Question

4.27- An experiment was conducted to investigate the filling capability of packaging equipment at a winery in Newberg, Oregon. Twenty bottles of Pinot Gris were randomly selected and the fill volume (in ml) measured. Assume that fill volume has a normal distribution. The data are as follows: 753, 751, 752, 753, 753, 753, 752, 753, 754, 754, 752, 751, 752, 750, 753, 755, 753, 756, 751, and 750.

(a) Do the data support the claim that the standard deviation of fill volume is less than 1 ml? Use alpha = 0.05
(b) Find a 95% two-sided confidence interval on the standard deviation of fill volume.

(c) Does it seem reasonable to assume that fill volume has a normal distribution?

Answer 1

Solution:

Part a

Here, we have to use Chi square test for population standard deviation.

H₀: σ = 1 versus H_a: σ < 1

This is a lower tailed test.

We are given

α = 0.05

n = 20

From given data, we have

Sample standard deviation = S = 1.5381

Degrees of freedom = n – 1 = 19

The test statistic formula is given as below:

Chi square = (n – 1)*S^2/σ^2

Chi square = (20 - 1)* 1.5381^2/1^2

Chi square = 44.94928

P-value = 0.9993

(by using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the standard deviation of fill volume is less than 1 ml.

Part b

Confidence interval for population standard deviation is given as below:

Sqrt(n – 1)*S² / χ²_{α/2, n – 1} ] < σ² < Sqrt[(n – 1)*S² / χ²_{1 -α/2, n– 1}]

We are given

Confidence level = 95%

Sample size = n = 20

Degrees of freedom = n – 1 = 19

Sample standard deviation = S = 1.5381

χ²_{α/2, n – 1} = 32.8523

χ²_{1 -α/2, n– 1} = 8.9065

(By using chi square table)

Sqrt(20 – 1)* 1.5381^2 / 32.8523] < σ² < Sqrt[(20 – 1)* 1.5381^2 / 8.9065]

Sqrt(1.3682) < σ < Sqrt(5.0468)

1.1697 < σ < 2.2465

Lower limit = 1.1697

Upper limit = 2.2465

Part c

Yes, it seems reasonable to assume that fill volume has a normal distribution, because for the large production process of filling volume follows an approximate normal distribution.