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?

So the p value obtained was a very high value and using the significance level given we have 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

estimated proportion of adults who said they were more likely to buy a product when there are free samples

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic  

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

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

Since is left tailed the p value would be:  

So the p value obtained was a very high value and using the significance level given we have 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

lbc corporation makes and sells a product called product wz. each unit of product wz requires 4.1 hours of direct labor at the rate of $20.00 per direct labor-hour. management would like you to prepare a direct labor budget for june. the company plans to sell 28,000 units of product wz in june. the finished goods inventories o

credit history is used by lenders to determine whether you are a 'risky' person to give their money to.

having a short history provide very little data for the lenders. they wouldn't be able to determine your credit behavior from such a small samples. even if they are perfect.

having long credit history with few blemishes provide larger and more accurate data for the lenders to predict your credit behavior.