight in a sample of 14 smoking mothers were 3.20 and 0.492 and in
a sample of 15 non-smoking mothers were 3.59 and 0.364. Can the birth weight of
the two groups be equal?
Solution: The following information is provided in the example:
n
1
= 14, x
1
= 3.20, S
1
= 0.492, and n
2
= 15, x
2
= 3.59, S
2
= 0.364
The pooled standard deviation and standard error of two groups can be obtained
by the following formula:
https://t.me/med1917
236 11 Hypothesis Testing
S
p
=
(
n
1
− 1
)
× S
2
1
+
(
n
2
− 1
)
× S
2
2
n
1
+ n
2
− 2
=
(
14 − 1
)
× 0.492
2
+
(
15 − 1
)
× 0.364
2
14 + 15 − 2
= 0.430
Se = S
p
×
1
n
1
+
1
n
2
= 0.430 ×
1
14
+
1
15
= 0.160
The variance of the population is unknown and the sample size is small, so instead
of the z distribution, we should use the t distribution with degree of freedom df =
n
1
+ n
2
− 2 = 14 + 15 − 2 = 27. Its critical value from the table is t
(
0.975,27
)
= 2.05.
The 95% confidence interval for mean difference in the newborn weights between
two groups is:
CI95% =
(
x
1
− x
2
)
± t
(
0.975,27
)
× Se =
(
3.59 − 3.2
)
± 2.05 × 0.16 = 0.39
± 0.328 =
(
0.062 to 0.717
)
The estimated interval does not include the zero value. As a result, based on this
example, the null hypothesis is rejected. That is, the mean weight newborns in the
two groups (smoker mothers and non-smoker mothers) are not equal.
Using the Test:
The null and alternative hypotheses are
H
0
: μ
1
= μ
2
H
A
: μ
1
/= μ
2
or
H
0
: μ
1
− μ
2
= 0
H
A
: μ
1
− μ
2
/= 0
:
Sample size in first and second groups is 14 and 15, respectively, and default
significant level is 0.05.
Test statistic is:
t =
(
x
1
− x
2
)
S
p
×
1
n
1
+
1
n
2
Critical value is t
(
0.975, 27
)
= 2.05. If the test statistic is greater than the critical
value of 2.05, then we reject the null hypothesis (
H
0
).
The test statistic according to sample data in two groups is:
t =
(
3.59 − 3.20
)
0.43 ×
1
14
+
1
15
=
0.39
0.16
= 2.44.
https://t.me/med1917
11.3 Comparing the Mean of Two Populations 237
Since 2.44 > 2.05, then we reject the null hypothesis (H
0
).
This interval tells us that the weight difference between the infants of the two
groups is between 62 and 717 g. The 95% confidence interval does not contain zero,
so the mean weight of infants in these two groups is not equal, and women smokers
give birth to infants with less weight.
To run this significant test in SPSS, it should be noted that the weight data of
both groups should be entered in only one column and the information about the
group of smoking and non-smoking mothers should be entered in a separate column.
Then select Analyze\Compare Means\ Independent Samples T Test to open following
window.
In this window, from the list on the left, we first select the name of the variable
whose mean in the two groups should be compared and shift it to the Test Variable (s)
box. This variable must be on a quantitative scale. In the box below, titled Grouping
Variable, enter the name of the variable that separates the data into two categories
(such as gender or smoker and non-smoker). This variable must be on categorical
scale. By entering the name of the grouping variable, the “Define Groups” button is
activated. Pressing it will open another sub-window into which to enter the codes
of the two groups being compared. Then, by pressing the “Continue” button and
confirming “OK”, the program will run.
https://t.me/med1917
238 11 Hypothesis Testing
The output of the SPSS will be as follows:
Group statistics
Smoking N Mean Std. deviation Std. error mean
Infant weight Yes 14 3.2000 0.49236 0.13159
No 15 3.5900 0.36401 0.09399
Levene’s test
for equality of
variances
t-test for equality of means
f Sig t df Sig.
(two-tailed)
Mean
difference
Std. error
difference
95% CI of the
difference
Lower Upper
2.127 0.156 −2.437 27 0.022 −0.39000 0.16002 −0.71833 −0.06167
11.4 Comparison of the Mean of the Two Paired Samples
Paired samples (dependent samples or related sample) are two samples
(measurements) drawn from the same coupled, matched, or paired individuals. In
this type of sample, we have a dataset in which each data point in one sample is
uniquely paired to a data point in another sample. Paired samples may include:
1—pre-test/post-test samples, 2—duplicate measurements on the same samples,
and 3—any situation in which each data point in one sample is uniquely matched to
https://t.me/med1917
Соседние файлы в папке Библиотека им академика М.И. Перельмана