Question

You want to know if, on average, households have more cats or more dogs. You take an SRS of 8 households and find the data below:

# of cats	# of dogs
2	1
0	1
3	3
2	4
0	0
4	2
0	2
2	1

A) Determine the population(s) and parameter(s) being discussed.

B) Determine which tool will help us find what we need (one sample z test, one sample t test, two sample t test, one sample z interval, one sample t interval, two sample t interval).

C) Check if the conditions for this tool hold.

D) Whether or not the conditions hold, use the tool you choose in part B. Use C=95% for all confidence intervals and alpha=5% for all significance tests.

* Be sure that all methods end with a sentence describing the results *

Answer 1

a)

We will be using sample estimates as we don't have population parameter i.e.

Sample mean and Sample standard deviation

b)

We will use two sample t test

c)

1)Normal Distribution

2)Using SRS method

3)Large sample size

4)homogeneity of variance.

d)

Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   1.63                  
standard deviation of sample 1,   s1 =    1.51                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   1.75                  
standard deviation of sample 2,   s2 =    1.28                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    1.6250   -   1.8   =   -0.13  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.3983                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.6992                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.1250   -   0   ) /    0.70   =   -0.179
                          
Degree of freedom, DF=   n1+n2-2 =    14                  
t-critical value , t* =        2.1448   (excel formula =t.inv(α/2,df)              

Decision:   | t-stat | < | critical value |, so, Do not Reject Ho                      

conclusion : There is not enough evidence to say that on average, households have more cats or more dogs

...........

95% CI-----------------

Degree of freedom, DF=   n1+n2-2 =    14              
t-critical value =    t α/2 =    2.1448   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.3983              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.6992              
margin of error, E = t*SE =    2.1448   *   0.6992   =   1.4996  
                      
difference of means =    x̅1-x̅2 =    1.6250   -   1.750   =   -0.1250
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.1250   -   1.4996   =   -1.6246
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.1250   +   1.4996   =   1.3746

CI ( -1.62 , 1.37)

CI contains 0 , so do not reject Ho

..............

thanks

revert back for doubt

rate please