A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The owner of the store has determined that home delivery will be successful if the average time spent on the deliveries does not exceed 36 minutes. The owner has randomly selected 20 customers and has delivered pizzas to their homes in order to test if the mean delivery time actually exceeds 36 minutes. Suppose the P-value for the test was found to be 0.0275. State the correct conclusion.


A. At α=0.05, we fail to reject H0.
B. At α=0.02, we reject H0.
C.At α=0.03, we fail to reject H0.
D. At α=0.025, we fail to reject H0.

Respuesta :

Answer:

d) At α=0.025, we fail to reject H0.

True. Since the p value is higher than the significance level 0.0275>0.025 we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation    

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

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

[tex]n=20[/tex] sample size    

[tex]\mu_o =36[/tex] represent the value that we want to test  

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

t would represent the statistic (variable of interest)    

[tex]p_v=0.0275[/tex] represent the p value for the test  

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is higher than 36 minutes :    

Null hypothesis:[tex]\mu \leq 36[/tex]    

Alternative hypothesis:[tex]\mu > 36[/tex]    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic    

We don't have info provided to calculate it but we can assume that is [tex]t_0[/tex]    

4) P-value    

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since is a right tailed test the p value would be:    

[tex]p_v =P(t_{(19)}>t_0)=0.0275[/tex]    

5) Conclusion    

a) At α=0.05, we fail to reject H0.

False. On this case since the p value is lower than the significance level 0.0275<0.05 we have enough evidence to reject the null hypothesis not the opposite.

b) At α=0.02, we reject H0.

False. Since the p value is higher than the significance level because 0.0275>0.02 we FAIL to reject the null hypothesis not the opposite.

c)  At α=0.03, we fail to reject H0.

False, since the p value is lower than the significance level because 0.0275<0.03 we have enough evidence to reject the null hypothesis not the opposite.

d) At α=0.025, we fail to reject H0.

True. Since the p value is higher than the significance level 0.0275>0.025 we have enough evidence to FAIL to reject the null hypothesis.