Question

There are two methods A and B for studying the latent heat of ices. The following are the collected data of the heat absorption when the ices absorbed heat and their temperature increased from initial -0.72℃ to temperature 0℃: Method A : 79.98, 80.04, 80.02, 80.03, 80.03, 80.04, 80.04 79.97, 80.05, 80.03, 80.02, 80.00, 80.02 Method B: 80.02, 79.94, 79.97, 79.98, 79.97, 80.03, 79.95, 79.97 Assume these data follow normal distribution with equal variance. Test the hypothesis H0“the mean ? of these two methods are the same" at the level ? = 0.05

Answer 1

For Method A :

∑x = 1040.27

∑x² = 83243.2

n1 = 13

Mean , x̅1 = Ʃx/n = 1040.27/13 = 80.0208

Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(83243.2125-(1040.27)²/13)/(13-1)] = 0.0240

For Method B :

∑x = 639.83

∑x² = 51172.8

n2 = 8

Mean , x̅2 = Ʃx/n = 639.83/8 = 79.9788

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(51172.8105-(639.83)²/8)/(8-1)] = 0.0314

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((13-1)*0.024² + (8-1)*0.0314²) / (13+8-2) = 0.0007

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (80.0208 - 79.9788) / √(0.0007*(1/13 + 1/8)) = 3.4722

df = n1+n2-2 = 19

Critical value :

Two tailed critical value, t crit = T.INV.2T(0.05, 19) = 2.093

Reject Ho if t < -2.093 or if t > 2.093

p-value :

Two tailed p-value = T.DIST.2T(ABS(3.4722), 19) = 0.0026

Decision:

p-value < α, Reject the null hypothesis.