Question

In a recent survey, 10 percent of the participants rated Pepsi as being "concerned with my health." PepsiCo's response included a new "Smart Spot" symbol on its products that meet certain nutrition criteria, to help consumers who seek more healthful eating options. Suppose a follow-up survey shows that 29 of 200 persons now rate Pepsi as being "concerned with my health." What is the p-value for the test that would show whether or not there is sufficient evidence that the percentage has increased?

Answer 1

`z=\frac{0.145 -0.1}{\sqrt{(0.1(1-0.1))/(200)}}=2.121`

`p_v =P(z>2.121)=0.0170`

Explanation:

Data given and notation

n=200 represent the random sample taken

X=29 represent the persons who now rate Pepsi as being "concerned with my health

`\hat p=(29)/(200)=0.145` estimated proportion of persons who now rate Pepsi as being "concerned with my health

`p_o=0.10` is the value that we want to test

`\alpha` represent the significance level

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 true percentage is higher than 10% or 0.1.:

Null hypothesis:
`p\leq 0.1`

Alternative hypothesis:
`p > 0.1`

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{0.145 -0.1}{\sqrt{(0.1(1-0.1))/(200)}}=2.121`

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 next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(z>2.121)=0.0170`