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.

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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

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

[tex]p_o=0.098[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level assumed

Confidence=95% or 0.95 assumed

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]p \geq 0.098[/tex]

Alternative hypothesis:[tex]p < 0.098[/tex]

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

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

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

Calculate the statistic

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

[tex]z=\frac{0.08 -0.098}{\sqrt{\frac{0.098(1-0.098)}{100}}}=-0.605[/tex]

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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.

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

[tex]p_v =P(z<-0.605)=0.273[/tex]

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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%)