Question

Summary statistics are given for independent simple random samples from two populations. Use the pooled t-test to conduct the required hypothesis test.

₁ = 13, s ₁ = 5, n ₁ = 10,   ₂ = 21, s ₂ = 4, n ₂ = 14
Perform a left-tailed hypothesis test using a significance level of α = 0.05.

	a	Test statistic: t = -1.526526 Critical value = -1.717 0.05 < P < 0.10 Do not reject H0
	b	Test statistic: t = -0.916.916 Critical value = -2.074 P-value > 0.100.100.005 Do not reject H0
	c	Test statistic: t = -4.355 Critical value = -1.717 P < 0.005 Reject H0
	d	Test statistic: t = -4.355 Critical value = -2.074 P < 0.005 Reject H0

can you please show the math? thank u!

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   13.00                  
standard deviation of sample 1,   s1 =    5.00                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   21.00                  
standard deviation of sample 2,   s2 =    4.00                  
size of sample 2,    n2=   14                  
                          
difference in sample means =    x̅1-x̅2 =    13.0000   -   21.0   =   -8.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.4364                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.8369                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -8.0000   -   0   ) /    1.84   =   -4.355
                          
Degree of freedom, DF=   n1+n2-2 =    22                  
t-critical value , t* =        -1.7171   (excel function: =t.inv(α,df)             
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.000127   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      

Please revert back in case of any doubt.

Please upvote. Thanks in advance.