The manufacturer of hardness testing equipment uses​ steel-ball indenters to penetrate metal that is being tested.​...
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The manufacturer of hardness testing equipment uses​ steel-ballindenters to penetrate metal that is being tested.​ However, themanufacturer thinks it would be better to use a diamond indenter sothat all types of metal can be tested. Because of differencesbetween the two types of​ indenters, it is suspected that the twomethods will produce different hardness readings. The metalspecimens to be tested are large enough so that two indentions canbe made.​ Therefore, the manufacturer uses both indenters on eachspecimen and compares the hardness readings. Construct a​ 95%confidence interval to judge whether the two indenters result indifferent measurements.
​Note: A normal probability plot and boxplot of the dataindicate that the differences are approximately normallydistributed with no outliers.
Specimen  Steel ball  Diamond
1Â Â 51Â Â 53
2Â Â 57Â Â 55
3Â Â 61Â Â 63
4Â Â 71Â Â 74
5Â Â 68Â Â 69
6Â Â 54Â Â 55
7Â Â 65Â Â 68
8Â Â 51Â Â 51
9Â Â 53Â Â 56
Construct a​ 95% confidence interval to judge whether the twoindenters result in different​ measurements, where the differencesare computed as​ 'diamond minus steel​ ball'.
The lower bound is __?__ .
The upper bound is __?__. ​
(Round to the nearest tenth as​ needed.)
The manufacturer of hardness testing equipment uses​ steel-ballindenters to penetrate metal that is being tested.​ However, themanufacturer thinks it would be better to use a diamond indenter sothat all types of metal can be tested. Because of differencesbetween the two types of​ indenters, it is suspected that the twomethods will produce different hardness readings. The metalspecimens to be tested are large enough so that two indentions canbe made.​ Therefore, the manufacturer uses both indenters on eachspecimen and compares the hardness readings. Construct a​ 95%confidence interval to judge whether the two indenters result indifferent measurements.
​Note: A normal probability plot and boxplot of the dataindicate that the differences are approximately normallydistributed with no outliers.
Specimen  Steel ball  Diamond
1Â Â 51Â Â 53
2Â Â 57Â Â 55
3Â Â 61Â Â 63
4Â Â 71Â Â 74
5Â Â 68Â Â 69
6Â Â 54Â Â 55
7Â Â 65Â Â 68
8Â Â 51Â Â 51
9Â Â 53Â Â 56
Construct a​ 95% confidence interval to judge whether the twoindenters result in different​ measurements, where the differencesare computed as​ 'diamond minus steel​ ball'.
The lower bound is __?__ .
The upper bound is __?__. ​
(Round to the nearest tenth as​ needed.)
Answer & Explanation Solved by verified expert
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 ,   DÌ… =ΣDi / n = Â
1.444
std dev of difference , Sd =   √ [ (Di-Dbar)²/(n-1) =
  1.667
sample size , Â Â n = Â Â 9Â Â
     Â
Degree of freedom, DF=Â Â n - 1 = Â Â
8  and α =   0.05 Â
t-critical value =   t α/2,df =  Â
2.3060  [excel function: =t.inv.2t(α/2,df) ] Â
  Â
       Â
     Â
std dev of difference , Sd =   √
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