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"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 them 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 programX 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?

Step by step solution

01

Given information

Stephen Chen runs the test program $50$times and generated the error twice. The error generated is$1\%$.

02

Explanation

State the hypothesis:

The null hypothesis states that the error generated by the test program is $1\%$and the alternate hypothesis states that the error generated by the test program is more than $1\%$.

$$\begin{gathered} H_0:p=0.01 \\ {H}_0:p>0.01 \end{gathered}$$

The normal distribution is:

$N\left(0.01,\sqrt{\frac{\left(0.01\right)\left(0.99\right)}{50}}\right)$

The Z test statistic is calculated as:

$z=\frac{p^{鈭�}-p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}}$

Where n is the sample size of $50$programs,

$$\begin{gathered} p^{鈭�}=\frac{x}{n} \\ =\frac{2}{50} \\ =0.04 \end{gathered}$$

Substitute the $p鈭�$value in the Z test statistic formula.

$$\begin{gathered} z=\frac{p^{鈭�}-p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{0.04-0.01}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{0.03}{0.0141} \\ =2.132 \end{gathered}$$

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