Question

QUESTION 16

1. A researcher wanted to see whether there was a difference between NYC citizens and Boston citizens in the amount of time (in hours) they spent outside during April 2020. He collected two samples – one of NYC citizens and the other of Boston citizens, and measured the amount of time each subject spent outside during April 2020. Sample #1 (of NYC citizens) contained 20 subjects (n₁=24), the mean number of hours spent outside was M₁=40, and the sample’s variance was S₁²=64. Sample #2 (of Boston citizens) contained 24 subjects (n₂=24), the mean number of hours spent outside was M₂=28, and the sample’s variance was S₂²=25.

   a. Which test should be used to analyze your data? Explain your answer. (3 points)

   b. Conduct the appropriate nondirectional hypothesis test on these data. Use α=0.05. You must submit your calculations (if you do not submit your calculations, you will lose points). (8 points)

   c. Calculate the appropriate effect size measure. You must submit your calculations (if you do not submit your calculations, you will lose points). (4 points)

   d. Compute a 95% confidence interval for the mean difference. You must submit your calculations (if you do not submit your calculations, you will lose points). (5 points)

Answer 1

a)

2 sample independent samples for means is used

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

b)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   40.00                  
standard deviation of sample 1,   s1 =    64.00                  
size of sample 1,    n1=   20                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   28.00                  
standard deviation of sample 2,   s2 =    25.00                  
size of sample 2,    n2=   24                  
                          
difference in sample means =    x̅1-x̅2 =    40.0000   -   28.0   =   12.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    46.8531                  
std error , SE =    Sp*√(1/n1+1/n2) =    14.1855                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   12.0000   -   0   ) /    14.19   =   0.846
                          
Degree of freedom, DF=   n1+n2-2 =    42                  

p-value =        0.402383   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis  

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

c)

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    0.256
.........

d)

                    Degree of freedom, DF=   n1+n2-2 =    42              
t-critical value =    t α/2 =    2.0181   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    46.8531              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    14.1855              
margin of error, E = t*SE =    2.0181   *   14.1855   =   28.6275  
                      
difference of means =    x̅1-x̅2 =    40.0000   -   28.000   =   12.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    12.0000   -   28.6275   =   -16.6275
Interval Upper Limit=   (x̅1-x̅2) + E =    12.0000   +   28.6275   =   40.6275

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

THANKS

revert back for doubt

please upvote