Elasticity is an important parameter of kernmantle ropes (used inclimbing, caving, mountaineering, etc.). One of the ways of describing it is static elongation, measuredas amount (percentage) of rope stretch under a resting load of 80 kg. Static elongation was measuredfor two random rope samples of two different types of weaving. The observed sample means, standarddeviations, and sample sizes were the following. Type 1: Type 2: Ti = 6.2888 $i = 0.1352 ni = 17 T2 = 6.4007 S2 = 0.1070 n1 = 14
Is there evidence to support the claim that type-2- ropes are more elastic than type 1? Use a = 0.05 and assume that the populations variances are equal. Remember for state the null and alternative hypotheses, the decision rule, and your conclusion.

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

Step-by-step 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.

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.

step-by-step explanation:

in which figure is point g a centroid