Question

In intaglio printing, a design or figure is carved beneath the surface of hard metal or stone. The business objective of an intaglio printing company is to determine whether there are differences in the mean surface hardness of steel plates, based on two different surface conditions – untreated and treated by lightly polishing with emery paper. An experiment is designed in which 40 steel plates are randomly assigned – 20 plates are untreated and 20 plates are treated, The results of the experiment are as follows:

Untreated	Treated
164.368          177.135	158.239          150.223
159.018          163.903	138.216          155.620
153.871          167.802	168.006          151.233
165.096          160.818	149.654          158.653
157.184          167.433	145.456          151.204
154.496          163.538	168.178          150.689
160.920          164.525	154.321          161.657
164.917          171.230	162.763          157.016
169.091          174.964	161.020          156.670
175.276          166.311	167.706          147.920

(a) Assuming that the population variances from both conditions are equal, is there evidence of a difference in the mean surface hardness between untreated and treated steel plates? (Use α = 0.05.) (Include: null and alternate hypotheses, value of test statistic, decision rule, conclusion in a complete sentence that includes why in terms of the decision rule.)

(b) Determine the p-value in (a) and interpret its meaning.

(c) In addition to equal variances, what other assumption is necessary in (a)?

(d) Construct and interpret a 95% confidence interval estimate of the difference between the population means from treated and untreated steel plates.

Answer 1

a) Assuming that the population variances from both conditions are equal, is there evidence of a difference in the mean surface hardness between untreated and treated steel plates? (Use α = 0.05.) (Include: null and alternate hypotheses, value of test statistic, decision rule, conclusion in a complete sentence that includes why in terms of the decision rule.)

A t-test is used when you're looking at a numerical variable - for example, height - and then comparing the averages of two separate populations or groups (e.g., males and females).

Requirements

• Two independent samples
• Data should be normally distributed
• The two samples should have the same variance

Null Hypothesis

H0: u1 - u2 = 0,

H1= u1 - u2 0

where u1 is the mean of first population and u2 the mean of the second.

As above, the null hypothesis tends to be that there is no difference between the means of the two populations; or, more formally, that the difference is zero (so, for example, that there is no difference between the average heights of two populations of males and females).

Equation

Untreated	Diff (X - M)	Sq. Diff (X - M)2			Treated	Diff (X - M)	Sq. Diff (X - M)2
164.368	-0.73	0.53			158.239	2.52	6.34
159.018	-6.08	36.93			138.216	-17.51	306.46
153.871	-11.22	125.97			168.006	12.28	150.9
165.096	0	0			149.654	-6.07	36.82
157.184	-7.91	62.58			145.456	-10.27	105.39
154.496	-10.6	112.34			168.178	12.46	155.15
160.92	-4.17	17.43			154.321	-1.4	1.96
164.917	-0.18	0.03			162.763	7.04	49.58
169.091	4	15.97			161.02	5.3	28.07
175.276	10.18	103.66			167.706	11.98	143.62
177.14	12.05	145.09			150.22	-5.5	30.27
163.90	-1.19	1.43			155.62	-0.1	0.01
167.80	2.71	7.32			151.23	-4.49	20.18
160.82	-4.27	18.27			158.65	2.93	8.57
167.43	2.34	5.45			151.20	-4.52	20.45
163.54	-1.55	2.42			150.69	-5.03	25.32
164.53	-0.56	0.32			161.66	5.94	35.26
171.23	6.14	37.64			157.02	1.3	1.68
174.96	9.87	97.32			156.67	0.95	0.9
166.31	1.22	1.48			147.92	-7.8	60.87
	M: 165.09	SS: 792.17				M: 155.72	SS: 1187.80

Difference Scores Calculations

Untreated
N₁: 20
df₁ = N - 1 = 20 - 1 = 19
M₁: 165.09
SS₁: 792.17
s²₁ = SS₁/(N - 1) = 792.17/(20-1) = 41.69


Treated 2

N₂: 20
df₂ = N - 1 = 20 - 1 = 19
M₂: 155.72
SS₂: 1187.8
s²₂ = SS₂/(N - 1) = 1187.8/(20-1) = 62.52


T-value Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂) = ((19/38) * 41.69) + ((19/38) * 62.52) = 52.1

s²_M1 = s²_p/N₁ = 52.1/20 = 2.61
s²_M2 = s²_p/N₂ = 52.1/20 = 2.61

t = (M₁ - M₂)/√(s²_M1 + s²_M2) = 9.37/√5.21 = 4.11

The t-value is 4.10617. The p-value is .000206. The result is significant at p < .05.

p value calculated in excel; = =T.DIST.2T(4.11,38) = 0.000206

(b) Determine the p-value in (a) and interpret its meaning.

The t-value is 4.10617. The p-value is .000206. The result is significant at p < .05. We reject the null hypothesis and conclude that there is significance difference between the untreated and treated group

(c) In addition to equal variances, what other assumption is necessary in (a)?

Normality of the data is the additional assumption for the test.

(d) Construct and interpret a 95% confidence interval estimate of the difference between the population means from treated and untreated steel plates.

his simple confidence interval calculator uses a t statistic and two sample means (M₁ and M₂) to generate an interval estimate of the difference between two population means (μ₁ and μ₂).

The formula for estimation is:

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

where:

M₁ & M₂ = sample means
t = t statistic determined by confidence level
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

Calculation

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 1983.79 / 38 = 52.2

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((52.2/20) + (52.2/20)) = 2.28

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 9.37 ± (2.02 * 2.28) = 9.37 ± 4.6254

Result

μ₁ - μ₂ = (M₁ - M₂) = 9.37, 95% CI [4.7446, 13.9954].

You can be 95% confident that the difference between your two population means (μ₁ - μ₂) lies between 4.7446 and 13.9954.

Thanks