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), It (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 It 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) The manufacturer of the ropes would be interested in testing whether the strength of the rope is appropriate given the standards. For this, he would focus only on ropes that have a lower breaking strength and not the higher breaking strength ones.

Thus, the alternative hypothesis for this particular situation is:

H₁ : The average breaking strength of the ropes is less than 237kg or .

2) The critical value for this particular case will follow normal distribution.

Hence,

3) The lower boundary of the region of non-rejection terms of the sample mean used in testing the claim is:

which can be solved as:

which can be solved to find the lower bound as:

kg

where n is 40 and sigma is 19.8kg

4) Since we have established that the following test is a one sided test there is no need for an upper boundary of the region of the non-rejection of the sample mean. This is because, the manufacturer has no problem if the breaking strength is high, it is the weakness of the breaking strength or the lower bound of the sample mean that is of the concern.