Question

1. A child wants to know if there is a difference between the two colonies that he is breeding.
   1. The kid samples his primary carpenter ants and receives the following 5 measurements: 5mm 8mm 7.3mm 6.8mm 9mm. The kid has a secondary, backup carpenter ant hive. They take the following 6 measurements: 3mm 6.2mm 7mm 11.3mm 6.1mm 6.5mm
      1. What is the null hypothesis?
      2. Compute a 95% confidence interval for this situation.
      3. Get the t-value and p-value for this situation What does it mean?
      4. Based on the p-value do we have enough evidence to reject the null hypothesis? (Does this match your answer to 2?)
2. After feeding their primary ant hive a special diet, the kid measures the same random sample of ants from #1 (previously it was 5mm 8mm 7.3mm 6.8mm 9mm) in the same order. Their measurements are:
   6mm 8.1mm 7.5mm 6.8mm 9mm
   1. We want to know if this new diet means anything, is statistically significant or if it is essentially. For the first step, for each ant, subtract the new score from the old score and record this value. Be sure to keep track of your positives and negatives. What are your 5 measurements?
   2. The null hypothesis becomes that the new mean (of your new data set computed in 5a) is 0. Find the p-value associated with this situation. Can we reject the null hypothesis? Explain, show work, describe your process, and site resources used.

Answer 1

a)

Ho :   µ1 - µ2 =   0

b)

Degree of freedom, DF=   n1+n2-2 =    9              
t-critical value =    t α/2 =    2.2622   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.2230              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    1.3461              
margin of error, E = t*SE =    2.2622   *   1.3461   =   3.0451  
                      
difference of means =    x̅1-x̅2 =    7.2200   -   6.683   =   0.5367
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    0.5367   -   3.0451   =   -2.5084
Interval Upper Limit=   (x̅1-x̅2) + E =    0.5367   +   3.0451   =   3.5818

c)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   successful                  
mean of sample 1,    x̅1=   7.22                  
standard deviation of sample 1,   s1 =    1.49                  
size of sample 1,    n1=   5                  
                          
Sample #2   ---->   unsuccessful                  
mean of sample 2,    x̅2=   6.68                  
standard deviation of sample 2,   s2 =    2.67                  
size of sample 2,    n2=   6                  
                          
difference in sample means =    x̅1-x̅2 =    7.2200   -   6.7   =   0.54  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.2230                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.3461                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.5367   -   0   ) /    1.35   =   0.399
                          
Degree of freedom, DF=   n1+n2-2 =    9                  

p-value =        0.6994 (excel function: =T.DIST.2T(t stat,df) )  

d)
       
Conclusion:     p-value>α , Do not reject null hypothesis     

Answer match with part b)

e)

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
5	6	-1.00	0.55
8	8.1	-0.10	0.03
7.3	7.5	-0.20	0.00
6.8	6.8	0.00	0.07
9	9	0.00	0.07

Ho :   µd=   0                  
Ha :   µd <   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    5                  
                          
mean of sample 1,    x̅1=   7.220                  
                          
mean of sample 2,    x̅2=   7.480                  
                          
mean of difference ,    D̅ =ΣDi / n =   -0.260                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    0.4219                  
                          
std error , SE = Sd / √n =    0.4219   / √   5   =   0.1887      
                          
t-statistic = (D̅ - µd)/SE = (   -0.26   -   0   ) /    0.1887   =   -1.378
                          
Degree of freedom, DF=   n - 1 =    4                  

                          
p-value =        0.120135   [excel function: =t.dist(t-stat,df) ]              
Conclusion:     p-value>α , Do not reject null hypothesis                      

Thanks in advance!

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