Question

In a random sample of 400 items where 84 were found to be​ defective, the null hypothesis that 20​% of the items in the population are defective produced Upper Z Subscript STATequalsplus 0.50. Suppose someone is testing the null hypothesis Upper H 0​: piequals0.20 against the​ two-tail alternative hypothesis Upper H 1​: pinot equals0.20 and they choose the level of significance alphaequals0.10. What is their statistical​ decision?

Answer 1

Answer:

$z=\frac{0.21 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=0.5$  

$p_v =2*P(Z>0.5)=0.617$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the true proportion of defectives it's not significant different from 0.2.  

Step-by-step explanation:

1) Data given and notation

n=400 represent the random sample taken

X=84 represent the number of items defective

$\hat p=\frac{84}{400}=0.21$ estimated proportion of defectives

$p_o=0.2$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.2 or 20%:  

Null hypothesis:$p=0.2$  

Alternative hypothesis:$p \neq 0.2$  

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.21 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=0.5$  

4) 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 provided $\alpha=0.1$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(Z>0.5)=0.617$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the true proportion of defectives it's not significant different from 0.2.