เตรียมนำเสนอโครงร่างวิจัย: Part Statistic component

From previous study, in the control group reported (mean 9.04, SD 5.22) when received music therapy treatment group, expecting treatment effect had reduced, anxiety score estimate 50% =4 Thus in treatment group estimate mean =5, in treatment group SD 4 P1= 9, P2= 5, S₁ = 5, S₂ = 4

We determine sample size for estimating hypothesis tests for two populations mean (two-sided test) with specified absolute precision as

below

n/ group = 2 [ (Z _α+ Z _β ) σ] ² / D²

σ = √ σ ₁²+ σ ₂² / √ 2 = 4.53

n/ group = 2 [ (1.96+ 0.842 ) 4.53] ² / (9-5)²

Then we can select 20 samples for each group in this study

Sampling technique and Randomized allocation

Cancer patient who had pain will be approached to determine interest in participation. Inform consent will be obtained from patients who agreed to participate in the study. This study use simple random sampling (SRS) technique for selection samples which inclusion criteria. The method of block randomization, with block size of four will be used. The allocation sequence will be generated by computer.

Based on the obtained schedule, the number of sealed envelopes, which are equal to the required blocks, will be prepared and labeled. Each envelope contains the corresponding sealed allocation cards that will allocate treatments to patients. The people, who prepare the randomized scheme, will not involve in the trial. The allocation schedule will be concealed until the end of trial.

The researcher need to know what type of approach to be provided to a certain patient, however, the sequence list will keep confidentially and not exposed to outcome assessors.

Outcome variable

Primary outcome variable:

Pain

Anxiety

Secondary outcome variable:

Satisfaction for using music therapy

Data analysis

Chi-square analysis will be used to determine between experimental and control groups according demographic data.

Dependent t-test between groups for mean of pain and anxiety: if normal distribution. I will be used ANCOVA for help reduce residual variation; a source of variation that affects the response variable, but was not controlled for during study design especially; baseline pain and anxiety. If there are not normal distribution data, will be used non parametric method by Mann-Whitney U test.

คนเดินดิน

เขียนเมื่อ 2 พฤศจิกายน 2008 13:53 น. ()

lสู้ต่อไปครับ

KitTyjump

เขียนเมื่อ 2 พฤศจิกายน 2008 13:54 น. ()

มาเป็นกำลังใจให้พี่แก้วค่ะ สู้ๆๆ

แก้ว..อุบล จ๋วงพานิช

เขียนเมื่อ 2 พฤศจิกายน 2008 14:02 น. ()

สวัสดีค่ะคนเดินดิน

ขอบคุณค่ะ ต้องสู้ๆ

สวัสดีค่ะน้อง kittyjump~เลขา~natadee

สู้ๆๆๆๆๆค่ะ ขอบคุณนะคะ

Kesanee Sottas

เขียนเมื่อ 3 พฤศจิกายน 2008 07:40 น. ()

Hi PiiKaw,

I am Mock presentation at home. Wiil be there. See you soon. I will present the 3rd one in this morning. Trying appropriate manage for 15 minutes for presentation...rrrhhh

LOL...KAde

แก้ว..อุบล จ๋วงพานิช

เขียนเมื่อ 3 พฤศจิกายน 2008 18:53 น. ()

วันนี้สรุปประเด็นที่ต้องแก้งานยวิจัของตัวเอง

Sample size calculation
Random allocation: Block of four
Internal valid
Test reliability
Compare mean difference between two group(Pain& Anxiety)

แก้ว..อุบล จ๋วงพานิช

เขียนเมื่อ 4 พฤศจิกายน 2008 06:14 น. ()

คุณเกศนี บุณยวัฒนางกุล

You had present very good, I think you will sucsess your study soon.

แก้ว..อุบล จ๋วงพานิช

เขียนเมื่อ 4 พฤศจิกายน 2008 06:18 น. ()

Sample size calculation

Sample size calculation based on the main objective of the study to determine

the effects of preferred music therapy on anxiety and pain in cancer patients who experience pain. The subjects in this trial will be divided into 2 independent groups and calculated by using the formula.

The information from previous research

From previous study, in the control group reported mean 9.04, SD 5.22 when received music therapy treatment group, expecting treatment effect had reduced, anxiety score estimate 50% =4 Thus in treatment group estimate mean =5, in treatment group SD 4

We determine sample size for estimating hypothesis tests for two populations mean (two-sided test) with specified absolute precision as

below

n/ group = 2 [ (Z _α+ Z _β ) σ] ² / (m₁-m ₂) ²

σ = √ σ ₁²+ σ ₂² / √ 2 = 4.53

m₁ = Mean anxiety score in control group = 9

m₂ = Anticipated mean of anxiety score in treatment group = 5

Zα = 1.96

Zβ = 0.842

σ ₁ = Estimate from SD in control group =5

σ ₂= Estimate from SD in treatment group =4

p value 0.05 (two-tails) and 80% statistical power.

σ = √ 5²+ 4 ² / √ 2 = 4.53

n/ group = 2 [ (1.96+ 0.842 ) 4.53] ² / (9-5)²

n/ group= 20.14

Then we can select 20 samples for each group in this study

If we expect the mean difference of control and treatment group at 80% statistic power and p value 0.05. Sample sizes can be calculated to detect the difference mean between two groups for perform estimation in the sample size determination in health studies, as Table

m1 control	m2experiment	Power	a	n
9	4	80%	0.05	16
9	4.5	80%	0.05	20
9	5	80%	0.05	25
9	5.5	80%	0.05	33
9	6	80%	0.05	44

@..สายธาร..@

เขียนเมื่อ 4 พฤศจิกายน 2008 06:19 น. ()

ผ่านพ้นไปแล้ว

พี่แก้วคงโล่งแล้วนะคะ

มีความสุข

สุขภาพแข็งแรงค่ะ

แก้ว..อุบล จ๋วงพานิช

เขียนเมื่อ 4 พฤศจิกายน 2008 06:20 น. ()

Sample size of 25 will be chosen due to its feasibility, such sample size can detect mean anxiety score in control group = 9 Anticipated mean of anxiety score in treatment group = 5, at 95%significant level, with the power of 80%.

Investigator attempt to prevent for dropt out, advice about an importance of following the treatment