Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

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Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

process 1, the sample size was 10 and had a mean and standard deviation of 200 and 15, respectively. From process 2, the sample size was 4 with a mean and standard deviation of 300 and 50, respectively.

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-04-11T13:00:38+03:00

The response is:

The 95% CI for the variance in means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Here’s a detailed explanation:

The inquiry lacks sufficient information:

"Determine a 95% confidence interval on the difference in absorbency means from two manufacturing methods. Assume standard deviations derived from the data. Round to two decimal places."

For process 1:

- Sample size: 10

- Mean: 200

- SD: 15

For process 2:

- Sample size: 4

- Mean: 300

- SD: 50

The difference between sample means is:

$M_d=M_1-M_2=200-300=-100$

The estimated standard deviation is:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom calculated are:

$df=n_1+n_2-2=10+4-2=12$

For a 95% confidence interval, the t-value for 12 degrees of freedom is t=±2.179.

So, the confidence interval can be expressed as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$