Question

A broth used to manufacture a pharmaceutical product has its sugar content, in mg/mL, measured several times on each of three successive days. Day 1: 5.0, 4.8, 5.1, 5.1, 4.8, 5.1, 4.8, 4.8, 5.0, 5.2, 4.9, 4.9, 5.0 Day 2: 5.8, 4.7, 4.7, 4.9, 5.1, 4.9, 5.4, 5.3, 5.3, 4.8, 5.7, 5.1, 5.7 Day 3: 6.3, 4.7, 5.1, 5.9, 5.1, 5.9, 4.7, 6.0, 5.3, 4.9, 5.7, 5.3, 5.6

a) Using α = 0.05, can you conclude that the variability of the process is greater on the second day than on the first day?b) Using the f-distribution, see if you can attempt a p-value for this problem

Answer 1

Answer:

Step-by-step explanation:

Hello!

The study variable is:

Xi: sugar level of a broth used to manufacture a pharmaceutical product (mg/mL) measured in day Di

Where i= 1, 2, 3

a) Using a significance level of 5% you need to test if the variability of the process on day 1 is greater than the variability of the process on day 2

The parameter that measures the variability of a given variable is the population variance δ², so in this item, you have to compare both population variances to test if the claim is correct. You have to do an F-ratio test, using the Snedecor-F distribution:

H₀: δ₁² ≤ δ₂²

H₁: δ₁² > δ₂²

α: 0.05

$F_{H_0}= \frac{S^2_1}{S^2_2} *\frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1; n_2-1}$

This test is one-tailed to the right this means that you will reject the null hypothesis to high values of the statistic:

$F_{n_1-1;n_2-1;1-\alpha }= F_{12;12;0.95}= 2.69$

The rejection region is F ≥ 2.69.

Using software I've calculated both sample variances:

Day 1: S₁²= 0.02

Day 2: S₂²= 0.15

$F_{H_0}= \frac{0.02}{0.15} * 1 = 0.13$

Remember, this is an F-ratio test, under the null hypothesis we assume that δ₁²/δ₂²= 1 that is why I've replaced the variance ratio with 1.

b) The p-value for this test is:

P($F_{12;12}$ ≥ 0.13)= 1 - P($F_{12;12}$ < 0.13) = 1 - 0.0006= 0.9994

The p-value is greater than α, the decision is to not reject the null hypothesis.

I hope this helps!