Question

Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This year, based on a sample of 50 business owners, a survey your team conducted indicated that 19 businesses were highly profitable.

Would you conclude that the proportion of businesses that are highly profitable increased or decreased from last year with a statistical significance?
Use a statistical hypothesis test to determine this, using a 1% significance level.

Answer 1

`z=\frac{0.38 -0.28}{\sqrt{(0.28(1-0.28))/(50)}}=1.575`

`p_v =2*P(z>1.575)=0.115`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.01` 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

Explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

`\hat p=(19)/(50)=0.38` estimated proportion of businesses were highly profitable

`p_o=0.28` 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:

Null hypothesis:
`p=0.28`

Alternative hypothesis:
`p \\eq 0.28`

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

`z=\frac{0.38 -0.28}{\sqrt{(0.28(1-0.28))/(50)}}=1.575`

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=0.01`. 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>1.575)=0.115`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.01` 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 businesses were highly profitable is not significantly different from 0.28 or 28%.