Question 1171439: "Untitled," by Stephen Chen
"I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes it difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.
So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?"
Conduct a hypothesis test at the 5% level.
Note: If we are using a Student's t-distribution for the problem, we may assume that the underlying population is normally distributed. (In general, we must first prove that assumption, though.)"
I need help stating the distribution to use for the test (I have to round it to four decimal places)
P' ~ N (? , ?)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this hypothesis test step-by-step.
**1. Define the Problem**
* **Null Hypothesis (H0):** The new test program (Y) has the same error generation rate as the original program (X), which is 1% (0.01).
* **Alternative Hypothesis (H1):** The new test program (Y) has a higher error generation rate than the original program (X).
* **Significance Level (α):** 5% or 0.05.
* **Sample Data:**
* Sample size (n) = 50
* Number of errors (x) = 2
* Sample proportion (p') = x/n = 2/50 = 0.04
**2. Choose the Appropriate Test**
Since we're dealing with proportions and a relatively small sample size, we'll use a one-proportion z-test. However, because the sample size is relatively small, and the proportion is also small, we need to check if the normal approximation will hold.
* n * p = 50 * 0.01 = 0.5
* n * (1 - p) = 50 * 0.99 = 49.5
Since n*p < 10, the normal approximation to the binomial distribution is not ideal. However, because this is for a software company, and it is a test run, we will proceed with the Z test.
**3. Calculate the Test Statistic (Z-score)**
The formula for the z-score in a one-proportion z-test is:
* Z = (p' - p) / sqrt(p(1 - p) / n)
Where:
* p' = sample proportion (0.04)
* p = population proportion (0.01)
* n = sample size (50)
Let's plug in the values:
* Z = (0.04 - 0.01) / sqrt(0.01(1 - 0.01) / 50)
* Z = 0.03 / sqrt(0.01(0.99) / 50)
* Z = 0.03 / sqrt(0.0099 / 50)
* Z = 0.03 / sqrt(0.000198)
* Z = 0.03 / 0.01407
* Z ≈ 2.1322
**4. Determine the P-value**
Since this is a right-tailed test (we're looking for a higher error rate), we need to find the probability P(Z > 2.1322).
Using a standard normal distribution table or a calculator, we find:
* P(Z > 2.1322) ≈ 0.0165
**5. Make a Decision**
* Compare the P-value (0.0165) to the significance level (0.05).
* Since 0.0165 < 0.05, we reject the null hypothesis.
**6. Conclusion**
There is sufficient evidence to conclude that the new test program (Y) has a significantly higher error generation rate than the original program (X) at the 5% significance level.
**7. Stating the Distribution**
* P' ~ N (0.01 , 0.000198)
* The mean is the original population proportion.
* The variance is p(1-p)/n.
* sqrt(p(1-p)/n) = sqrt(.01*.99/50) = sqrt(.000198) = 0.01407.
* 0.01407^2 = 0.000198
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