Question

1. The plant manager of an apple juice bottling facility is interested in comparing the performance of two different production lines in her plant. A random sample of 10 hours from Production Line #1 produced an average of 326.8 bottles with a standard deviation of 5.3 bottles. A random sample of 12 hours from Production Line #2 produced an average of 303.6 bottles with a standard deviation of 7.9 bottles. Assume that the number of bottles produced per hour on each production line follows a normal distribution.

(a) Calculate the ratio of the maximum sample variance to the minimum sample variance. Does it appear that the population variances are equal or unequal? Explain.

(b) Construct the appropriate 95% confidence interval for the difference in the average hourly production between the two production lines and interpret your results.

For parts (c) through (e), explain how you would calculate the interval differently in part (b). Do not calculate the interval, just write out the formula you would use and give the numerical value of the corresponding critical value from the z or t distribution.

(c) The sample standard deviations are 4.2 and 8.1 instead of 5.3 and 7.9.

(d) The standard deviations of 5.3 and 7.9 are the true population standard deviations, not sample results.

(e) The sample sizes are each 100 and 120 instead of 10 and 12.

Answer 1

Solution

Let

X = number of bottles produced per hour on Production Line #1

Y = number of bottles produced per hour on Production Line #2

We are given, X, Y ~ Normal Distribution, say X ~ N(µ₁, σ₁²) and Y ~ N(µ₂, σ₂²). ........................................ (1)

Part (a)

Ratio of larger variance to smaller variance

=(7.9²/5.3²)

= 2.2219 Answer 1

Hypotheses:

Null H₀: σ₁² = σ²₂   Vs Alternative H_A: σ₁² ≠ σ²₂

Test statistic:

F= s₂²/s₁² =

where s₁and s₂ are standard deviations based on n1 observations on X and n2 observations on Y.

[Note: F is defined as above since s₂² > s₁².

Calculations

Given
n2	12
n1	10
s2	7.9
s1	5.3
Fcal	2.221787
Assuming
α	0.05
Fcrit	3.102485
n1 - 1	11
n2 - 1	9
p-value =	0.120385

Distribution, Critical Value and p-value

Under H₀, F~ F_{n2 – 1, n1– 1}

Critical value = upper α% point of F_{n2 – 1, n1 – 1}

p-value = P(F_{n2 – 1, n1 – 1} > F_cal)

Assuming α = 0.05 , Fcrit and and p-value are found to be as shown in the above table [using Excel Function: Statistical   FINV and FDIST respectively]

Decision:

Since Fcal < Fcrit, or equivalently, since p-value > α. H₀ is accepted.

Conclusion:

There is not sufficient evidence to conclude that the population variances are different. Answer 2

Part (b)

100(1 - α) % Confidence Interval for (μ₁ - μ₂) is:

(Xbar – Ybar) ± MoE,

where

MoE{(t_{2n – 2, α/2})(s)√{(1/n₁) + (1/n₂)}

with

Xbar and Ybar as sample means,

s = pooled sample estimate of σ given by

s = sqrt[{(n₁ – 1)s₁² + (n₂ – 1)s₂²)}/(n₁ + n₂ – 2)];

s₁, s₂ being respective the sample standard deviations;

t_{n1 + n2 – 2, α/2} = upper (α/2) percent point of t-distribution with n₁ + n₂ – 2 degrees of freedom

n₁ , n₂ being the sample sizes.

So, 95% confidence interval for the difference in average production of the two production lines is:

[17.08, 29.32] Answer 3

Calculation details

Given	n1	10
	n2	12
	Xbar	326.8
	Ybar	303.6
	s1	5.3
	s2	7.9
s^2		46.966
s		6.853174447
α		0.05
tα/2		2.085963441
MoE		6.120955145
Lower Bound		17.07904485
Upper Bound		29.32095515

Interpretation

We can be 95% confident that the true difference between the two populations means would lie in the interval [17.07, 29.32] Answer 4

Common back-up concepts for Parts (c) to (e)

All the variations cited in these parts affect only the MoE and the new MoE can be directly derived from the MoE derived under Part (b). Let for convenience, let M represent MoE derived under Part (b) and Mn represent the new MoE.

Part (c)

Critical value and the distribution remains the same.

Mn = M x (s/s_n) Answer 5

where s_n is pooled standard deviation with the new standard deviations.

Part (d)

Here the distribution will be Z and not t. Accordingly, the critical value will be the upper 2.5% of N(0, 1).

Mn = M x (s/σ) Answer 6

where σ = √{(σ₁²/n₁)+ (σ₂²/n₂)} = √{(5.3²/10)+ (7.9/12)}

Part (e)

Here the distribution continues to be t but with degrees of freedom 218 which will be almost equal to 1.96, the upper 2.5% of N(0, 1).

Mn = M x (2.086/1.96) Answer 7

where 2.086 is the critical value for M.

DONE