Question

An organization published an article stating that in any one-year period, approximately 9.8 percent of adults in a country suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, eight of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult population in the country.

Answer 1

Answer:

$z=\frac{0.08 -0.098}{\sqrt{\frac{0.098(1-0.098)}{100}}}=-0.605$  

$p_v =P(z$  

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 of people with depression or a depressive illness is not significantly lower than 0.098 (9.8%)

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=8 represent the people with depression or a depressive illness

$\hat p=\frac{8}{100}=0.08$ estimated proportion of people with depression or a depressive illness

$p_o=0.098$ is the value that we want to test

$\alpha=0.05$ represent the significance level  assumed

Confidence=95% or 0.95  assumed

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 sof people with depression or a depressive illness is lower than the general adult population value of 9.8% or 0.098.:  

Null hypothesis:$p \geq 0.098$  

Alternative hypothesis:$p < 0.098$  

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 requires we can replace in formula (1) like this:  

$z=\frac{0.08 -0.098}{\sqrt{\frac{0.098(1-0.098)}{100}}}=-0.605$  

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

Since is a left tailed test the p value would be:  

$p_v =P(z$  

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 of people with depression or a depressive illness is not significantly lower than 0.098 (9.8%)