Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

EXERCISES 1. Use technology to perform a one-sample z-test In Exercises 4 and 5,

ID: 3229750 • Letter: E

Question

EXERCISES 1. Use technology to perform a one-sample z-test In Exercises 4 and 5, use technology to perform a to test the hypothesis that the probability that a two-sample z-test to determine whether there is a found coin. will be lying heads up is 0.5. Use difference in the mint dates and in the values of coins a 0.01. Use Casey's data as your sample and found on a street from 1985 through 1996 for the two mint locations. Write your conclusion as a sentence. write your conclusion as a sentence. Use at 0.05. 2. Do Casey's data differ significantly from chance? If so, what might be the reason? 4. Mint dates of coins (years Philadelphia: X1 1984.8 s1 8.6 3. In the simulation shown above, what percent of the Denver: x2 1983.4 s2 8.4 trials had heads less than or equal to the number of heads in Casey's data? Use technology to repeat 5. Value of coins (dollars) the simulation. Are your results comparable? Philadelphia: T1 $0.034 S1 $0.054 Denver: x2 $0.033 S2 $0.052

Explanation / Answer

Data:    

n = 11902   

p = 0.5   

p' = 5772/11902 = 0.484960511   

Hypotheses:    

Ho: p = 0.5   

Ha: p 0.5   

Decision Rule:    

= 0.01   

Lower Critical z- score = -2.5758   

Upper Critical z- score = 2.5758   

Reject Ho if |z| > 2.5758   

Test Statistic:    

SE = {p (1 - p)/n} = (0.5 * (1 - 0.5)/11902) = 0.0046

z = (p'- p)/SE = (0.484960510838515 - 0.5)/0.00458310736624905 = -3.2815

p- value = 0.0010   

Decision (in terms of the hypotheses):    

Since 3.2815 > 2.5758 we reject Ho and accept Ha

Conclusion (in terms of the problem):

There is sufficient evidence that p 0.5    

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote