200k views
3 votes
A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and receive discounts on certain purchases. A typical Saturday morning shopper who does not participate in this program spends $120 on her or his order. In a sample of 80 shoppers participating in the loyalty program, each shopper spent $130 on average during a recent Saturday, with standard deviation $40. Is this statistical proof that the shoppers participating in the loyalty program spent more on average than typical shoppers?

User King Holly
by
5.1k points

1 Answer

3 votes

Answer:


t=(130-120)/((40)/(√(80)))=2.236

Now we can find the degrees of freedom:


df=n-1=80-1=79

Now we can calculate the p value with the alternative hypothesis using the following probability:


p_v =P(t_((79))>2.236)=0.0141

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

Explanation:

Information given


\bar X=130 represent the sample mean


s=40 represent the sample standard deviation


n=80 sample size


\mu_o =120 represent the value to verify

t would represent the statistic


p_v represent the p value

System of hypothesis

We want to verify if shoppers participating in the loyalty program spent more on average than typical shoppers (120) , the system of hypothesis would be:

Null hypothesis:
\mu \leq 120

Alternative hypothesis:
\mu > 120

The statistic for this case is given by:


t=(\bar X-\mu_o)/((s)/(√(n))) (1)

Replacing the info given we got:


t=(130-120)/((40)/(√(80)))=2.236

Now we can find the degrees of freedom:


df=n-1=80-1=79

Now we can calculate the p value with the alternative hypothesis using the following probability:


p_v =P(t_((79))>2.236)=0.0141

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

User Puradox
by
5.3k points