Question

It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. In a random sample of 3,000 individuals, it was determined that 17 developed schizophrenia. Is there evidence to support the claim that the true proportion of people who will develop schizophrenia is different from 0.004 at the ? = 0.05 level of significance?Step 1:Step 2:Step 3:Test statistic Z0P-valueStep 4:Step 5:

Answer 1

Answer:

Null hypothesis:$p=0.004$  

Alternative hypothesis:$p \neq 0.004$  

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$  

$p_v =2*P(z>1.149)=0.2506$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.  

Step-by-step explanation:

1) Data given and notation

n=3000 represent the random sample taken

X=17 represent the individuals developed schizophrenia in the sample

$\hat p=\frac{17}{3000}=0.00567$ estimated proportion of individuals developed schizophrenia in the sample

$p_o=0.004$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 true proportion of people who will develop schizophrenia is different from 0.004:  

Null hypothesis:$p=0.004$  

Alternative hypothesis:$p \neq 0.004$  

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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$  

4) P value

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 $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z>1.149)=0.2506$

5) Decision

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.