Question

ANSWER B) AND C) WITHOUT USING TECHNOLOGY.

Formulation 1 Formulation 2 206 188 205 187 193 207 185 189 192 210 194 178 197 206 201 176 185 200 197 198 188 189 203 (a) Construct normal probability plots for both samples Do these plots support assumptions of normality and equal variance for both samples? (b) Do the data support the claim that the mean deflection temperature under load for formulation 1 exceeds that of formulation 2? Use α 0.05 (c) What is the P-value for the test in part (a)?

Answer 1

Rank	Proportion	Z Value	data I
1	0.076923	-1.42608	178
2	0.153846	-1.02008	185
3	0.230769	-0.73632	187
4	0.307692	-0.5024	188
5	0.384615	-0.29338	189
6	0.461538	-0.09656	192
7	0.538462	0.096559	193
8	0.615385	0.293381	194
9	0.692308	0.502402	205
10	0.769231	0.736316	206
11	0.846154	1.020076	207
12	0.923077	1.426077	210

Rank	Proportion	Z Value	data II
1	0.076923	-1.42608	176
2	0.153846	-1.02008	177
3	0.230769	-0.73632	185
4	0.307692	-0.5024	188
5	0.384615	-0.29338	189
6	0.461538	-0.09656	197
7	0.538462	0.096559	197
8	0.615385	0.293381	198
9	0.692308	0.502402	200
10	0.769231	0.736316	201
11	0.846154	1.020076	203
12	0.923077	1.426077	206

yes,plot support assumptions of normality and equal variance for both samples.

b)

using excel addins tool data analysis tool for t test

t-Test: Two-Sample Assuming Equal Variances
	data I	data II
Mean	194.5	193.0833333
Variance	103.5454545	98.99242424
Observations	12	12
Pooled Variance	101.2689394
Hypothesized Mean Difference	0
df	22
t Stat	0.344830101
P(T<=t) one-tail	0.366748471
t Critical one-tail	1.717144374
P(T<=t) two-tail	0.733496943
t Critical two-tail	2.073873068

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 >   0

t-stat = 0.3448

Degree of freedom, DF=   n1+n2-2 =    22
t-critical value , t* =        1.717
Decision rule : reject Ho , if | t-stat | > t-critical,otherwise not  
so, conclusion is fail to reject Ho.

c)

p-value=0.366748471

excel formula =T.DIST.RT(0.34483,22)