Question

In: Statistics and Probability

You have been asked to evaluate a new raw material supplier. The most important specification for...

You have been asked to evaluate a new raw material supplier. The most important specification for this material entering your process is that it should have a consistent viscosity from lot to lot. The new supplier has given you samples from 5 lots, which had viscosities of 15.6, 15.2, 15.4, 15.4,and 15.7. Reviewing the last 5 lots of material from the current supplier showed viscosities of 12.3, 12.2, 12.3, 12.5, and 12.1. Based on this data, construct and evaluate a hypothesis test at 90% significance to determine if the new supplier is better than the current supplier.

Solutions

Expert Solution

Solution:

Here, we have to use F-test for two variances for the comparison of two different suppliers. Here, we have to check whether the new supplier is better than the current supplier or not. This means we have to check whether the variance of the viscosities of new supplier is less than the variance of the viscosities of current supplier or not. This means we have to check whether the variance of the viscosities of current supplier is more than the variance of the viscosities of new supplier. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The variances of the viscosities of new and current supplier are same.

Alternative hypothesis: Ha: The variance of the viscosities of current supplier is more than the variance of the viscosities of new supplier.

H0: σ12 = σ22 versus Ha: σ12 > σ22

This is an upper tailed test.

We are given confidence level = 90% = 0.90

So, the level of significance for this test is given as 1 – 0.90 = 0.10.

α = 0.10

The test statistic formula is given as below:

F = S1^2 / S2^2

From given data, we have

N1 = 5

N2 = 5

S1^2 = 0.038 (For new supplier)

S2^2 = 0.022 (For current supplier)

F = 0.038/0.22 = 0.172727273

Test statistic = F = 1.7273

DF1 = N1 – 1 = 5 – 1 = 4

DF2 = N2 – 1 = 5 – 1 = 4

P-value = 0.3047

(by using F-table or excel)

P-value > α = 0.10

So, we do not reject the null hypothesis.

There is not sufficient evidence to conclude that the variance of the viscosities of current supplier is more than the variance of the viscosities of new supplier.

This means, There is not sufficient evidence to conclude that the variance of the viscosities of new supplier is less than the variance of the viscosities of current supplier.

There is not sufficient evidence to conclude that the new supplier is better than the current supplier.


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