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 228.5 kilograms and a standard deviation of 27.3 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 234.9 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. Blank 6

Answer 1

Solution

Back-up Theory

Type I error probability of 0.01=> significance level, α = 0.01

Now, to work out the solution,

Q1

The direction of the alternative hypothesis for the test: lt (less than) Answer 1

Q2

Given Type I error probability of 0.01=> significance level, α = 0.01

So, the critical value = lower 1% point of N(0, 1) = - 2.326 [from Tables] Answer 2

[Since population standard deviation is known (to be 27.3), the test statistic follows a Normal distribution.]

Q3

The lower boundary of the region of non-rejection in terms of the sample mean used in testing the claim (to two decimal places) = 218.46 Answer 3

[Region of non-rejection is: Z ≥ - 2.326, where Z = (√n)(Xbar - µ₀)/σ is the test statistic and – 2.326 is the critical value as obtained in Answer 2.

i.e., (√40)(Xbar – 228.5)/27.3 ≥ - 2.326

Or, Xbar ≥ 218.46]    

Q4

The upper boundary of the region of non-rejection in terms of the sample mean used in testing the claim: gt Answer 4

Q5

Since the average breaking strength found from the sample is 234.9 kilograms and it is greater than the critical value (218.46 vide Answer 3) the null hypothesis cannot be rejected. No. Answer 5

DONE