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.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.

Respuesta :

Answer:

a) Null hypothesis:[tex]\mu \leq 120[/tex]  

Alternative hypothesis:[tex]\mu > 120[/tex]  

b) [tex]t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236[/tex]  

The degrees of freedom are given by:

[tex] df = n-1 = 80-1=79[/tex]

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

[tex]p_v =P(t_{79}>2.236)=0.0141[/tex]  

Step-by-step explanation:

Information given

[tex]\bar X=130[/tex] represent the sample mean for the amount spent each shopper

[tex]s=40[/tex] represent the sample standard deviation

[tex]n=80[/tex] sample size  

[tex]\mu_o =120[/tex] represent the value to verify

t would represent the statistic    

[tex]p_v[/tex] 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:[tex]\mu \leq 120[/tex]  

Alternative hypothesis:[tex]\mu > 120[/tex]  

The statistic for this case would be given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

Replacing the info given we got:

[tex]t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236[/tex]  

The degrees of freedom are given by:

[tex] df = n-1 = 80-1=79[/tex]

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

[tex]p_v =P(t_{79}>2.236)=0.0141[/tex]  

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