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

`z=\frac{0.46 -0.51}{\sqrt{(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

Step-by-step 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{(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{(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