Question

NOTE: NEED HELP FINDING TWO QUESTIONS FOR LOWER AND UPPER BOUND

1)

The manufacturer of hardness testing equipment uses​ steel-ball indenters to penetrate metal that is being tested.​ However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of​ indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made.​ Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a​ 95% confidence interval to judge whether the two indenters result in different measurements.

​Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

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Construct a​ 95% confidence interval to judge whether the two indenters result in different​ measurements, where the differences are computed as​ 'diamond minus steel​ ball'.

The lower bound is ??????????

The upper bound is???????

​(Round to the nearest tenth as​ needed.)

Specimen   Steel ball   Diamond
1   51   53
2   57   56
3   61   61
4   70   74
5   68   69
6   54   55
7   65   68
8   51   51
9   53   56

2)

To illustrate the effects of driving under the influence​ (DUI) of​ alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of special goggles to simulate the effects of alcohol on vision. For a random sample of nine​ teenagers, the time​ (in seconds) required to bring the vehicle to a stop from a speed of 60 miles per hour was recorded. Complete parts​ (a) and​ (b).

​Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

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​(a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the​ experiment?

A.

This is a good idea in designing the experiment because it controls for any​ "learning" that may occur in using the simulator.

Your answer is correct.

B.

This is a good idea in designing the experiment because reaction times are different.

C.

This is a good idea in designing the experiment because the sample size is not large enough.

​(b) Use a​ 95% confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as​ "impaired minus​ normal."

The lower bound is ???????

The upper bound is ????????

​(Round to the nearest thousandth as​ needed.)

Subject   Normal, Xi   Impaired, Yi
1   4.49   5.77
2   4.34   5.85
3   4.58   5.45
4   4.65   5.29
5   4.31   5.83
6   4.80   5.49
7   4.55   5.23
8   5.00   5.63
9   4.79   5.63

Answer 1

1)

sample size ,    n =    9          
Degree of freedom, DF=   n - 1 =    8   and α =    0.05  
t-critical value =    t α/2,df =    2.3060   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.6667          
                  
std error , SE = Sd / √n =    1.6667   / √   9   =   0.5556
margin of error, E = t*SE =    2.3060   *   0.5556   =   1.2811
                  
mean of difference ,    D̅ =   1.444          
confidence interval is                   
Interval Lower Limit= D̅ - E =   1.444   -   1.2811   =   0.1633
Interval Upper Limit= D̅ + E =   1.444   +   1.2811   =   2.7256
                  
so, confidence interval is (   0.16   < µd <   2.73   )  

2)

sample size ,    n =    9          
Degree of freedom, DF=   n - 1 =    8   and α =    0.05  
t-critical value =    t α/2,df =    2.3060   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    0.3714          
                  
std error , SE = Sd / √n =    0.3714   / √   9   =   0.1238
margin of error, E = t*SE =    2.3060   *   0.1238   =   0.2855
                  
mean of difference ,    D̅ =   0.962          
confidence interval is                   
Interval Lower Limit= D̅ - E =   0.962   -   0.2855   =   0.6767
Interval Upper Limit= D̅ + E =   0.962   +   0.2855   =   1.2477
                  
so, confidence interval is (   0.68   < µd <   1.25   )  

Please let me know in case of any doubt.

Thanks in advance!

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