Question

The output of a chemical process is monitored by taking a sample of 20 vials to determine the level of impurities. The desired mean level of impurities is 0.048 grains per vial. If the mean level of impurities in the sample is too high, the process will be stopped and purged; if the sample mean is too low, the process will be stopped and the values will be readjusted. Otherwise, the process will continue.
a)  Sample results provide sample mean to be equal to 0.057 gram with sample standard deviation equal to 0.018. At a significance level of 0.01, should the process be stopped? If so, what type of remedial action will be required?
b)  Assume that the mean level of impurities is within tolerable limits. If the maximum tolerable variability of the process is 0.0002, do the sample results verify the suspicion that the maximum tolerable variability has been exceeded? Use a 5% level of significance.

Answer 1

Solution

Let

X = the level of impurities (grains per vial).

Let µ and σ be respectively the mean and standard deviation of X

Part (a)

Claim: The mean level of impurities in the sample is too high

Hypotheses:

Null H₀: µ = µ₀ = 0.048   Vs Alternative H_A: µ > 0.048 [claim]

Test statistic:

t = (√n)(Xbar - µ₀)/s,

where

n = sample size;

Xbar = sample average;

s = sample standard deviation.

Summary of Excel Calculations is given below:

Given, n	20
µ0	0.048
Xbar	0.057
s	0.018
tcal	2.236068
Given α	0.01
tcrit	2.539483
p-value	0.01877

Distribution, Significance Level, α Critical Value and p-value

Under H₀, t ~ t_{n – 1}

Critical value = upper α% point of t_{n – 1}.

p-value = P(t > t_{n – 1})

Using Excel Function, Statistical TINV and TDIST, tcrit and p-value are found to be as shown in the above table.

Decision:

Since tcal < tcrit, or equivalently since p-value > α. H₀ is accepted.

Conclusion:

There is not sufficient evidence to support the claim and hence we conclude that the mean level of impurities is within acceptable level. Answer 1

Since the mean level of impurities is within acceptable level, the process need not be stopped. Answer 2

Part (b)

Here we want to test if the process variation represented by the standard deviation is within tolerable limits.

Claim: Process variation is above the tolerable limit

Hypotheses:

Null H₀: σ² = σ²₀ = 0.0002 Vs Alternative H_A: σ² > 0.0002[claim]

Test statistic:

χ² = (n - 1)s²/σ²₀

where

n = sample size

s = sample standard deviation  

Calculations

n	20
σ20	0.0002
s	0.018
s^2	0.000324
χ2(cal)	30.78
α	0.05
Critical value	30.14353
p-value	0.042673

Distribution, Significance level, α , Critical Value and p-value

Under H₀, χ² ~ χ²_{n - 1}

Critical value = upper α% point of χ²_{n - 1}.

p-value = P(χ²_{n – 1} > χ²_(cal))

Using Excel Function: Statistical CHIDIST and CHIINV, Critical Value and p-value are found to be as shown in the above table.

Decision:

Since χ²cal > χ²crit, or equivalently, since p-value < α. H₀ is rejected.

Conclusion:

There is sufficient evidence to support the claim and hence we conclude that the maximum tolerable variability has been exceeded. Answer 3

DONE