Question

The following data represent the muzzle velocity (in feet per second) of rounds fired from a 155-mm gun. For each round, two measurements of the velocity were recorded using two different measuring devices, resulting in the following data. Complete parts (a) through (d) below. 1 3 Observation A B 790.4 791.0 2 792.9 788.4 794.1 795.5 4 792.9 788.0 5 6 794.9 792.8 797.6 791.5 Click the icon to view the table of critical t-values.
(a) Is there a difference in the measurement of the muzzle velocity between device A and device B at the α = 0.01 level of​ significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Let d_i = A_i − B_i.

(i) Identify the null and alternative hypotheses.

(ii) Determine the test statistic for this hypothesis test (t₀ = ?). Round to two decimal places as needed.

(iii) Calculate avg(d) and s_d. Round to six decimal places as needed.

(iv) Find the P-value (P = ?). Round to six decimal places as needed.

(b) Identify the lower and upper bounds for the 99% confidence interval. Round to six decimal places as needed.

(c) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part​ (a)?

Answer 1

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
790.4	791	-0.6	2.560
792.9	788.4	4.5	12.250
794.1	795.5	-1.4	5.760
792.9	788	4.9	15.210
794.9	797.6	-2.7	13.690
792.8	791.5	1.3	0.090

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	4758	4752.00	6.000	49.560

Ho :   µd=   0                  
Ha :   µd ╪   0                  
                          
Level of Significance ,    α =    0.01                  
                          
sample size ,    n =    6                  
                          
mean of sample 1,    x̅1=   793.000                  
                          
mean of sample 2,    x̅2=   792.000                  
                          
mean of difference ,    D̅ =ΣDi / n =   1.0000                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.148333
                          
std error , SE = Sd / √n =    3.1483   / √   6   =   1.2853      
                          
t-statistic = (D̅ - µd)/SE = (   1   -   0   ) /    1.2853   =   0.78
                          
Degree of freedom, DF=   n - 1 =    5                  
t-critical value , t* =    ±   4.0321   [excel function: =t.inv.2t(α,df) ]               
                          
p-value =        0.471746 [excel function: =t.dist.2t(t-stat,df) ]               
Decision:   p-value>α , Do not reject null hypothesis  

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

sample size ,    n =    6          
Degree of freedom, DF=   n - 1 =    5   and α =    0.01  
t-critical value =    t α/2,df =    4.0321   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.1483          
                  
std error , SE = Sd / √n =    3.1483   / √   6   =   1.2853
margin of error, E = t*SE =    4.0321   *   1.2853   =   5.1825
                  
mean of difference ,    D̅ =   1.000          
confidence interval is                   
Interval Lower Limit= D̅ - E =   1.000   -   5.1825   =   -4.182520
Interval Upper Limit= D̅ + E =   1.000   +   5.1825   =   6.182520

Chart Title 6 5 4 3 2 1 0 -1 -2 -3 -4 1

                  
      

Please let me know in case of any doubt.

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