Question

The table below gives the pretest and posttest scores on the listening test in Spanish for 20 high school Spanish teachers who attended an intensive summer course in Spanish

Pretest 30 28 31 26 20 30 34 15 28 20 30 29 31 29 34 20 26 25 31 29

Posttest 29 30 32 30 16 25 31 18 33 25 32 28 34 32 32 27 28 29 32 32

a. At 5% significance level, test the hypothesis that attending an intensive summer course improves listening skills in Spanish.

b. Find a 90% confidence interval for the mean increase in listening score due to attending an intensive summer course in Spanish.

Answer 1

a)

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
30	29	1.00	6.00
28	30	-2.00	0.30
31	32	-1.00	0.20
26	30	-4.00	6.50
20	16	4.00	29.70
30	25	5.00	41.60
34	31	3.00	19.80
15	18	-3.00	2.40
28	33	-5.00	12.60
20	25	-5.00	12.60
30	32	-2.00	0.30
29	28	1.00	6.00
31	34	-3.00	2.40
29	32	-3.00	2.40
34	32	2.00	11.90
20	27	-7.00	30.80
26	28	-2.00	0.30
25	29	-4.00	6.50
31	32	-1.00	0.20
29	32	-3.00	2.40

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	546	575	-29.000	194.950

Ho :   µd=   0                  
Ha :   µd <   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    20                  
                          
mean of sample 1,    x̅1=   27.300                  
                          
mean of sample 2,    x̅2=   28.750                  
                          
mean of difference ,    D̅ =ΣDi / n =   -1.450                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.2032                  
                          
std error , SE = Sd / √n =    3.2032   / √   20   =   0.7163      
                          
t-statistic = (D̅ - µd)/SE = (   -1.45   -   0   ) /    0.7163   =   -2.024
                          
Degree of freedom, DF=   n - 1 =    19                     
                          
p-value =        0.028609   [excel function: =t.dist(t-stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis  

attending an intensive summer course improves listening skills in Spanish.                   

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

b)

sample size ,    n =    20          
Degree of freedom, DF=   n - 1 =    19   and α =    0.1  
t-critical value =    t α/2,df =    1.7291   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.2032          
                  
std error , SE = Sd / √n =    3.2032   / √   20   =   0.7163
margin of error, E = t*SE =    1.7291   *   0.7163   =   1.2385
                  
mean of difference ,    D̅ =   -1.450          
confidence interval is                   
Interval Lower Limit= D̅ - E =   -1.450   -   1.2385   =   -2.689
Interval Upper Limit= D̅ + E =   -1.450   +   1.2385   =   -0.211
                  
so, confidence interval is (   -2.6885   < µd <   -0.2115   )  

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

THANKS

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