Question

5. According to a survey conducted with 132 adults by a sociology student, 73.4 % of them indicated that they were satisfied with their jobs. Test the claim that more than 70% of the people are satisfied with their jobs. Use a significance level of 0.01.

For each question you need to find the following

1. a) < 1 point > Write the given .

Step 1

b) < 2 points > Write the claim and its opposite , then determine which is the null hypothesis and which is the alternative hypothesis.

c) < 0.5 point >Write the type of test you are going to use ( LT , RT , or 2 T )

d) < 0.5 point > what distribution you are going to use ( z or t )

Step 2

Use a graphing Calculator: Find the calculated test statistic and the P- Value

i) < 0.5 > what did you use to find the CTS :   Z- Test, T – Test , or 1-PropZtest ( choose one of them)

        ii ) < 2 >Write the value of the calculated test statistics :   C. T . S : ( choose z or t ) =   

        iii) < 1.5 point > P -value =

Step 3   Analyze

Traditional Method < 3 points> P-value Method. < 2 points >

For traditional , you need to Sketch the graph For P-value method, you need to do

Shade the required area P-value             α Need to write the value of each

label the critical value/s compare the values ( < or > )

Label the C.T.S                     

Step 4 : Write

1. i) < 1 point > Decision: (no credit if your answer is just reject or fail to reject only ; you need to write the complete expression)
2. ii) < 3 points > Conclusion ( fill in) : There ___________ enough evidence to _____________

                         the ______________ that _______________

Answer 1

Null

Ho :   p =    0.7            

Alternate:
H1 :   p >   0.7       (Right tail test)         

Z distribution
                          
Level of Significance,   α =    0.01                  
Number of Items of Interest,   x =   96.888                  
Sample Size,   n =    132                  
                          
Sample Proportion ,    p̂ = x/n =    0.7340                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0399                  
Z Test Statistic = ( p̂-p)/SE = (   0.7340   -   0.7   ) /   0.0399   =   0.8524
                          
critical z value =        2.326   [Excel function =NORMSINV(α)              
                          
p-Value   =   0.197 [Excel function =NORMSDIST(-z)             
Decision:   p value>α ,do not reject null hypothesis

                      
There is not enough evidence to conclude that more than 70% of the people are satisfied with their jobs

THANKS

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