A local pizza store knows the mean amount of time it takes them to deliver an order is 45 minutes after the order is placed. The manager has a new system for processing delivery orders, and they want to test if it changes the mean delivery time. They take a sample of delivery orders and find their mean delivery time is 48 minutes and the standard deviation is 10 minutes. Significance level (a=0.05)
Is there any difference of the pizzas delivery time between the sample average and company's claim?

Respuesta :

Answer:

We need to conduct a hypothesis in order to check if the true mean is different from 45(alternative hypothesis), and then the system of hypothesis would be:  

Null hypothesis:[tex]\mu =45[/tex]  

Alternative hypothesis:[tex]\mu \neq 45[/tex]  

Since we don't know the sample size we can't calculate the statistic and the only thing that we can do is create the sytem of hypothesis.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=48[/tex] represent the sample mean    

[tex]s=10[/tex] represent the sample standard deviation

[tex]n[/tex] sample size  

[tex]\mu_o =45[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is different from 45(alternative hypothesis), and then the system of hypothesis would be:  

Null hypothesis:[tex]\mu =45[/tex]  

Alternative hypothesis:[tex]\mu \neq 45[/tex]  

Since we don't know the sample size we can't calculate the statistic and the only thing that we can do is create the sytem of hypothesis.

Answer:

Yes, because 0.023<0.05

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