Question

A group of Brigham Young University—Idaho students (Matthew Herring, Nathan Spencer, Mark Walker, and Mark Steiner) collected data in November 2005 on the speed of vehicles traveling through a construction zone on a state highway, where the posted speed was 25 mph. The recorded speed (in mph) of 15 randomly selected vehicles is given below. Assume speeds are approximately normally distributed with σ=5.95 and that the average speed of drivers is about 5 mph over the limit, i.e. µ = 30.

20, 24, 27, 28, 29, 30, 32, 33, 33, 34, 36, 38, 39, 40, 40

1. Describe the distribution of a single vehicle’s speed traveling through a construction zone (the population distribution).

2. Describe the distribution of the mean speed of 15 vehicles traveling through a construction zone (the sampling distribution).

3. Did you need to apply the Central Limit theorem in 2. above? Explain why or why not.

4. What is the probability that a single vehicle will be traveling faster than 33 mph through the construction zone?

5. What is the probability that the average speed of 15 vehicles is faster than 33 mph through the construction zone?

6. Construct a 95% confidence interval for the true mean speed of drivers in this construction zone. Interpret the interval.

7. Construct a 99% confidence interval for the true mean speed of drivers in this construction zone. Interpret the interval.

8. Compare the widths of the 95% and 99% confidence intervals. Note which one is wider/narrower and explain why.

9. Based on your intervals in 6. and 7., do you think that it is possible that the true average speed of all drivers is 25 mph (the actual speed limit)?

There is a rumor that most police officers will give some leeway in citations for speeding tickets of about 5 mph. Officers familiar with this construction zone have told you that they believe the average speed of drivers is 30 mph. We would like to use the sample above to test the notation that the mean speed of drivers is over 30 mph at the α = 0.05 level of significance. Use σ=5.95

10. State the Null and Alternative hypothesis for this test of significance.

11. Calculate the test statistic and interpret the result. (You may use your calculator to check this value, but for this problem you need to find this value “by hand” and show the work here.)

12. State the P-value for this test of significance and provide a short explanation.

13. Do we reject or fail to reject the Null Hypothesis? Write a complete statement here as modeled in your notes.

14. Is there sufficient evidence that the mean speed of drivers in the construction zone is over 30 mph? Again, write a complete statement here as modeled in your notes.

Answer 1

population distribution, X~N(30,5.95)

distribution of sample mean , Xbar~N(30,5.95/√15) = N(30,1.5363)

there is no need to apply central limit theorem as population is already normally distributed

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probability that a single vehicle will be traveling faster than 33 mph through the construction zone

Z =   (X - µ )/(σ/√n) = (   33   -   30   ) / (    5.95   / √   1   ) =   0.504  
                                          
P(X ≥   33   ) = P(Z ≥   0.50   ) =   P ( Z <   -0.504   ) =    0.3071           (answer)
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probability that the average speed of 15 vehicles is faster than 33 mph through the construction zone

Z =   (X - µ )/(σ/√n) = (   33   -   30   ) / (    5.95   / √   15   ) =   1.953  
                                          
P(X ≥   33   ) = P(Z ≥   1.95   ) =   P ( Z <   -1.953   ) =    0.0254           (answer)
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Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.96   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   5.9500   / √   15   =   1.5363
margin of error, E=Z*SE =   1.9600   *   1.5363   =   3.0111
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    32.20   -   3.011060   =   29.1889
Interval Upper Limit = x̅ + E =    32.20   -   3.011060   =   35.2111
95%   confidence interval is (   29.18894   < µ <   35.21106   )
======================

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.58   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   5.9500   / √   15   =   1.5363
margin of error, E=Z*SE =   2.5758   *   1.5363   =   3.9572
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    32.20   -   3.957204   =   28.2428
Interval Upper Limit = x̅ + E =    32.20   -   3.957204   =   36.1572
99%   confidence interval is (   28.24280   < µ <   36.15720   )
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99% CI is wider

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it is not possible that the true average speed of all drivers is 25 mph (the actual speed limit)

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Ho :   µ =   30                  
Ha :   µ >   30       (Right tail test)          
                          
Level of Significance ,    α =    0.010                  
population std dev ,    σ =    5.9500                  
Sample Size ,   n =    15                  
Sample Mean,    x̅ = ΣX/n =    32.2000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   5.9500   / √    15   =   1.5363      
Z-test statistic= (x̅ - µ )/SE = (   32.200   -   30   ) /    1.5363   =   1.432
                          

                          
p-Value   =   0.0761   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value>α, Do not reject null hypothesis                       
Conclusion: There is not enough evidence to support that true mean is over 30