Question

The Manager at Rainbow Valley II advertises that the typical family visiting the park spends at least one hour in the park during weekends. A sample of 25 visitors during the weekends in the month of July revealed that the mean time spent in the Park was 63 minutes with a standard deviation of 8 minutes.

1. Using the 0.01 significance level and a one-tailed test, is it reasonable to conclude that the mean time in the Park is greater than 60 minutes? Show all steps in your test of hypothesis.

2. Repeat the analysis at the 0.05 significance level. (You may show only the calculations that change).

3. Repeat with a survey mean of 64 minutes at the .01 significance level. (You may show only the calculations that change).

4. What do you conclude from your analysis?

Answer 1

Ho :   µ =   60                  
Ha :   µ >   60       (Right tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s =    8.0000                  
Sample Size ,   n =    25                  
Sample Mean,    x̅ =   63.0000                  
                          
degree of freedom=   DF=n-1=   24                  
                          
Standard Error , SE = s/√n =   8.0000   / √    25   =   1.6000      
t-test statistic= (x̅ - µ )/SE = (   63.000   -   60   ) /    1.6000   =   1.88
                            
p-Value   =   0.0365   [Excel formula =t.dist(t-stat,df) ]             
Decision:   p-value>α, Do not reject null hypothesis

it is not reasonable to conclude that the mean time in the Park is greater than 60 minutes

....

for 0.05

p value < 0.05, reject Ho

it is reasonable to conclude that the mean time in the Park is greater than 60 minutes

........

for survay mean 64 and at 0.01

Sample Mean,    x̅ =   64.0000                  
                          
degree of freedom=   DF=n-1=   24                  
                          
Standard Error , SE = s/√n =   8.0000   / √    25   =   1.6000      
t-test statistic= (x̅ - µ )/SE = (   64.000   -   60   ) /    1.6000   =   2.50   
                          
p-Value   =   0.0098   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
it is reasonable to conclude that the mean time in the Park is greater than 60 minutes

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