Amanufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 439.0 gram setting. it is believed that the machine is underfilling the bags. a 47 bag sample had a mean of 433.0 grams. a level of significance of 0.05 will be used. determine the decision rule. assume the standard deviation is known to be 21.0. enter the decision rule.

Let be the average weight of chocolate chips in each bag.

As per given , we have

Since is left-tailed , so we perform a left-tailed test.

Also, the population standard deviation is known .

So we use z-test.

Test statistic :

Substitute given value ,

Significance level = 0.05

Decision rule : Reject when p-value < 0.05.

By z-table , P- value for left-tailed test  = P(z<-1.96)= 1- P(z<1.96)

[∵ P(Z<-z)=1-P(Z<z)]

= 1- 0.9750

=0.025

Since , P-value(0.025) < 0.05 , so we reject the null hypothesis.

We support the claim that the machine is underfilling the bags at 0.05 significance level .

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

step-by-step explanation:

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