Question

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)].

Answer 1

`p_v =2*P(z<-0.85)=0.395`

Failure to Reject (FRH0)

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

Explanation:

1) Data given and notation

n=750 represent the random sample taken

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

`\hat p=(266)/(750)=0.355` estimated proportion of people that regularly take a multiple vitamin

`p_o=0.37` 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)

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:
`p=0.37`

Alternative hypothesis:
`p \\eq 0.37`

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`.

3) Calculate the statistic

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

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

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
`\alpha=0.05`. 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<-0.85)=0.395`

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