Question

A newspaper poll asked respondents if they trusted "eco friendly" labels on cleaning products. Out of 1000 adults surveyed, 498 responded "yes." We would like to test if the proportion of respondents that trust these labels is at least 50%. The calculated test statistic value is

Answer 1

Answer:

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ 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 respondents that trust these labels is not significanlty less than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=498 represent the adults that trust these labels

$\hat p=\frac{498}{1000}=0.498$ estimated proportion of respondents that trust these labels

$p_o=0.5$ 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of respondents that trust these labels is at least 50%:  

Null hypothesis:$p\geq 0.5$  

Alternative hypothesis:$p < 0.5$  

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.498 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-0.126$  

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 is not provided but we can assume it as $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z$  

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ 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 respondents that trust these labels is not significanlty less than 0.5 or 50%.