a potato chip manufacturer has been accused of shortchanging their customers with 10 ouce of bags potato chips. a sample of 50 of that company's 10 ounce bags was take, and it was found that the average weight was 9.8 ounces with a standard deviation of 0.3 ounces.can we conclude that the company's bags of potato chips are less than 10 ounces?use 5%significance level

Question

a potato  chip  manufacturer  has  been  accused  of  shortchanging  their  customers  with  10  ouce  of  bags  potato  chips. a  sample  of  50  of  that  company's  10  ounce  bags  was  take, and  it  was  found  that  the  average  weight  was  9.8 ounces  with  a  standard  deviation  of 0.3 ounces.can  we  conclude  that  the  company's  bags  of  potato  chips  are  less  than  10  ounces?use  5%significance level

PsyDAG · Answer

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.