Question

Suppose that an independent research company was tasked with testing the validity of complaints against a pesticide manufacturing firm for under-filling their 20 lb bags of pesticide. Past experience has shown that the amount of pesticide dispensed by the machines that fill the bags follows a normal distribution with a mean of 20 lb and a standard deviation of 0.6 lb. To verify the validity of the complaints, a researcher randomly selected 16 of the firm’s 20 lb pesticide bags and recorded the following weights. 18.59,19.74,19.18,20.09,18.54,20,19.04,19.24,19.23,20.31,19.74,18.69,18.61,20.18,18.68,20.28 If you wish, you may download the data in your preferred format. CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc The mean weight of the 16 bags collected was 19.384 lb. Calculate the value of the one-sample ? ‑ statistic. Give your answer precise to three decimal places.

Answer 1

Ho :   µ =   20                  
Ha :   µ <   20       (Left tail test)          
                          
Level of Significance ,    α =    0.01                  
population std dev ,    σ =    0.6000                  
Sample Size ,   n =    16                  
Sample Mean,    x̅ =   19.3840                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   0.6000   / √    16   =   0.1500      
Z-test statistic= (x̅ - µ )/SE = (   19.384   -   20   ) /    0.1500   =   -4.107
                          
critical z value, z* =       -2.3263   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
p-Value   =   0.000020   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value<α, Reject null hypothesis