Question

Macarthurs, a manufacturer of ropes used in abseiling, wished to determine whether changing the fibre used in the production of the ropes had affected their average breaking strength. It was known that ropes manufactured using the old fibre had an average breaking strength of 228.5 kilograms and a standard deviation of 27.3 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 0.05.

1. State the direction of the alternative hypothesis for the test e.g. gt (greater than), lt (less than), or ne (not equal to)

2. State, in absolute terms, the critical value

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

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

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

Answer 1

The manufacturer wishes to determine whether changing the fibre used in the production of the ropes had "affected"  their average breaking strength. This means that they want to know if thebreaking strength has changed (higher or lower ).

If $\mu=228.5 \text{ kilograms}$ is the average breaking strength and $\sigma=27.3 \text{ kilograms}$ is the standard deviation of breaking strength

we want to test the hypotheses

$\begin{align*} &H_0: \mu = 228.5\leftarrow\text{null hypothesis}\\ &H_a:\mu \ne 228.5\leftarrow\text{alternative hypothesis}\\ \end{align*}$

1) The direction of the alternative hypothesis for the test is ne (not equal to)

2. The probability of type 1 error is $\alpha=0.05$ . Since the sample size n=40 is greater than 30, we will use a large sample analysis. That means the sample mean is normally distributed and the test statistics is z

Since this is a 2 tail test, this indicates the total area under both the tails. That is the acceptance region is if the test statistics Z is between $-z_{\alpha/2}<Z<z_{\alpha/2}$

That means area under each tail is half of 0.05.

The area under the right tail is $P(Z>z_{\alpha/2})=\alpha/2=0.025$

The above can be written as

$\begin{align*} &P(Z>z_{\alpha/2})=1-P(Z<z_{\alpha/2})=0.025\\ \implies&P(Z<z_{\alpha/2})=1-0.025=0.975 \end{align*}$

Using the standard normal tables we can get that for $z_{\alpha/2}=1.96$ we can get $P(Z>z_{\alpha/2})=0.025$

Similarly for the left tail we need $P(Z<-z_{\alpha/2})=0.025$ . Due to the symmetry of normal distribution, the left tail critical value is -1.96

This means that the abolute values of the critical value is 1.96

3) the lower boundary of the region of non-rejection is

$\mu-z_{\alpha}\sigma=228.5-1.96\times 27.3 =174.99\text{ kilograms}$

4) the upper boundary of the region of non-rejection is

$\mu+z_{\alpha}\sigma=228.5+1.96\times 27.3 =282.01 \text{ kilograms}$

That means the region of non-rejection is

$174.99<\bar{X}<282.01$

5) the sample average breaking strength is $\bar{x}=234.9$ . This value falls in the region of non rejection calculated above.

Hence we cannot reject the null hypothesis.

is the null hypothesis rejected for this test?

ans: No

6)  if the null hypothesis was rejected, it implies that the sample evidence supports the calim at 0.05 level of significane that he new fibre has affected the breaking strength of the rope.

Hence 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?

Ans: Yes