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

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

`p_v =P(z<-0.605)=0.273`

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%)

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=(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{(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{(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<-0.605)=0.273`

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%)