Question

A test of breaking strengths of 6 ropes manufactured by a company showed a (sample) mean breaking strength of 7750 lbs and a (sample) standard deviation of 145 lbs, whereas the manufacturer claimed a true mean breaking strength of 8000 lbs. You think that the manufacturer is wrong and that the true mean is not 8000 lbs, but you are agnostic as to whether it is larger or smaller. Can we support the manufacturer’s claim at a level of significance equal to 0.05? (Be sure to write down your null and alternative hypothesis. If you do not, you will lose many points

Answer 1

To determine whether we can support the manufacturer's claim about the true mean breaking strength of the ropes, we need to conduct a hypothesis test at a significance level of 0.05. The null hypothesis states that the true mean breaking strength is 8000 lbs, while the alternative hypothesis states that the true mean breaking strength is not 8000 lbs. We can use the t-distribution to calculate the t-value and compare it to the critical t-value to make a decision.

Step-by-step explanation:

To determine whether we can support the manufacturer's claim about the true mean breaking strength of the ropes, we need to conduct a hypothesis test at a level of significance equal to 0.05.

The null hypothesis (H0) states that the true mean breaking strength is 8000 lbs, while the alternative hypothesis (Ha) states that the true mean breaking strength is not 8000 lbs.

To conduct the hypothesis test, we can use the t-distribution since we have a small sample size (n = 6) and the population standard deviation is unknown. With a significance level of 0.05 and 5 degrees of freedom (n-1), the critical t-value is approximately ±2.571. We can calculate the t-value using the formula:

t = (sample mean - population mean) / (sample standard deviation/√n)

After calculating the t-value, we compare it to the critical t-value to make a decision. If the calculated t-value falls within the range of ±2.571, we fail to reject the null hypothesis. If the calculated t-value is outside the range, we reject the null hypothesis in favor of the alternative hypothesis.

Answer 2

`t=(7750-8000)/((145)/(√(6)))=-4.22`

`p_v =2*P(t_((5))<-4.22)=0.0083`

If we compare the p value and the significance level for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean breaking strenghths is different from 8000 at 5% of significance.

Step-by-step explanation:

Data given and notation

`\bar X=7750` represent the mean breaking strength value for the sample

`s=145` represent the sample standard deviation

`n=6` sample size

`\mu_o =8000` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

Is a two tailed test.

What are H0 and Ha for this study?

We want to test is the true mean is equal to 8000 or not.

Null hypothesis:
`\mu = 8000`

Alternative hypothesis :
`\mu \\eq 800`

Compute the test statistic

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(7750-8000)/((145)/(√(6)))=-4.22`

Give the appropriate conclusion for the test

First we need to find the degrees of freedom given by:

`df=n-1=6-1=5`

Since is a two tailed test the p value would be:

`p_v =2*P(t_((5))<-4.22)=0.0083`

Conclusion

If we compare the p value and the significance level for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean breaking strenghths is different from 8000 at 5% of significance.