Question

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Question: The quality control manager for a filling operation in a bottling plant is concerned with the variability in the volume of milk dispensed into gallon jugs. The filling process results in jugs whose volumes are normally distributed with a mean of 1.02 gallons. In order for this process to be in-control, the standard deviation should be less than 0.05 gallons (otherwise milk spills all over the production line). The manager takes a sample of 25 jugs of milk off the production line and determines the standard deviation to be 0.03 gallons. Is the process in-control with respect to the variability in the volume of milk dispensed into the jugs? Use the P-value approach with a 2.5% level of significance.

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis H₀: σ > 0.05

Alternative hypothesis H_A: σ < 0.05

Formulate an analysis plan. For this analysis, the significance level is 0.025.

Analyze sample data. Using sample data, the degrees of freedom (DF), and the test statistic (X²).

DF = n - 1 = 25 -1

D.F = 24

$\chi ^{2}=\frac{(n-1)s^{2}}{\sigma ^{2}_{0}}$

$\chi ^{2}=\frac{(25-1)\times (0.03)^{2}}{(0.05)^2}$

$\chi ^{2}= 8.64$

We use the Chi-Square Distribution Calculator to find P(Χ² < 8.64) = 0.002

Interpret results. Since the P-value (0.002) is less than the significance level (0.025), we have to reject the null hypothesis.

From the above test we do have sufficient evidence in the favor of the claim that process in-control with respect to the variability in the volume of milk dispensed into the jugs.