Question

.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?a.State the null and the alternative hypotheses.b.Find the p-value of the test.

Answer 1

Answer:

a) Null hypothesis:$\mu \leq 120$  

Alternative hypothesis:$\mu > 120$  

b) $t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$  

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$  

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean for the amount spent each shopper

$s=40$ 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 f

Part a

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

Null hypothesis:$\mu \leq 120$  

Alternative hypothesis:$\mu > 120$  

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$  

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$