Question

Type or paste question hereAn experiment was conducted to examine if people living in a big city had different stress levels (0-100) than those living in rural areas. Five people living in Cleveland and six people living in Vermilion were asked to report their stress levels each day for a week. The data below reports their average daily stress across that week. Using the .01 significance level and the data below carry out the 5 steps of null hypothesis testing. Show all your work

city=50,60,40,70,40

rural=60,70,55,45,70,60

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.01                  
                          
Sample #1   ----> City
mean of sample 1,    x̅1=   52.00                  
standard deviation of sample 1,   s1 =    13.04                  
size of sample 1,    n1=   5                  
                          
Sample #2   ---->   rural   
mean of sample 2,    x̅2=   60.00                  
standard deviation of sample 2,   s2 =    9.49                  
size of sample 2,    n2=   6                  
                          
difference in sample means =    x̅1-x̅2 =    52.0000   -   60.0   =   -8.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    11.2052                  
std error , SE =    Sp*√(1/n1+1/n2) =    6.7851                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -8.0000   -   0   ) /    6.79   =   -1.17906
                          
Degree of freedom, DF=   n1+n2-2 =    9                  
  
p-value =        0.268604   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis             

       
There is sufficient evidence that people living in a big city had different stress levels (0-100) than those living in rural areas.

Please revert back in case of any doubt.

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