Question

Problem 3: Metal Hardness Testing

The manufacturer of hardness testing equipment uses steel-ball indenters to indent metal that is being tested. However, the manufacturer thinks there might be a difference in hardness reading when using a diamond indenter. The metal specimens to be tested are large enough so that two indentations can be made. Therefore, the manufacturer wants to use both indenters on each specimen and compare the readings. The order of the indentations will be random. This particular design is called the paired design (or matched pairs design or dependent samples design). Assume all conditions are satisfied in this problem. The data set used for this problem is called “Metal Hardness Testing”.

1. Calculate the difference between specimens by subtracting Steel Ball – Diamond. For example, the first difference is 51 – 52 = -1. List the difference for each of the 14 pairs in your document.

2. For the first piece of metal, which indenter produced the larger hardness reading? Answer this question in a complete sentence.

3. Obtain the mean of these differences and the standard deviation of these differences in StatCrunch. You may copy and paste the box that you obtain from StatCrunch or list the values. Please round these values to four decimal places.

4. Construct a 95% confidence interval using the above data. Please do this “by hand” using the formula and showing your work (please type your work). Use your t-table (found in the last page of our formula packet) to obtain your t* critical value needed for the confidence interval. Present this confidence as (lower limit, upper limit)

5. Use StatCrunch to obtain a 95% confidence interval for the above data by selecting:

Stat à T Stats à Paired. Enter Steel Ball for Sample 1 and Diamond for Sample 2. Copy and paste your output into your document.

6. Does your confidence interval capture 0? Answer this question and briefly explain what this implies in one or two sentences in the context of the question.

g. Using your answer to part (g), imagine you were using a hypothesis test to determine if a significant difference exists in mean hardness reading between the two indenters (the hypotheses would be H₀: m_D = 0 vs H_a: m_D ≠ 0). What decision and conclusion can be made in this case? Provide an answer and a reason for your choice in one or two sentences. Please only use your confidence interval to answer this question (i.e. do not run this hypothesis test).

Specimen   Steel Ball   Diamond
1 51   52
2   57   56
3   61   61
4   71   72
5   68   69
6   53   55
7   67   68
8   51   51
9   54   56
10   89   87
11   51   53
12   70   73
13   56   61
14   77   77

Answer 1

a) Data table with differences between hardness of specimens

Specimen	Steel Ball	Diamond	Difference
1	51	52	-1
2	57	56	1
3	61	61	0
4	71	72	-1
5	68	69	-1
6	53	55	-2
7	67	68	-1
8	51	51	0
9	54	56	-2
10	89	87	2
11	51	53	-2
12	70	73	-3
13	56	61	-5
14	77	77	0

b) For the first piece of metal the diamond indenter produces more hardness. (The hardness value = 52)

c) Mean = = -1.0714

std dev S_d = = 1.7305

d) confidence level = 95%

alpha = 5%

CI =

t_0.025,13 = 2.160

95% CI = (-1.0714 - 2.160*1.7305/sqrt(14) , -1.0714 - 2.160*1.7305/sqrt(14) ) = (-2.0704 , -0.0724)

e)

Estimation for Paired Difference

Mean	StDev	SE Mean	95% CI for μ_difference
-1.0714	1.7305	0.4625	(-2.0706, -0.0723)

µ_difference: mean of (Steel Ball - Diamond)

f) 95% confidence interval doesnot capture 0, which means there exist a significant difference between hardness results measured by Steel Ball - Diamond.

g) Hypothesis

H₀: m_D = 0 vs H_a: m_D ≠ 0

Since zero is not captured in 95% CI, we reject null hypothesis and conclude that there is a significant difference between hardness results measured by Steel Ball - Diamond, since CI consists in complete negative range wee conclude that Diamond is giving higher hardness results.