Question 1171439
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