Question

1.       High Mountain produces a variety of climbing and mountaineering equipment. One of the products is a traditional three-stranded climbing robe. An important characteristic of any climbing rope is its tensile strength. High Mountain produces the three-strand rope on two separate production lines: one in Bozeman and the other in Challis. The Bozeman line has recently installed new production equipment. High Mountain regularly tests the tensile strength of its ropes by randomly selecting ropes from production and subjecting then to various test. The most recent random sample of ropes, taken after the new equipment was installed at the Bozeman plant, reveled the following:

Bozeman	Challis
mean=7,200lb	mean= 7,087 lb
s1 = 425 lb	s2 = 415 lb
n1 = 25	n2 = 20

High Mountain’s production managers are willing to assume that the population of tensile strength for each plant is approximately normally distributed with equal variances. Based on the sample results, can High Mountain’s managers conclude that there is a difference between the mean tensile strengths of ropes produced in Bozeman and Challis? Conduct the appropriate hypothesis test at the 0.05 level of significance.

a.       Specify the population parameter(s) of interest (proportion or mean):

b.       Formulate the null (H0) and alternate (HA) hypotheses:

c.       Specify the level of significance (alpha, a, probability of a Type I error):

d.       What is/are the critical value(s)?

e.       What is/are the value(s) of the test statistic?

f.        Based on your graph, do you reject H0 or do not reject H0? Explain

g.       To validate your results, we’ll also check our p-value. What is the p-value?

h.       Based on your p-value, do you reject H0 or do not reject H0? Explain

i.         State your summary statement of the conclusion in non-technical terms.

Answer 1

For Bozeman :

x̅1 = 7200, s1 = 425, n1 = 25

For Challis :

x̅2 = 7087, s2 = 415, n2 = 20

a) population parameter(s) of interest : mean

b) Null and Alternative hypothesis:  

Ho : µ1 = µ2  

H1 : µ1 ≠ µ2  

c) level of significance = 0.05

d) df = n1+n2-2 =    43

Critical value :  

Two tailed critical value, t crit = T.INV.2T(0.05, 43) =    2.017

Reject Ho if t < -2.017 or if t > 2.017  

e) Pooled variance :  

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((25-1)*425² + (20-1)*415²) / (25+20-2) = 176913.3721

Test statistic:  

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (7200 - 7087) / √(176913.3721*(1/25 + 1/20)) = 0.8955

f) Decision:

Do not reject the null hypothesis

g) p-value = T.DIST.2T(ABS(0.8955), 43) = 0.3755

h) Decision:

p-value > α, Do not reject the null hypothesis

i) There is not enough evidence to conclude that there is a difference between the mean tensile strengths of ropes produced in Bozeman and Challis.