Rutter Nursery Company packages its pine bark mulch in 50pou

Rutter Nursery Company packages its pine bark mulch in 50-pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution and the standard deviation of this process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today\'s production.

45.0

47.7

47.6

46.3

46.2

47.4

49.2

55.8

47.5

48.5

Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .03 significance level. In answering, you have to show: null & alternative hypotheses, testing rule, test statistic, & conclusion.

In a brief report, tell why Mr. Rutter can use the z distribution as the test statistic.

Compute the p-value.

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   50  
Ha:    u   <   50  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.03   ,      
alpha =    0.03          
zcrit =    -   1.880793608      

Thus, we reject Ho if z < -1.8808.
              
Getting the test statistic, as              
              
X = sample mean =    48.12          
uo = hypothesized mean =    50          
n = sample size =    10          
s = standard deviation =    2.951007361          
              
Thus, z = (X - uo) * sqrt(n) / s =    -2.014594094                  
Comparing z < -1.8808, we   REJECT THE NULL HYPOTHESIS.          
              
Thus, there is significant evidence that the mean weight of the bags is less than 50 pounds. [CONCLUSION]

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b)

He can use the z statistic because the distirbution is normal here.

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c)

Getting the left tailed area of z = -2.0146,

p =    0.021973598   [ANSWER]