Macarthurs, a manufacturer of ropes used in abseiling, wished to determine wheth

Question

Macarthurs, a manufacturer of ropes used in abseiling, wished to determine whether changing the fibre used in the production of the ropes had increased their average breaking strength. It was known that ropes manufactured using the old fibre had an average breaking strength of 232.0 kilograms and a standard deviation of 22.7 kilograms. They planned to test the breaking strength of the new ropes using a random sample of forty ropes and also indicated they 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 240.7 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 improved the breaking strength of the rope at the 1% level of significance? Type yes or no.________

Explanation / Answer

Null Hyp: The average breaking strength of the fibre is 232.0 kilograms

H0: mu = 232

Alt Hyp: The average breaking strength of the fibre is greater than 232.0 kilograms

Ha: mu > 232

1) Ans ge

2) alpha = 0.01

Absolute Critical value for z at 0.01 sig level for a one tailed test is 2.33

3) Lower boundary for non rejection = -infinity

4) Upper boundary = +2.33

5) sampe mean = 240.7

sample size n = 40

Test stat Z = (240.7 - 232)/(22.7/sqrt(40)) = 2.42

SInce test stat > critical vlaue it lies in the rejection region

Ans) Yes

6) Yes

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