A local bottler in Hawaii wishes to ensure that an average o

A local bottler in Hawaii wishes to ensure that an average of 24 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 50 bottles. The mean weight of the passion fruit juice in the sample is 23.69 ounces. Assume that the population standard deviation is 0.87 ounce. Use Table 1.

a.

Select the null and the alternative hypotheses for the test.

b-1.

Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)

b-2.

Find the critical value(s). (Round your answer to 2 decimal places.)

b-3.

What is the conclusion?

c.

Make a recommendation to the bottler.

The accuracy of the bottling process is:

a. compromised or

b. not compromised

Solution

Set Up Hypothesis
Null H0: U=24
Alternate H1: U!=24
Test Statistic
Population Mean(U)=24
Given That X(Mean)=23.69
Standard Deviation(S.D)=0.87
Number (n)=50
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=23.69-24/(0.87/Sqrt(50)
Zo =-2.5196
| Zo | =2.5196
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.5196 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value : Two Tailed ( double the one tail ) - Ha : ( P != -2.5196 ) = 0.0117
Hence Value of P0.05 > 0.0117, Here we Reject Ho

ANSWERS:
H0: = 24; HA: 24
Zo =-2.5196
| Z | =1.96
Reject H0 since the value of the test statistic is less than the negative critical value.
b. not compromised