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

`z=\frac{0.21 -0.2}{\sqrt{(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.

Explanation:

1) Data given and notation

n=400 represent the random sample taken

X=84 represent the number of items defective

`\hat p=(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 \\eq 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{(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{(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.