Question

Macarthurs, a manufacturer of ropes used in abseiling, wished to determine if the production of their ropes was performing according to their specifications. All ropes being manufactured were required to have an average breaking strength of 237.0 kilograms and a standard deviation of 19.8 kilograms. They planned to test the breaking strength of their ropes using a random sample of forty ropes and were prepared to accept a Type I error probability of 0.01.

1. State the direction of the alternative hypothesis for the test. Type gt (greater than), ge (greater than or equal to), lt (less than), le (less than or equal to) or ne (not equal to) as appropriate in the box.

2. State, in absolute terms, the critical value as found in the tables in the textbook.

3. Determine the lower boundary of the region of non-rejection in terms of the sample mean used in testing the claim (to two decimal places). If there is no (theoretical) lower boundary, type lt in the box.

4. Determine the upper boundary of the region of non-rejection in terms of the sample mean used in testing the claim (to two decimal places). If there is no (theoretical) upper boundary, type gt in the box.

5. If the average breaking strength found from the sample is 242.7 kilograms, is the null hypothesis rejected for this test? Type yes or no.

6. Disregarding your answer for 5, if the null hypothesis was rejected, would it appear that the new fibre has affected the breaking strength of the rope at the 5% level of significance? Type yes or no.

Answer 1

1)

Ho :   µ =   237
Ha :   µ ╪   237

2) α=0.01

critical z value, z* =      2.576

3)

hypothesis mean,   µo =    237              
significance level,   α =    0.01              
sample size,   n =   40              
std dev,   σ =    19.8000              
                      
δ=   µ - µo =    -237              
                      
std error of mean,   σx = σ/√n =    19.8000   / √    40   =   3.13065

Zα/2   = ±   2.576   (two tailed test)                      

Zα/2   = ±   2.576   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -2.576   and   2.576
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-2.576   ≤(x̄ - µo)/σx≤   2.576                          
228.936   ≤ x̄ ≤   245.064                          

lower boundary of the region of non-rejection in terms of the sample mean is x̄≥ 228.94

4)

Upper boundary of the region of non-rejection in terms of the sample mean is x̄ ≤ 245.064

5)

Ho :   µ =   237                  
Ha :   µ ╪   237       (Two tail test)          
                          
Level of Significance ,    α =    0.010                  
population std dev ,    σ =    19.8000                  
Sample Size ,   n =    40                  
Sample Mean,    x̅ =   242.7000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   19.8000   / √    40   =   3.1307      
Z-test statistic= (x̅ - µ )/SE = (   242.700   -   237   ) /    3.1307   =   1.82
                          
critical z value, z* =   ±   2.576   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
Decision: z stat <2.576, Do not reject null hypothesis                       

answer: NO

6)

p-Value   =   0.0687
Decision:   p-value>α=0.05, Do not reject null hypothesis