Question

A sample of 1500 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 51% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

Answer:

$z=\frac{0.53 -0.51}{\sqrt{\frac{0.51(1-0.51)}{1500}}}=1.55$  

Step-by-step explanation:

Data given and notation  

n=1500 represent the random sample taken

$\hat p=0.53$ estimated proportion of chips that do not fail in the first 1000 hours of use

$p_o=0.51$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.51 or 51%:  

Null hypothesis:$p=0.51$  

Alternative hypothesis:$p \neq 0.51$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info required we can replace in formula (1) like this:  

$z=\frac{0.53 -0.51}{\sqrt{\frac{0.51(1-0.51)}{1500}}}=1.55$  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed on this case is $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(Z>1.55)=0.121$  

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that do not fail in the first 1000 hours of use is not significatnly different from 0.51 .