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. Observation 1 2 3 4 5 6 A 793.1 790.6 791.8 793.5 794.7 790.2 B 797.9 786.7 794.4 792.8 796.9 789.8 ​(a) Why are these​ matched-pairs data? b. Is there a difference in the measurement of the muzzle velocity between device A and device B at the alpha equals 0.01α=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 di=Ai−Bi. Identify the null and alternative hypotheses. Upper H 0H0​: Upper H 1H1​: Determine the test statistic for this hypothesis test. t0= ​(Round to two decimal places as​ needed.) Find the​ P-value. ​P-valuee= ​(Round to three decimal places as​ needed.)What is your conclusion regarding Upper H 0= α=0.01 level of significance to conclude that there is a difference in the measurements of velocity between device A and device B. ​(c) Construct a​ 99% confidence interval about the population interval Interpret the confidence interval draw a boxplot

Answer 1

a)

these are matched-pairs data becaue  For each​ round, two measurements of the velocity were recorded using two different measuring​ devices .hence, data is dependent

b)

Ho :   µd=   0
Ha :   µd ╪   0

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
793.1	797.9	-4.800	16.2678
790.6	786.7	3.900	21.7778
791.8	794.4	-2.600	3.3611
793.5	792.8	0.700	2.1511
794.7	796.9	-2.200	2.0544
790.2	789.8	0.400	1.3611

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	4753.9	4758.5	-4.6	47.0

mean of difference ,    D̅ =ΣDi / n =   -0.767                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.0651                  
                          
std error , SE = Sd / √n =    3.0651   / √   6   =   1.2513      
                          
t-statistic = (D̅ - µd)/SE = (   -0.7667   -   0   ) /    1.2513   =   -0.61

Degree of freedom, DF=   n - 1 =    5      
  
p-value =        0.567 [excel function: =t.dist.2t(t-stat,df) ]   

p-value>α , Do not reject null hypothesis  
there is not enough evidence to  conclude that there is a difference in the measurements of velocity between device A and device B

c)

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) =    3.0651          
                  
std error , SE = Sd / √n =    3.0651   / √   6   =   1.2513
margin of error, E = t*SE =    4.0321   *   1.2513   =   5.0455
                  
mean of difference ,    D̅ =   -0.767          
confidence interval is                   
Interval Lower Limit= D̅ - E =   -0.767   -   5.0455   =   -5.8121
Interval Upper Limit= D̅ + E =   -0.767   +   5.0455   =   4.2788
                  
so, confidence interval is (   -5.8121   < µd <   4.2788   )