Question

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A manufacturer of liquid detergent is interested in the uniformity of the machine used to fill bottles. Specifically, it is desirable that the standard deviation σof the filling process be less than 0.25 fluid ounces; otherwise, there will be a higher than desirable percentage of bottles that are underfilled. We will assume that fill volume is approximately normally distributed. A random sample of 20 bottles in a sample variance of s2= 0.0153 (fluid ounces)2. At a 95% upper-confidence interval will the target of σ= 0.25 oz be met? Explain by using the P-value

.H0: σ2=0.0625

H1: σ2>0.0625

χ2= (20-1)(0.0153)/0.0625 = 4.6512

χ2.95,19= 30.14

Since χ2.95,19> χ2, fail to reject the null hypothesis that σ2= 0.0625The actual p-value is < 0.005.

Answer 1

To test against

Here

sample variance

and sample size

The test statistic can be written as

which under H₀ follows a chi square distribution with n-1 df.

We reject H₀ at 95% confidence level if P-value < 0.05

Now,

The value of the test statistic

and P-value =

Since P-value < 0.05, so we reject H₀ at 95% confidence level and we can conclude that  the standard deviation σof the filling process is significantly less than 0.25.