Hippocrates magazine states that 37 percent of all Americans take multiple vitamins regularly. Suppose a researcher surveyed 750 people to test this claim and found that 266 did regularly take a multiple vitamin. Is this sufficient evidence to conclude that the actual percentage is different from 37% at the 5% significance level? Select the [p-value, Decision to Reject (RH0) or Failure to Reject (FRH0)].

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 that regularly take a multiple vitamin it's not significantly different from 0.37 or 37% .

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=266 represent the people that regularly take a multiple vitamin

[tex]\hat p=\frac{266}{750}=0.355[/tex] estimated proportion of people that regularly take a multiple vitamin

[tex]p_o=0.37[/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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the actual percentage is different from 37%.:

Null hypothesis:[tex]p=0.37[/tex]

Alternative hypothesis:[tex]p \neq 0.37[/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].

3) Calculate the statistic

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

[tex]z=\frac{0.355 -0.37}{\sqrt{\frac{0.37(1-0.37)}{750}}}=-0.85[/tex]

4) 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=0.05[/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<-0.85)=0.395[/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 that regularly take a multiple vitamin it's not significantly different from 0.37 or 37% .