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Home - ← → O ώ 凸 mathx com Student PlayerHomework.aspx?homeworkid=5134701948 questionid=5&flushed=false&cld- ☆ 左 久 MAT-240-T3349 Applied Statistics 19EW3 sey Charlesworth1 1/31/19 12:32 PM Homework: 5-2 MyStatLab: Module Five Problem Set Save Score: 0 of 5 pts 9 of 14 (8 complete) HW Score: 33.16%, 21.55 of 65 pts 11.2.13 Question Help 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 probablty plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Ell Click the icon to view the data table 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.) Enter your answer in the edit fields and then click Check Answer part Clear All Check Answer 10:32 AM 1/31/201 CLIP Type here to search

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'.

Upper Bound

Lower Bound

Answer 1

Let X will be denoted by steel ball

Y will be denoted by diamond

Let us find the mean of both

Mean of steel ball data $=\overline{x}=\frac{50+57+.....+51+53}{9}=\frac{530}{9}=58.89$

Mean of diamond data=$=\overline{y}=\frac{52+56+.....+51+56}{9}=\frac{543}{9}=60.33$

Difference of both mean will be denoted by d=60.33-58.89=1.44

Now let us find the sample variance for steel ball & diamond  which is denoted by $S_{1}^{2}$and $S_{2}^{2}$ respectively

Sample Variance $S_{1}^{2}$is given as below & calculation are shown in below table

Specimen	Steel Ball (X)	Xbar	x-xbar	(x-xbar)^2	Diamond (Y)	Ybar	y-ybar	(y-ybar)^2
1	50	58.89	-8.89	79.0321	52	60.33	-8.33	69.3889
2	57	58.89	-1.89	3.5721	56	60.33	-4.33	18.7489
3	61	58.89	2.11	4.4521	61	60.33	0.67	0.4489
4	71	58.89	12.11	146.6521	74	60.33	13.67	186.8689
5	68	58.89	9.11	82.9921	69	60.33	8.67	75.1689
6	54	58.89	-4.89	23.9121	56	60.33	-4.33	18.7489
7	65	58.89	6.11	37.3321	68	60.33	7.67	58.8289
8	51	58.89	-7.89	62.2521	51	60.33	-9.33	87.0489
9	53	58.89	-5.89	34.6921	56	60.33	-4.33	18.7489
Xbar	58.89	Total of ((x-xbar)^2)		474.88	Total of (y-ybar)^2			534
Ybar	60.33	Sample variance 1		59.36	Sample Variance 2			66.75

$S_{1}^{2}=\sum_{i=1}^{n}\frac{(X_{i}-X_{avg})^2}{n-1}=59.36$

And Sample standard deviation of steel ball will be $S_{1}=\sqrt{59.36}=7.70$

N=number of observations

Here n-1=9-1=8

Similarly we will find the sample variance for diamond specimen

$S_{2}^{2} =\sum_{i=1}^{n}\frac{(Y_{i}-Y_{avg})^2}{n-1}=66.75$

Sample standard deviation of diamond will be $S_{2}=\sqrt{66.75}=8.17$

Now we will find the difference of sample standard deviation using the below formula

$S=\sqrt{(n_{1}-1)s_{1}^2+(n_{2}-1)s_{2}^2/(n_{1}+n_{2}-2)$

=$S=\sqrt{\frac{(9-1)\times 59.22+(9-1)\times 66.75}{(9+9-2)}}=\sqrt{\frac{1000.76}{16}}=7.93$

Now we will find $S_{d}=S\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}=7.93\times \sqrt{\frac{1}{9}+\frac{1}{9}}=7.93\times 0.47=3.74$

As we know that Value of $Z_{\alpha /2}$ for 95% Confidence interval will be 1.96

So confidence interval is given by for finding the upper and lower bound

$d\pm Z_{\alpha /2}\times S_{d}=1.44\pm 1.96\times 3.74$

$(1.44-7.35,1.44+7.35)=(-5.91,8.79)$

Where d is difference of sample means

Sd is Sample standard deviation difference

There is sufficient proof so that we can conclude that two indenters produce different hardness readings

Note :Please vote if any doubt do not hesitate to comment