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   A B
1 792.4   797.8
2    792.6   790.7
3 793.5   796.9
4 793.9   792.1
5 793.7   794.9
6   792.7   789.3

​a) Why are these​ matched-pairs data?

A.The same round was fired in every trial.

B.All the measurements came from rounds fired from the same gun.

C.The measurements​ (A and​ B) are taken by the same instrument.

D.Two measurements​ (A and​ B) are taken on the same round.

b) Is there a difference in the measurement of the muzzle velocity between device A and device B at the a=0.0 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.

H0:μd (<,>,=,≠) BLANK

H1:μd​​​​​​​ (<,>,=,≠) BLANK

Determine the test statistic for this hypothesis test.

t0= BLANK

Find the critical​ value(s) for this hypothesis test.

BLANK

What is your conclusion regarding H0​?

BLANK H0.There BLANK sufficient evidence at the a=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 mean difference. Compute the difference as device A minus device B. Interpret your results.

The confidence interval is=

Interpret the confidence interval. Choose the correct answer below.

A. One can be​ 99% confident that the mean difference in measurement is 0.01.

B. One can be​ 99% confident that the mean difference in measurement is 0.

C. One can be​ 1% confident that the mean difference in measurement lies in the interval found above.

D. One can be​ 99% confident that the mean difference in measurement lies in the interval found above.

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

Does this visual evidence support the results obtained in part​ (b)?​​​​​​​

A. ​No, because the boxplot is too large.

B. ​No, because 0 is contained in the boxplot.

C. ​Yes, because 0 is contained in the boxplot.

D. ​Yes, because the boxplot shows no outliers.

Answer 1

B.All the measurements came from rounds fired from the same gun.

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

Ho :   µd=   0
Ha :   µd ╪   0
Level of Significance ,    α =    0.01   
sample size ,    n =    6                  
                          
mean of sample 1,    x̅1=   793.133                  
                          
mean of sample 2,    x̅2=   793.617                  
                          
mean of difference ,    D̅ =ΣDi / n =   -0.483                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.4400                  
                          
std error , SE = Sd / √n =    3.4400   / √   6   =   1.4044      
                          
t-statistic = (D̅ - µd)/SE = (   -0.483333333   -   0   ) /    1.4044   =   -0.344
                          
Degree of freedom, DF=   n - 1 =    5                  
t-critical value , t* =    ±   4.032   [excel function: =t.inv.2t(α,df) ]   
                          
Conclusion:     |test stat| < |critical value| , Do not reject null hypothesis          

.There is not sufficient evidence at the a=0.01 level of significance to conclude that there is a difference in the measurements of velocity between device A and device B.           

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

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) =    3.4400          
                  
std error , SE = Sd / √n =    3.4400   / √   6   =   1.4044
margin of error, E = t*SE =    4.0321   *   1.4044   =   5.6627
                  
mean of difference ,    D̅ =   -0.483          
confidence interval is                   
Interval Lower Limit= D̅ - E =   -0.483   -   5.6627   =   -6.146
Interval Upper Limit= D̅ + E =   -0.483   +   5.6627   =   5.179
                  
so, confidence interval is (   -6.1460   < µd <   5.1793   )  

D. One can be​ 99% confident that the mean difference in measurement lies in the interval found above.

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

difference , Di =sample1-sample2

-5.40

1.90

-3.40

1.80

-1.20

3.40

C. ​Yes, because 0 is contained in the boxplot.

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

THANKS

revert back for doubt

please upvote