Question

Stet by step in R and attach R file and R codes too - Thanks

Use one of the real-world example data sets from R (not previously used in the R practice assignment) or a dataset you have found, and at least two of the tests and R functions covered in the practice assignment to conduct a hypothesis test then report your findings and give proper conclusion(s).

Use the following supporting materials for R syntax, data sets and tools, along with other resources found in this module or that you find on your own.

• Using T-Tests in R from the Department of Statistics at UC Berkley

• Test of equal or given proportions from R Documentation

• F-Test: Compare Two Variances in R from STHDA (Statistical tools for high-throughput data analysis)

Please answer step by step with R files attached and R codes

Answer 1

To Use one of the real-world example data sets from R

Using T-Tests in R from the Department of Statistics at UC Berkley

Let us import data from library MASS

We will use "Cats ' data ,

Total Row = 144 , Colums = 3

ibrary(MASS)
> D=(cats)         # to import data of "Cats"

> head(D,10)              # to show only first 10 values of data set
   Sex Bwt Hwt
1    F 2.0 7.0
2    F 2.0 7.4
3    F 2.0 9.5
4    F 2.1 7.2
5    F 2.1 7.3
6    F 2.1 7.6
7    F 2.1 8.1
8    F 2.1 8.2
9    F 2.1 8.3
10   F 2.1 8.5

> D[44:54,]           # between 44 and 54 observation
   Sex Bwt Hwt
44   F 2.9 10.1
45   F 2.9 10.1
46   F 3.0 10.6
47   F 3.0 13.0
48   M 2.0 6.5
49   M 2.0 6.5
50   M 2.1 10.1
51   M 2.2 7.2
52   M 2.2 7.6
53   M 2.2 7.9
54   M 2.2 8.5

In tese data set we have Gender of cat

Column 1 -    "Feamle" = F ,   Male =M

Column 2 -    "Weight of cats " = Bwt ,             

Column 3 -    "Height of cats " = Hwt ,  

Now we wish to cheak weater weight of Female cats is less than Meal cats or not

I.e let u1 be Weight of Female cats , and u2 be weight of Males cats

Hypothesis to test are

H0 : u1 = u2     ( weight of female and Male cat is same )

H1 : u1 < u2     ( weight of female is less than weight of Male cat )

R functions - t.test()

First we will sort data for Male and Female cats

> weight_Female=D[c(Female),2]          # only observation which represents Femal cats
> weight_Male=D[c(Male),2]                 # only observation which represents Femal cats

> weight_Female                # Weight of female cat
[1] 2.0 2.0 2.0 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.2 2.2 2.2 2.2 2.2 2.2 2.3
[20] 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.4 2.4 2.4 2.4 2.5 2.5 2.6 2.6
[39] 2.6 2.7 2.7 2.7 2.9 2.9 2.9 3.0 3.0

> weight_Male                       # Weight of male cat
[1] 2.0 2.0 2.1 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.3 2.4 2.4 2.4 2.4 2.4 2.5 2.5
[20] 2.5 2.5 2.5 2.5 2.5 2.5 2.6 2.6 2.6 2.6 2.6 2.6 2.7 2.7 2.7 2.7 2.7 2.7 2.7
[39] 2.7 2.7 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.9 2.9 2.9 2.9 2.9 3.0 3.0 3.0 3.0 3.0
[58] 3.0 3.0 3.0 3.0 3.1 3.1 3.1 3.1 3.1 3.1 3.2 3.2 3.2 3.2 3.2 3.2 3.3 3.3 3.3
[77] 3.3 3.3 3.4 3.4 3.4 3.4 3.4 3.5 3.5 3.5 3.5 3.5 3.6 3.6 3.6 3.6 3.7 3.8 3.8
[96] 3.9 3.9

# noww we will test our hypothesis

# to test alternative hypothsis " < " , use comand t.test (x,y, )

> t.test(weight_Female,weight_Male,)

        Welch Two Sample t-test

data: weight_Female and weight_Male
t = -8.7095, df = 136.84, p-value = 4.416e-15
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
       -Inf -0.4376663
sample estimates:
mean of x mean of y
2.359574 2.900000

Since P-value = 4.416e-15 is very small , i.e P-value <<<0.05 ,so we reject null hypothesis at 5% of level of significance .

Hence Mean weight of Female cats is less than Male cats

• Test of equal or given proportions from R Documentation

Suppose we wish to test that in town the proportion of males cats in population 65%

Than is we wish to cheak that in every 100 cats there are 65 Male cats .

But someone says that Proportion of males cats is more than 65%

So ,

Hypothesis to test are

H0 : po = 0.65     ( proportion of male cats in town is 65% )

H1 : po > 0.65     ( proportion of male cats in town is more than 65% )

Now we have sample of size 144 , so we will test weater proportion of cats in sample is 65% or more than 65%

we will use here prop.test (x,n,po=)  

                                                                   

> n=length(D[,1])
> n                 # total samples
[1] 144            

> x=length(D[c(Male),2])
> x                    # number of Male Cats in sample
[1] 97      

   

> prop.test(x,n,0.65,)

        1-sample proportions test with continuity correction

data: x out of n, null probability 0.65
X-squared = 0.25672, df = 1, p-value = 0.3062
alternative hypothesis: true p is greater than 0.65
95 percent confidence interval:
0.6030755 1.0000000
sample estimates:
        p
0.6736111

       

Here we have , p-value = 0.3062 > 0.05

So at 5% of level of significane we do not reject null hypothesis

Hence Proportion of Male cats in town may be equal to 65%

•         F-Test: Compare Two Variances in R from STHDA (Statistical tools for high-throughput data analysis)                 

{ Note - In STHDA am not getting any usefull dataset since in testing F-Test: Compare Two Variances we requires two variable which have atrribute like gender , Height ect . So if any dataset like avilable you can use}

Since Here we are suppose to perform F-Test: Compare Two Variances we can use previous data set of cats only ,and can cheack weathere Feamle and Male Cats ahve same variability in their weight or not :

To test

H0 : 1 = 2    ( No varibility between weigths of diferent gender of cats )

H1 : 1 2   ( varibility between weigths of male and female cats differs)

    

> weight_Female        # already obtain in first part i.e weight of female cats
[1] 2.0 2.0 2.0 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.2 2.2 2.2 2.2 2.2 2.2 2.3
[20] 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.3 2.4 2.4 2.4 2.4 2.5 2.5 2.6 2.6
[39] 2.6 2.7 2.7 2.7 2.9 2.9 2.9 3.0 3.0

> weight_Male          # already obtain in first part i.e weight of male cats
[1] 2.0 2.0 2.1 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.3 2.4 2.4 2.4 2.4 2.4 2.5 2.5
[20] 2.5 2.5 2.5 2.5 2.5 2.5 2.6 2.6 2.6 2.6 2.6 2.6 2.7 2.7 2.7 2.7 2.7 2.7 2.7
[39] 2.7 2.7 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.9 2.9 2.9 2.9 2.9 3.0 3.0 3.0 3.0 3.0
[58] 3.0 3.0 3.0 3.0 3.1 3.1 3.1 3.1 3.1 3.1 3.2 3.2 3.2 3.2 3.2 3.2 3.3 3.3 3.3
[77] 3.3 3.3 3.4 3.4 3.4 3.4 3.4 3.5 3.5 3.5 3.5 3.5 3.6 3.6 3.6 3.6 3.7 3.8 3.8
[96] 3.9 3.9

# to Compare Two Variances we use var.test(x,y,conf.level="")

> var.test(weight_Female,weight_Male,conf.level=0.95)

        F test to compare two variances

data: weight_Female and weight_Male
F = 0.3435, num df = 46, denom df = 96, p-value = 0.0001157
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.2126277 0.5803475
sample estimates:
ratio of variances
         0.3435015

Conclusion - Since P-value = 0.0001157 < 0.05

We reject null hypothesis at 5% of level of significance.

Hence variability in weights of male and female cats may differ .