Question

1.A manufacturer of mountaineering equipment produces traditional​ three-strand climbing rope on two separate production​ lines, line 1 and line 2. The manufacturer regularly tests the tensile strength of its ropes by randomly selecting ropes from production and subjecting them to various tests. The results from the most recent random sample of ropes are shown below.

Assuming the population of tensile strengths for each line is approximately normally distributed with equal​ variances, can the manufacturer conclude there is a difference between the mean tensile strengths of ropes produced on the two​ lines? Conduct the appropriate hypothesis test at the 0.01 level of significance.

Line 1 Line 2

x overbar x1=7,248 lb x overbar x2=7,725 lb

s1=405 s2=430

n1=25 n2=20

What are the appropriate hypotheses to​ test?

A. H0​: μ1−μ2≥0

HA​: μ1−μ2<0

B. H0​: μ1−μ2≤0

HA​: μ1−μ2>0

C. H0​: μ1−μ2<0

HA​:μ1−μ2≥0

D. H0​:μ1−μ2>0

HA​: μ1−μ2≤0

E. H0​: μ1−μ2≠0

HA​: μ1−μ2=0

F. H0​: μ1−μ2=0

HA​: μ1−μ2≠0

Determine the rejection region for the test statistic t. Select the correct choice below and fill in the answer box to choose choice.

​(Round to four decimal places as​ needed.)

A.

t>?

B.

t<?

C.

t<? or t>?

Calculate the value of the test statistic.

t=?(Round to four decimal places as​ needed.)

Since the test statistic ▼ (choose is/ is not )in the rejection​ region, ▼(do not reject/reject)the null hypothesis. There is ▼(sufficient /insufficient) evidence to conclude that the two population means are different.

Answer 1

We are given the following information:

,   ,   ,   ,   ,  

a)

We try to test if there is a difference in the means. Thus, we basically test if the means are equal to each other or are different.

Thus the appropriate hypothesis would be:

Hence, option (F) is the correct answer.

b)

Next, we are asked to test it at 0.01 (or 1% ) level of significance.

Note that the sample mean is estimated in each of the two groups, which leads to the loss of 2 degrees of freedom.

Hence, the degrees of freedom = n1-1+n2-1 = 25+20-2 = 43

We need to check the critical value of t at 1% level and 43 degrees of freedom

This gives a t value +/- 2.6951

Hence,rejection region is given as:

t< -2.6951 and t>2.6951

Hence, option (C) is the correct option form.

c)

The general form of the test statistic is:

where

is the difference between sample means

is the difference between population means

is the standard error =

Now,

SE

=

=

Hence, test statistic :

Under, null hypothesis, the differences between population mean is 0

Hence,

or,

from the problem,

and , also = 125.722

Thus,

SInce the test statistic is in the rejection region, reject the null hypothesis. There is sufficient evidence to conclude that the two population means are different.