Answer:
We reject H₀ . We found that at significance level of 0,05 the mean of the type 2 rope is greater than that of rope 1
Explanation:
Type 1 rope information:
x₁ = 6.2888
s₁ = 0.1352
n₁ = 17
Type 2 rope Information:
x₂ = 6.400
s₂ = 0.1070
n₂ = 14
We don´t know population standard deviation
We assume that populations follow normal distribution
Hypothesis Test:
Null Hypothesis H₀ x₂ = x₁
Alternative Hypothesis Hₐ x₂ > x₁
The alternative hypothesis indicates a one-tail test on the right
Significance level α = 0,05 confidence Interval 95 %
Degrees of freedom for t(c) is n₁ + n₂ - 2 = 17 + 14 - 2
df = 31 - 2 df = 29
Then from t- student table we get t(c) = 2,045
To calculate t(s)
t(s) = ( x₂ - x₁ ) / sp √ ( 1/n₁ + 1/n₂
Doing calculations:
√ (1/n₁ + 1/n₂ = √ 1/17 + 1/14 = √ 0,0588 + 0,0714 = √ 0,1302
√ (1/n₁ + 1/n₂ = 0,3608
sp² = ( n₁ - 1 )* s₁² + ( n₂ - 1 )*s₂² / n₁ + n₂ - 2
sp² = 16* (0,1352)² + 13* ( 0,1070)² / 29
sp² = 16*0,0183 + 13* 0,0114 / 29 = 0,2928 + 0,1482 / 29
sp² = 0,01520
sp = 0,1233
By substitution
t(s) = ( 6,400 - 6,2888 ) / 0,1233*0,3608
t(s) = 0,1112 / 0,0444
t(s) = 2,5045
Comparing t(s) and t(c)
t(s) > t(c) 2,50 > 2,045
Then t(s) is in the rejection region we reject H₀ . We found differences between the means of the two groups.