Question

In one study (n=72) of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one year after the treatment and 32 were not smoking after one year of treatment. Use a 0.05 significance level to test the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment.

1. State the claim using words (i.e., “The claim is [insert the English words here]”.)

2. Write the null (H₀) and alternative (H₁) hypotheses using symbols.

3. Indicate whether H₀ orH₁ is the claim by writing “(claim)” next to whichever one it is.

4. State the tail of the hypothesis test: left-tailed/one-tailed, right-tailed/one-tailed, or two-tailed.

5. State the level of significance. (i.e., a = __)

6. State the degrees of freedom (i.e., DF = __) or write “not applicable” if not relevant.

7. State all the requirements with respect to the problem.

8. Verify any requirement that should require a calculation. If none, write “not applicable”.

9. Identify which test statistic to use and write the associated formula using symbols.

10. Compute the value of the test statistic. If using StatCrunch, circle it.

11. State the P-value. If using StatCrunch, circle it

12. Show and apply the P-value decision rules (with values substituted appropriately) that lead you to “rejectH₀”or “fail to rejectH₀”.

13. State the critical value(s). If using StatCrunch, circle it.

14. Show and apply the critical value decision rules (with values substituted appropriately) that lead you “rejectH₀”or “fail to rejectH₀”.

15. Write the conclusion utilizing the “correct” words using Table 8-3, p. 366. Make sure you insert the relevant portions of the claim into the conclusion.

16. Construct the associated confidence interval (CI).

17. State the CI using the correct notation using the correct notation.

18. State the margin of error (E = __).

19. State the point estimate using the correction notation.

20. Does the CI support the hypothesis test conclusion? Explain/interpret. This may require a sentence or two, not just a single word.

21. Extra Credit: Express the Type I error in the context of the problem using the words from the “helper” document. Make sure the conclusion is worded such that it addresses the claim (p. 368).

22. Extra Credit: Express the Type II error in the context of the problem using the words from the “helper” document. Make sure the conclusion is worded such that it addresses the claim (p. 368).
    

Answer 1

Ho :   p =    0.5                  
H1 :   p >   0.5 ( claim ) (Right tail test)         
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   39                  
Sample Size,   n =    72                  
                          
Sample Proportion ,    p̂ = x/n =    0.5417                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0589                  
Z Test Statistic = ( p̂-p)/SE = (   0.5417   -   0.5   ) /   0.0589   =   0.7071

df not requried

critical z value =        1.645   [Excel function =NORMSINV(α)              
                          
p-Value   =   0.2398 [Excel function =NORMSDIST(-z)             
Decision:   p value>α ,do not reject null hypothesis                      

...................

z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0587          
margin of error , E = Z*SE =    1.960   *   0.0587   =   0.1151
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.542   -   0.1151   =   0.4266
Interval Upper Limit = p̂ + E =   0.542   +   0.1151   =   0.6568
                  
95%   confidence interval is (   0.4266   < p <    0.6568   )

0.5 lies within the CI , so donot reject HO

so, CI agree with the hypothesis test