Question

1) The speed limit on College Terrace is 25 mph. Even after traffic-calming measures, a resident complains that cars still speed. 250 randomly selected cars were clocked with mean speed 25.55 mph and a standard deviation of 3.618. Is the mean speed of all cars greater than 25 mph? Use a significance level of a = 0.05. a) Write the hypotheses. (2 points) b) Perform the appropriate hypothesis test using a significance level of a = 0.05, report the test statistic and p-value, and write a full conclusion. Be sure to write out what you entered in your calculator. (5 points) c) Are the results statistically significant? Why or why not? (2 points) d) Find the appropriate 90% confidence interval. Be sure to write out what you entered in your calculator. (3 points) e) Are the results practically significant? Why or why not? (2 points) 2) New road signs are made with the intention of improving visibility for drivers. Highway safety engineers setup a test course that included both the old and new signs. Volunteers drove the course and rated the old and new signs in terms of visibility. (2 points each) a) Write the null and alternative hypotheses in words using "improved visibility" and "not improved visibility": b) Describe a Type I error in the context of the problem. c) Describe a Type Il error in the context of the problem.

Answer 1

1)

a)

Ho :   µ =   25
Ha :   µ >   25

b)

Level of Significance ,    α =    0.05                  
sample std dev ,    s =    3.6180                  
Sample Size ,   n =    250                  
Sample Mean,    x̅ =   25.5500                  
                          
degree of freedom=   DF=n-1=   249                  
                          
Standard Error , SE = s/√n =   3.6180   / √    250   =   0.2288      
t-test statistic= (x̅ - µ )/SE = (   25.550   -   25   ) /    0.2288   =   2.404
                          
  
p-Value   =   0.0085 [Excel formula =t.dist.rt(t-stat,df) ]              

c)

p-value<α, Reject null hypothesis   
so, test is statiscally significant

d)

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   249          
't value='   tα/2=   1.6510   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.6180   / √   250   =   0.2288
margin of error , E=t*SE =   1.6510   *   0.2288   =   0.3778
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    25.55   -   0.377785   =   25.1722
Interval Upper Limit = x̅ + E =    25.55   -   0.377785   =   25.9278
90%   confidence interval is (   25.17   < µ <   25.93   )

e)

yes