Question

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.

Specimen   Steel ball   Diamond
1   51   53
2   57   55
3   61   63
4   71   74
5   68   69
6   54   55
7   65   68
8   51   51
9   53   56

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.)

Answer 1

paired t test

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
53	51	2	0.31
55	57	-2	11.86
63	61	2	0.31
74	71	3	2.42
69	68	1	0.20
55	54	1	0.20
68	65	3	2.42
51	51	0	2.09
56	53	3	2.42

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	544	531	13	22.222

mean of difference ,    D̅ =ΣDi / n =   1.444
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.667

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   =   0.2
Interval Upper Limit= D̅ + E =   1.444   +   1.2811   =   2.7
                  
so, confidence interval is (   0.2 < Dbar <   2.7 )