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. At α = .05, would a follow-up survey showing that 18 of 100 persons now rate Pepsi as being "concerned with my health" prove that the percentage has increased?

Answer 1

Answer:

$z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67$  

$p_v =P(z>2.67)=0.0038$  

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=18 represent the adults who rate Pepsi as being "concerned with my health"

$\hat p=\frac{18}{100}=0.18$ estimated proportion of adults who rate Pepsi as being "concerned with my health"

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

$\alpha=0.05$ 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 the true proportion is higher than 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

$z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67$  

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:  

$p_v =P(z>2.67)=0.0038$  

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1