Julia thinks that she has a special relationship with the number 6. In particula

Question

Julia thinks that she has a special relationship with the number 6. In particular, Julia thinks that she would roll a 6 with a fair 6-sided die more often than you'd expect by chance alone. Suppose pp is the true proportion of the time Julia will roll a 6. (Note you will need to find the probability of getting a four when rolling a fair 6-sided die like we did in chapter 3 so that you can compare pp to this value.)

(a) State the null and alternative hypotheses for testing Julia's claim. (Type the symbol "p" for the population proportion, whichever symbols you need of "<", ">", "=", "not =", ">=", or "<=" and express any values as a fraction e.g. p = 1/3)
H0 =  
Ha =

(b) Now suppose Julia makes n = 42 rolls, and a 6 comes up 9 times out of the 42 rolls. Determine the P-value of the test:
P-value =

(c) Answer the question: Does this sample provide evidence at the 5 percent level that Julia rolls a 6 more often than you'd expect?
(Type: Yes or No)

Explanation / Answer

(a) Ho: p = 1/6 versus Ha: p > 1/6

(b)

Data:     

n = 42    

p = 0.1666667    

p' = 0.2142857    

Hypotheses:    

Ho: p 0.1666667    

Ha: p > 0.1666667    

Decision Rule:    

= 0.05    

Critical z- score = 1.644853627   

Reject Ho if z > 1.644853627   

Test Statistic:    

SE = {(p (1 - p)/n} = (0.166666666666667 * (1 - 0.166666666666667)/42) = 0.057505463

z = (p' - p)/SE = (0.214285714285714 - 0.166666666666667)/0.0575054632785295 = 0.828078671

p- value = 0.203813    

Decision (in terms of the hypotheses):   

Since 0.8280787 < 1.644853627 we fail to reject Ho

(c) Conclusion (in terms of the problem):   

There is no sufficient evidence that Julia rolls a six more often than expected.

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