Question 1185853: A company manufacture ropes. From a large number of tests over a long period of time, they have found a mean breaking strength of 300 lbs and a standard deviation of 24 lbs. Assume that these values are MU and SIGMA. It is believed that by a newly developed process, the mean breaking strength can be increased.
(a) Design a decision rule for rejecting the old process with an ALPHA error of 0.01 if it is agreed to test 64 ropes.
(b) Using the decision rule adopted in (1.3.1), what is the probability of accepting the old process when in fact the new process has increased the mean breaking strength to 310 lbs.? Assume SIGMA is still 24 lbs. Use a diagram to illustrate what you have done, (draw the reference distribution)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to approach this hypothesis testing problem:
**(a) Designing the Decision Rule:**
1. **Hypotheses:**
* Null Hypothesis (H0): The mean breaking strength is still 300 lbs (μ = 300).
* Alternative Hypothesis (H1): The mean breaking strength has increased (μ > 300). This is a one-tailed test.
2. **Significance Level (alpha):** α = 0.01. This is the probability of rejecting the null hypothesis when it is actually true (Type I error).
3. **Test Statistic:** Since we know the population standard deviation (σ), we use a z-test. The test statistic is:
```
z = (x̄ - μ) / (σ / √n)
```
Where:
* x̄ is the sample mean
* μ is the population mean under H0 (300 lbs)
* σ is the population standard deviation (24 lbs)
* n is the sample size (64)
4. **Critical Value:** For a one-tailed test with α = 0.01, we look up the z-score that corresponds to 0.99 in the standard normal distribution table (or use a calculator). The critical value is approximately z = 2.33.
5. **Decision Rule:** Reject H0 if the calculated z-score is greater than 2.33. In other words, if the sample mean breaking strength is high enough to produce a z-score greater than 2.33, we will reject the old process. We can also express the decision rule in terms of the sample mean:
```
x̄ > μ + z * (σ / √n)
x̄ > 300 + 2.33 * (24 / √64)
x̄ > 307
```
So, reject H0 if the sample mean breaking strength is greater than 307 lbs.
**(b) Probability of Accepting the Old Process When the New Process Has a Mean of 310 lbs (Beta - Type II Error):**
1. **New Mean (μ_new):** 310 lbs.
2. **Calculating Beta:** Beta is the probability of failing to reject the null hypothesis when it is false. In this case, it's the probability of accepting the old process when the new process has a mean of 310 lbs.
* First, we need to find the sample mean value (x̄) that corresponds to our critical z-score of 2.33. We already did this above: x̄ = 307 lbs. This is our "acceptance region boundary".
* Now, we calculate the z-score for this x̄ value, but using the *new* mean (310 lbs):
```
z_new = (x̄ - μ_new) / (σ / √n)
z_new = (307 - 310) / (24 / √64)
z_new = -1
```
* We want to find the probability that the sample mean falls *below* 307 lbs given the new mean is 310 lbs. This corresponds to the area to the *left* of z = -1 on the standard normal distribution.
* Using a z-table or calculator, we find the probability corresponding to z = -1 is approximately 0.1587.
3. **Beta:** β ≈ 0.1587. There's approximately a 15.87% chance of accepting the old process when the new process has a mean breaking strength of 310 lbs.
**Diagram:**
[Unfortunately, I can't directly draw diagrams here. However, I can describe what it should look like.]
1. **Reference Distribution (under H0):** Draw a normal distribution curve centered at 300. Mark the critical value (307) on the right tail. The area to the right of 307 represents the rejection region (alpha = 0.01).
2. **Distribution under H1:** Draw another normal distribution curve centered at 310 (the new mean). The area to the *left* of 307 under this curve represents beta (≈ 0.1587). This area is the probability of failing to reject H0 when H1 is true. It will overlap with the right tail of the first curve.
The diagram visually shows the overlap between the two distributions and how beta represents the probability of a Type II error.
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