Why? What?

This is based on a collection of videos provided by Openintro.org. They are available on youtube at

https://www.youtube.com/playlist?list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919.

Each video demonstrates the use of a TI calculator in performing a typical task for an introductory statistics course.

After each video, I have placed a link to one of my videos demonstrating how to do the task in R. The R work is always done on a freely avalaible cloud-based resource, which could be used by anyone with a chromebook. No installation on a student’s computer is required.

I have used https://rdrr.io/ but there are similar resources avalable.

For convenience, I have also provided a final version of the R code I used. Feel free to copy, modify, and use this code.

Video 1

Entering Data and 1 Variable Statistics

Here are the TI instructions.

https://www.youtube.com/playlist?list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919

Here is the corresponding R code.

# Enter the data using the c() function

L1 = c(2,4,6,8,40)

sum(L1)
## [1] 60
sum(L1^2)
## [1] 1720
mean(L1)
## [1] 12
sd(L1)
## [1] 15.81139
n = length(L1)

sd(L1) * sqrt((n-1)/n)
## [1] 14.14214
sqrt(mean((L1 - mean(L1))^2))
## [1] 14.14214

Video 2

Boxplot and five number summary

Here are the TI instructions.

https://www.youtube.com/watch?v=VvCw5MRo1P4&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=2

Here is the corresponding R code.

v = c(2,3,4,5,8,12)

boxplot(v,horizontal = TRUE)

summary(v)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   2.000   3.250   4.500   5.667   7.250  12.000

Video 3

Computing the Binomial Coefficient and Factorial

Here are the TI instructions.

https://www.youtube.com/watch?v=MgSitJ7Aqxg&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=3

Here is the corresponding R code.

n = 8
r = 2

factorial(n)
## [1] 40320
factorial(r)
## [1] 2
factorial(n)/(factorial(n-r)* factorial(r))
## [1] 28
choose(n,r)
## [1] 28

Video 4

Binomial Formula

Here are the TI instructions

https://www.youtube.com/watch?v=F6JBimUE43U&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=4

Here is the corresponding R code.

p = 6/10
n = 5
k = 3

dbinom(k,n,p)
## [1] 0.3456

Video 5

Binomial Cumulative Distribution

Here are the TI instructions.

https://www.youtube.com/watch?v=bGxo7wqKyFY&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=5

Here is the corresponding R code.

pbinom(2,5,.6)
## [1] 0.31744
dbinom(0,5,.6) + dbinom(1,5,.6) + dbinom(2,5,.6)
## [1] 0.31744

Video 6

Areas under the normal curve

Here are the TI instructions.

https://www.youtube.com/watch?v=o4bQYtsq6Ig&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=6

Here is the corresponding R code.

pnorm(1)
## [1] 0.8413447
pnorm(-1)
## [1] 0.1586553
pnorm(1) - pnorm(-1)
## [1] 0.6826895
1 - pnorm(1)
## [1] 0.1586553
pnorm(110,mean = 100, sd = 10)
## [1] 0.8413447
pnorm(90,mean = 100, sd = 10)
## [1] 0.1586553

Video 7

Here are the TI instructions

https://www.youtube.com/watch?v=7mkfGJO8ehM&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=7

Here is the corresponding R code.

qnorm(.84)
## [1] 0.9944579
qnorm(pnorm(1))
## [1] 1

Video 8

1-Proportion Z-Test

Here are the TI Instructions.

https://www.youtube.com/watch?v=-OjbnQGIA3Y&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=8

Here is the corresponding R code.

prop.test(260, 500, p =.5,
          alternative = "greater",
          conf.level = 0.95,
          correct = FALSE)
## 
##  1-sample proportions test without continuity correction
## 
## data:  260 out of 500, null probability 0.5
## X-squared = 0.8, df = 1, p-value = 0.1855
## alternative hypothesis: true p is greater than 0.5
## 95 percent confidence interval:
##  0.4832406 1.0000000
## sample estimates:
##    p 
## 0.52
p0 = .5
n = 500

se = sqrt(p0*(1-p0)/n)
z = .02/se
z
## [1] 0.8944272
1 - pnorm(z)
## [1] 0.1855467

Video 9

Here are the TI instructions.

https://www.youtube.com/watch?v=ZQu-EFcJ0R4&list=PLkIselvEzpM7N8zVRRUl7V8aTdoTsJ919&index=9

Here is the corresponding R code.

# Follow the formulas

x1 = 899
n1 = 1000
x2 = 958
n2 = 1000

phat = (x1 + x2)/(n1 + n2)

p1 = x1/n1
p2 = x2/n2

dp = p1 - p2
null = 0

se = sqrt(phat*(1-phat)) * sqrt(1/n1 + 1/n2)

z = (dp - null)/se
z
## [1] -5.120272
pnorm(z)
## [1] 1.525476e-07
# Use the built-in convenience function.

phat = (899 + 958)/(1000 + 1000)
prop.test(c(899,958),
          c(1000,1000),
          c(phat,phat),
          alternative = "less",
          correct = FALSE)
## 
##  2-sample test for given proportions without continuity correction
## 
## data:  c(899, 958) out of c(1000, 1000), null probabilities c(phat, phat)
## X-squared = 26.217, df = 2, p-value = 2.028e-06
## alternative hypothesis: two.sided
## null values:
## prop 1 prop 2 
## 0.9285 0.9285 
## sample estimates:
## prop 1 prop 2 
##  0.899  0.958