Population Sampling And Hypothesis Testing In R: An Essay.

a) Create a normally distributed ‘population’ of 100000 cases with a mean (μ) of 50 and a standard deviation (σ) of 8, and call it ‘pop’. This is your pretend population.
b) Take a sample of size N = 30 (without replacement) from the population, and call it ‘samp’.
c) Pretend to give your sample an intervention. More specifically, add a normally distributed variable with µ = 5 and σ = 5 (call it ‘inter’) to your sample (‘samp’). Call this new variable ‘inter_samp’.
d) Use null hypothesis testing (all steps, α = .05) to determine if your intervention was effective at increasing scores. Be sure to interpret (even if it is purely subjective) the magnitude of the effect size and the confidence interval width, and include a general summary of the results.

Dr. White asked 400 students to choose between five characteristics of an instructor that they find most important. The characteristics that they had to choose among were (dataset codes in parentheses): 1) enthusiasm (enthusiasm); 2) humour (humour); 3) level of difficulty (difficulty); 4) clarity (clarity); or 5) caring attitude (caring). The dataset (available on Moodle) is entitled ‘ass2_q2.csv’. Use α = .01 for the questions below.

a) Use null hypothesis testing (all steps) to determine if enthusiasm, humour, level of difficulty, clarity and caring attitude are not equally important in terms of being the most important characteristic of a good instructor. Be sure to interpret the effect size (again, even if it is purely subjective) and include a general summary regarding the results.
b) What Dr. Black was really interested in was whether enthusiasm or level of difficulty differed in terms of the frequency with which they are chosen as a characteristic that students value in an instructor. Using only students who chose one of these two options (enthusiasm, difficulty), use R to generate a table of frequencies for each of the two options. Then, BY HAND, use null hypothesis testing (all steps) to determine if students differed in their preference for enthusiasm or level of difficulty as their most important characteristic.Calculate the p-value and/or critical value using R. Be sure to (subjectively) interpret the effect size and include ageneral summary statement regarding your results.

Generating a normally distributed population in R

R Code

#creating random samples of n=100000 with a mean of 50 and standard deviation of 8

pop<-rnorm(100000,mean=50,sd=8)

#sampling data from “pop” with N=30 and without replacement

samp<-sample(pop,15)

#generating another sample data known as “inter” with mean=5 and sd=5

inter<-rnorm(1,mean=7,sd=5)

#combining both “inter” and “samp” data

inter_samp<-c(samp,inter)

#Hypothesis statement and hypothesis test

#Null hypothesis “H0:mu=mu0; alternative hypothesis “H1:mu>mu0”

t.test.right<-function(data,mu0,alpha)

{

#declaring and defining the t-statistic formula

t.stat<-(mean(data)-mu0)/(sqrt(var(data)/length(data)))

#degrees of freedom calculation

dof<-length(data)-1

#calculating t critical value

#Es alpha 0.05 -> 1.64(df = Inf)

t.critical<-qt(1-alpha,df=dof)

# Calculation of p-value

p.value<-1-pt(t.stat,df=dof)

#Decision making using test results

if(t.stat>t.critical)

{
print("Reject H0")
}
else
{
print("Accept H0")

}
print('T statistic')

print(t.stat)

print('T critical value')

print(t.critical)

print('P value')

print(p.value)
return(t.stat)

}

t.test.right(inter_samp,mu0=50,alpha= 0.05)

#summary

summary(inter_samp)

#calculation of 95 percent confidence interval

error<-qt(0.995,df=length(inter_samp)-1)*sd(inter_samp)/sqrt(length(inter_samp))

error

#Lower bound confidence interval calculation

Lower<-mean(inter_samp)-error

Lower

#Upper bound confidence interval calculation

Upper<-mean(inter_samp)+error

Upper

#End of program

Program Output

> #creating random samples of n=100000 with a mean of 50 and standard deviation of 8

> pop<-rnorm(100000,mean=50,sd=8)

> #sampling data from “pop” with N=30 and without replacement

> samp<-sample(pop,15)

> #generating another sample data known as “inter” with mean=5 and sd=5

> inter<-rnorm(1,mean=7,sd=5)

> #combining both “inter” and “samp” data

> inter_samp<-c(samp,inter)

> #Hypothesis statement and hypothesis test

> #Null hypothesis “H0:mu=mu0; alternative hypothesis “H1:mu>mu0”

> t.test.right<-function(data,mu0,alpha)

+ {

+     #declaring and defining the t-statistic formula

+     t.stat<-(mean(data)-mu0)/(sqrt(var(data)/length(data)))

+     #degrees of freedom calculation

+     dof<-length(data)-1

+     #calculating t critical value

+     #Es alpha 0.05 -> 1.64(df = Inf)

+     t.critical<-qt(1-alpha,df=dof)

+     # Calculation of p-value

+     p.value<-1-pt(t.stat,df=dof)

+     #Decision making using test results

+     if(t.stat>t.critical)

+     {

+         print("Reject H0")

+     }

+     else

+     {

+         print("Accept H0")

+     }

+     print('T statistic')

+     print(t.stat)

+     print('T critical value')

+     print(t.critical)

+     print('P value')

+     print(p.value)

+     return(t.stat)

+ }

> t.test.right(inter_samp,mu0=50,alpha= 0.05)

[1] "Accept H0"

[1] "T statistic"

[1] -3.043226

[1] "T critical value"

[1] 1.75305

[1] "P value"

[1] 0.9958919

[1] -3.043226

> #summary

> summary(inter_samp)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.

  11.32   39.54   44.94   42.33   48.95   52.16

> #calculation of 95 percent confidence interval

> error<-qt(0.995,df=length(inter_samp)-1)*sd(inter_samp)/sqrt(length(inter_samp))

> error

[1] 7.424564

> #Lower bound confidence interval calculation

> Lower<-mean(inter_samp)-error

> Lower

[1] 34.9077

> #Upper bound confidence interval calculation

> Upper<-mean(inter_samp)+error

> Upper

[1] 49.75682

> #End of program

Interpretation

The results shows that 95% confidence interval is (34.91, 49.76) with a mean of 42.33 and standard error of 7.42

Reject null hypothesis is the observed t statistic is greater than the value of t critical (Gentleman, 2009; Braun & Murdoch, 2012; Baker & Trietsch).The test statistic = -3.04 which is greater than t critical (1.79) therefore, we fail to reject null hypothesis. We do not have sufficient evidence thus we accept null hypothesis and conclude that intervention was effective in increasing the scores.

The five number summaries are 11.32, 39.54, 44.94, 42.33, 48.95, and 52.16.

Null hypothesis: Enthusiasm, humour, difficulty, and clarity characteristics are equally important for a good instructor

Alternative hypothesis: At least of the characteristics is not important for a good instructor

We are dealing with categorical variables therefore chi-square test is the most appropriate statistical test for this problem. Next we create a frequency table as follows:

	Observed , O	Expected ,E	O-E	=
Enthusiasm	121	80	41	= 21.0125
Humour	45	80	-35	= 15.3125
Difficulty	92	80	12	= 1.8
Clarity	87	80	7	= 0.6125
Caring	55	80	-25	= 7.8125
	= 400			= 46.55

n=5 variables

Mean,  =  =  = 80

Observed Chi-test value,  = 46.55

Degrees of freedom = n-1 = 5-1 = 4

Chi-test critical value () = 9.4877 (from chi-square table)

Interpretation

The observed chi-square statistic is less than the chi-square critical value thus the test results are statistically significant. We therefore reject null hypothesis and conclude that at least of the four characteristics (enthusiasm, humour, difficulty, and clarity) is not important for a good instructor.

R code

#Attaching dataset

R_order <- read_excel("C:/Users/User/Downloads/Desktop/R order.xlsx")

View(R_order)

attach(R_order)

#Creating frequency table

table<-table(enthusiasm,difficulty)

table

R output

>attach(R_order)

> table<-table(enthusiasm,difficulty)

> table

          difficulty

enthusiasm  1

          121     92

Null hypothesis: Students differed in their preference for enthusiasm or level of difficulty

Alternative hypothesis: Students did not differ in their preference for enthusiasm or level of difficulty

Chi-square calculation by hand

	Observed , O	Expected ,E	O-E	=
Enthusiasm	121	106.5	121-106.5 =14.5	= 1.974
Difficulty	92	106.5	92-106.5= -14.5	= 1.974
	= 213			= 3.948

n=2 variables

Mean,  =  =  = 106.5

Observed Chi-test value,  = 3.948

Degrees of freedom = n-1 = 2-1 = 1

Chi-test critical value () = 3.84 (from chi-square table)

# R code to calculate p-value at alpha = 0.01

pchisq(3.948, df=1, lower.tail=FALSE)

[1] 0.04692711

The observed chi-square statistic is greater than the chi-square critical value thus the test results are statistically significant. We therefore reject null hypothesis and conclude that Students did not differ in their preference for enthusiasm or level of difficulty. Similarly, using P-value techniques, we reject null hypothesis since the observed p-value is less than 0.01 thus we reject null hypothesis.

References

Baker, K., & Trietsch, D.(2009) Principles of sequencing and scheduling.

Braun, J., & Murdoch, D.(2012). A first course in statistical programming with R.

Gentleman, R. (2009). R programming for bioinformatics. Boca Raton: CRC Press.