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.

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

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

[tex]\hat p=\frac{19}{50}=0.38[/tex] estimated proportion of businesses were highly profitable

[tex]p_o=0.28[/tex] is the value that we want to test

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

Confidence=95% or 0.95

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 of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:

Null hypothesis:[tex]p=0.28[/tex]

Alternative hypothesis:[tex]p \neq 0.28[/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 required we can replace in formula (1) like this:

[tex]z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575[/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 provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

[tex]p_v =2*P(z>1.575)=0.115[/tex]

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