Question

Testing a claim about a population proportion:

According to the Gallup poll results of 2014, support for the U.S. death penalty for convicted murderers was stable at 63%. (After several years of diminishing support, American support had leveled out in the low 60s.) In a survey of 150 randomly selected 18-20 year olds, 75 said they were in favor of the death penalty for those convicted of murder. Does this sample appear to come from a population with a lower proportion in favor of the death penalty? Use α = .05

1.Are the conditions for a binomial distribution met?

                -Fixed number of trials?

                -Exactly two outcomes?

                -Trials independent?

                -Probability of success the same for each trial?

2. Check to see if np and nq are greater than 5. (Then we can use the normal approximation to the binomial.)

3. Write the claim in words:

4. Translate the claim into symbols:

5. Create the null and alternative hypotheses:

6. Identify the type of test (left-tailed, right-tailed, two-tailed):

7. Compute the test statistic z=p-ppqn                  p=                  q=                   n=           

8. Substitute and simplify    z =

9. Critical Value Method: Compare the test statistic with the critical value . Find by locating the area .05 in the body of the z-score table A-2. Note that we use α since it is a single-tailed test.

10. Draw a normal distribution, locate the critical value, and shade the critical region. Plot the test statistic z on the horizontal axis of your normal distribution.

11. Is the test statistic in the critical region?

If yes, we reject the null hypothesis. If no, we fail to reject the null hypothesis.

12. Form a conclusion:

If yes – There is sufficient sample evidence to support the claim ----- ------

If no –There is not sufficient sample evidence to support the claim ---- ------                                                                                  

(to reject the claim ----- ------)    

13. P-value method: Compare the P-value with the level of significance to make a decision about rejecting . Draw a normal distribution and plot the test statistic on the horizontal axis. Find the area to the left using table A-2. This is our P-value. Compare it to α.

14. If P-value α, we reject the null hypothesis. If P-value > α, we fail to reject the null hypothesis.

15. Form a conclusion as above.

Answer 1

Ho :   p =    0.63                  
H1 :   p <   0.63       (Left tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   75                  
Sample Size,   n =    150                  
                          
Sample Proportion ,    p̂ = x/n =    0.5000      

q = 0.5            
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0394                  
Z Test Statistic = ( p̂-p)/SE = (   0.5000   -   0.63   ) /   0.0394   =   -3.2978
                          
critical z value =        -1.645   [excel function =NORMSINV(α)]         

test stat > crtical vaue , reject Ho

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

                          
p-Value   =   0.000487313   [excel function =NORMSDIST(z)]             
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to support the claim

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

  

THANKS

revert back for doubt

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