Question

You interview a random sample of 50 adults. the results of the survey show that 46​% of the adults said they were more likely to buy a product when there are free samples. at alphaequals0.05​, can you reject the claim that at least 51​% of the adults are more likely to buy a product when there are free​ samples?

Answer 1

Answer:

$z=\frac{0.46 -0.51}{\sqrt{\frac{0.51(1-0.51)}{50}}}=-0.707$  

$p_v =P(z<-0.707)=0.240$  

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 adults who said they were more likely to buy a product when there are free samples is higher or equal than 0.51

Explanation:

Data given and notation

n=50 represent the random sample taken

$\hat p==0.46$ estimated proportion of adults who said they were more likely to buy a product when there are free samples

$p_o=0.51$ 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is at least 0.51 or no.:  

Null hypothesis:$p \geq 0.51$  

Alternative hypothesis:$p < 0.51$  

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:  

$z=\frac{0.46 -0.51}{\sqrt{\frac{0.51(1-0.51)}{50}}}=-0.707$  

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 left tailed the p value would be:  

$p_v =P(z<-0.707)=0.240$  

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 adults who said they were more likely to buy a product when there are free samples is higher or equal than 0.51