Question

5. A manufacturing company produces electric insulators. To test the strength of the insulators, you apply a force to an insulator until it breaks. You measure the force by observing! how many pounds are required to break the insulator. You collect the force data for a random sample of 30 insulators selected for this experiment. The data is in the Excel file for this assignment in the tab Force. a. Construct a 95% confidence interval estimate for the population mean force. b. What assumption must you make about the population distribution in order to construct the confidence interval estimate in (a)? C. Do you think the assumption reference is part (b) is valid for this experiment? Explain your answer. 6. If n - 200 and X 50, construct a 95% confidence interval for the population proportion. 7. In a survey of 1003 adults concerning complaints about restaurants, 732 complained about dirty bathrooms and 381 complained about loud or distracting diners at other tables. a. Construct a 95% confidence interval estimate of the population proportion of adults who complained about dirty bathrooms. b. Construct a 95% confidence interval estimate of the population proportion of adults who complained about loud or distracting diners at other tables. c. How would the manager of a chain of restaurants use the results of parts (a) and (b)? 8. If you want to be 95% confident when estimating the population mean to within a sampling error of t 20 and the standard deviation is assumed to be 100, what sample size is required? 9. If you want to be 95% confident of estimating the population proportion to within a sampling error of t 0.02 and there is historical evidence that the population proportion is approximately 0.40, what sample size is needed?

Answer 1

6)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   50          
Sample Size,   n =    200          
                  
Sample Proportion ,    p̂ = x/n =    0.2500          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0306          
margin of error , E = Z*SE =    1.960   *   0.0306   =   0.0600
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.250   -   0.0600   =   0.1900
Interval Upper Limit = p̂ + E =   0.250   +   0.0600   =   0.3100
                  
95%   confidence interval is (   0.1900   < p <    0.3100   )

8)

Standard Deviation ,   σ =    100                  
sampling error ,    E =   20                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   100   /   20   ) ² =   96.036
                          
                          
So,Sample Size needed=       97                  

9)

sample proportion ,   p̂ =    0.4                          
sampling error ,    E =   0.02                          
Confidence Level ,   CL=   0.95                          
                                  
alpha =   1-CL =   0.05                          
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.960   /   0.02   ) ² *   0.40   * ( 1 -   0.40   ) =    2304.88
                                  
                                  
so,Sample Size required=       2305