Question

This is the remaining question for the previous one I just posted

3. 10pts. Form the hypotheses (null and alternative) and identify the claim Claim: The Null hypothesis in this case is that the tuition cost is less than or equal to the amount mentioned in the College Board’s website ( i.e. $9410.00). Ho: u=9410 Hα:u>9410 Alternative:greater Ho: u>9410 Ha:u<9410 4. 6pts. State the distribution and test statistic to be used. Normal distribution A random sample of 16 states tuitions, the mean was 9521 and standard deviation was 9410 5. 4pts. Find the sample mean to the nearest whole number. 6. 4 pts. Find the sample standard deviation to the nearest who le number. 7. 6pts. Find the value of the test statistic. To 3 decimal places 8. 8pts. Compute and interpret the p-value. To 2 decimal places 9. 4pts. Use the previous informat ion to sketch a picture of this situation. CLEARLY, label and sca le the horizonta l ax is and shade the region(s) corresponding to the p-value. 10. 6pts. Make the decision to reject or to not reject the null hypothesis. Explain. 11. 6pts. Form a conclusion written in terms of the original claim. 12. 6 pts. Ide ntify any Type I or Type II errors. Type 1: Type II: 13) 8pts. Construct a 95 % confidence interval. Explain what this means.

Answer 1

3)

Ho :   µ =   9410                  
Ha :   µ >   9410       (Right tail test)      

4)

t distribution with 1 mean test

5)

Sample Mean,    x̅ =   9521

6)   
                             
sample std dev ,    s =    9410.0000                  
Sample Size ,   n =    16                  
          
                          
degree of freedom=   DF=n-1=   15                  
                          
Standard Error , SE = s/√n =   9410.0000   / √    16   =   2352.5   

7)

t-test statistic= (x̅ - µ )/SE = (   9521.000   -   9410   ) /    2352.5000   =   0.047

8)

                                
                          
p-Value   =   0.48 [Excel formula =t.dist(t-stat,df) ]     

10)

  
Decision:   p-value>α, Do not reject null hypothesis

11)

  
Conclusion: There is not enough evidence that  tuition cost is greater than the amount mentioned in the College Board’s website

12)

Type 2 error

13)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   15          
't value='   tα/2=   2.1314   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   9410.0000   / √   16   =   2352.500000
margin of error , E=t*SE =   2.1314   *   2352.50000   =   5014.235056
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    9521.00   -   5014.235056   =   4506.764944
Interval Upper Limit = x̅ + E =    9521.00   -   5014.235056   =   14535.235056
95%   confidence interval is (   4506.76   < µ <   14535.24   )

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