Question

UTRGV spring 2019 Statistics Students are interested in understanding the education level (1 means having a higher education and 0 is the opposite) in RGV, they collected data from the population of RGV:

1 0 0 0 1 1 1 0 1 1 1 0 0 1 1 0
1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1

a. Use R to provide a frequency and relative frequency table.


b. Use prop.test to for the following hypothesis testing.

?₀: ? = 0.6

?₁: ? > 0.6




c. Determine the p-value.


d. Can you reject the null hypothesis, justify your answer?


e. Provide your R coding

Answer 1

frequency = total number of observations of type.

relative frequency = frequency / sample size

a)

> x<-c(1, 0 ,0 ,0, 1, 1 ,1 ,0, 1, 1, 1 ,0 ,0, 1 ,1, 0,1 ,1 ,1 ,0 ,0 ,1 ,1 ,0 ,1 ,0 ,1 ,0 ,0 ,1 ,1 ,1)
> n<-length(x)
> freq<-table(x)
> freq
x
0 1
13 19
>
> relative.frequencies<-freq/n
> round(relative.frequencies,2)
x
0 1
0.41 0.59

b)

?₀: ? = 0.6

?₁: ? > 0.6

> p<-0.59 # relative freq corresponding to 1
> x<-n*p # number of observation corresponding to 1
> prop.test(x,n,p=0.60,alternative = "greater")

   1-sample proportions test with continuity correction

data: x out of n, null probability 0.6
X-squared = 0, df = 1, p-value = 0.5
alternative hypothesis: true p is greater than 0.6
95 percent confidence interval:
0.4358169 1.0000000
sample estimates:
p
0.59

C. p-value = 0.5

(from above output)

D.Reject Ho.

If P- value < alpha=0.05

Here P-value = 0.5 > 0.05

So we fail to reject to Ho.

we may conclude that ? = 0.6.

E.

>x<-c(1, 0 ,0 ,0, 1, 1 ,1 ,0, 1, 1, 1 ,0 ,0, 1 ,1, 0,1 ,1 ,1 ,0 ,0 ,1 ,1 ,0 ,1 ,0 ,1 ,0 ,0 ,1 ,1 ,1)
>n<-length(x)
>freq<-table(x)
>freq

>relative.frequencies<-freq/n
>round(relative.frequencies,2)

>p<-0.59 # relative freq corresponding to 1
>x<-n*p # number of observation corresponding to 1
>prop.test(x,n,p=0.60,alternative = "greater")