Question

A company produces “52-gram” snack pack of candies. A random sample of 10 snack packs yielded the following weights:

55.95 56.54 57.58 55.13 57.48 56.06 59.93 58.30 52.57 58.46

a) Calculate and interpret a 95% confidence interval for the true mean weight of snack packs produced by the company.

b) [MINITAB]Whenfindingtheconfidenceintervalina),what assumption did you make about the population of snack pack weights? Use the appropriate Minitab plot to verify the assumption for this dataset. Include the plot, explain how you use it, and your conclusion.

c) The company would consider changes in the production process if the true average weight of snack packs is below the claimed 52 grams. Should they introduce changes in the production process? Explain.

Answer 1

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.0519
Sample Size ,   n =    10
Sample Mean,    x̅ = ΣX/n =    56.8000

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   2.262   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.0519   / √   10   =   0.6489
margin of error , E=t*SE =   2.2622   *   0.6489   =   1.4678
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    56.80   -   1.467844   =   55.3322
Interval Upper Limit = x̅ + E =    56.80   -   1.467844   =   58.2678
95%   confidence interval is (   55.33   < µ <   58.27   )

b) Assumptions: data is normally distributed

since, p value >0.05 , fail to reject Ho

and hence, data is normally distributed

c)

since, both limits of CI is greater than 52

so, they should not introduce changes in the production process