R language-Select the number of samples

Efficacy analysis: Can help in the case of a given degree of confidence in the detection of a given effect value of the required sample amount, but also at a given level of confidence in the case of a certain sample amount can be detected in the probability of a given effect value

1.t Inspection

Case study: Experiment using cell phone and driver response time

1 Library (PWR)2 #n indicates sample size3 #d indicates the difference between standardized mean values4 #Sig.level represents a significant level of5 #power for efficacy level6 #Type refers to the test types7 #alternative refers to a two-sided test or a single-sided test.8Pwr.t.test (D=.8,sig.level =. 05,power =. 9,type ='Two.sample', alternative ='two.sided')

Conclusion: A total of 34 samples (68) per group is required to ensure that the 0.8 effect value is detected by 90% and a maximum of 5% is error.

2. Variance analysis

Case: Analysis of 5 groups of data, to achieve 0.8 of the efficacy, the effect is 0.25, select a significant level of 0.5. Calculate the size of the population sample

# k indicates the number of groups # F represents the effect value pwr.anova.test (k=5,f=.25,sig.level =. 05,power =. 8)

Conclusion: It is necessary to 39*5,195 subjects to participate in the experiment in order to obtain the above results

3. Relevance

Case: The relationship between depression and loneliness, 0 hypothesis and study hypothesis for

h0:p<=0.25 and h1:p>0.25

Set a significant level of 0.05, slap rejection 0 hypothesis, want to have 90% of the information rejected H0, how many testers

1 # R indicates the effect value 2 ' Greater ')

Conclusion: 134 subjects were required to participate in the experiment

4. Linear model

Case: The boss's leadership style affects employee satisfaction, salaries and gratuities explain 30% employee satisfaction Variance, leadership style can explain 35% variance,

To reach 90% confidence, the significant level is 0.05, and how many subjects are required to reach the variance contribution rate

F2 = (0.35-0.3)/(1-0.35) =0.0769

1 # u represents molecular degrees of freedom 2 # v represents the denominator degrees of freedom 3 # F2 represents the effect value 4 pwr.f2.test (u=3,f2=0.0769,sig.level =. 05,power =. 90)

Conclusion: v= Overall sample-predictor variable-1, so n=v+7+1=187+7+1=193

5. Proportional inspection

Case: A drug with a 60% cure rate, the new drug has a 65% cure rate, now how many subjects can appreciate the difference between the two drugs

1 ' Greater ')

Conclusion: Using unilateral test in this case, it was concluded that 1605 subjects were required to obtain the difference between the two drugs.

6. Chi-Square inspection

The chi-square test is used to evaluate the relationship between the two variables, 0 is assumed to be independent of the variables, and the reject 0 assumes that the variable is not independent

Case study: The relationship between promotion and race: 70% of the sample was white, 10% Black, 20% Hispanic, and 50% of whites were more likely to get promoted than 20% blacks and 60% Hispanics.

1 prob <-matrix (C (. 42,.28,.03,.07,.10,.10), Byrow = T,nrow = 3)2#  Calculates the effect value of the alternative hypothesis in a two-factor list 3es.w2 (prob)4#  W is the effect value,5#  DF is the degree of Freedom 6 pwr.chisq.test (w=0.1853198,df=2,sig.level =. 05,power =. 90)

Conclusion: This experiment requires 369 testers to prove the association between promotion and race.

7. Select the appropriate effect value in the new case

7.1 Single Factor

1Es <-seq (. 1,.5,.01)2NES <-Length (es)3Samsize <-NULL4  for(Iinch1: NES) {5Result <-pwr.anova.test (k=5,f=es[i],sig.level =. 05,power =. 90)6Samsize[i] <-Ceiling (result$n)7 }8Plot (samsize,es,type='L', lwd='2', col='Red',9Ylab ='Effect Size',TenXlab ='Sample Szie', OneMain ='One- Anova with power=.90 and alpha=.05')

Conclusion: When the number of Ben Ben is higher than 200, the effect is not obvious when the sample is increased

7.2 Drawing Efficiency Analysis diagram

1 #1. Generate a series of correlation coefficients and efficacy values2R <-seq (. 1,.5,.01)3NR <-Length (R)4 5P <-seq (. 4,.9,.1)6NP <-Length (p)7 8 #2. Get Sample size9Samsize <-Array (numeric (NR*NP), Dim =C (NR,NP))Ten  One  for(Iinch1: NP) { A    for(jinch1: NR) { -Result <-pwr.r.test (n=null,r=r[j],sig.level =. 05,power = P[i],alternative ='two.sided') -Samsize[j,i] <-Ceiling (result$n) the   } - } -  - #3. Create a graphic +Xrange <-Range (r) -Yrange <-Round (range (samsize)) +Colors <-Rainbow (Length (p)) APlot (xrange,yrange,type='N', atXlab ='corrlation coefficient', -Ylab ='Sample Size') - #4. Add Efficiency curve -  for(Iinch1: NP) { -Lines (r,samsize[,i],type='L', lwd=2,col=Colors[i]) - } in #5. Grid lines -Abline (V=0,h=seq (0,yrange[2],50), lty=2,col='grey89') toAbline (H=0,v=seq (xrange[1],xrange[2],.02), lty=2,col='grey89') + #6. Title and Notes -Title'Sample Size estimation for corrlation\nsig=0.05') theLegend'TopRight', title ='Power', As.character (P), fill=colors)

Conclusion: At 40% confidence level, it is necessary to detect that 0.2 of the correlation requires about 75 samples, and at 90% confidence level, it takes about 260 samples to detect the same correlation.

R language-Select the number of samples