Question

In a study to compare two different corrosion inhibitors, specimens of stainless steel were immersed for four hours in a solution containing sulfuric acid and a corrosion inhibitor. Forty- seven specimens in the presence of inhibitor A had a mean weight loss of 242 mg and a standard deviation of 20 mg, and 42 specimens in the presence of inhibitor B had a mean weight loss of 220 mg and a standard deviation of 31 mg. Find a 95% confidence interval for the difference in mean weight loss between the two inhibitors.

Answer 1

Answer:

$(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862$

$(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138$

And the 95% confidence interval for the difference in the means is given by: $10.862 \leq \mu_A -\mu_B \leq 33.138$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_A = 242$ sample mean for inhibitor A

$s_A = 20$ sample standard deviation for inhibitor A

$n_A = 47$ sample size for A

$\bar X_B = 220$ sample mean for inhibitor B

$s_B = 31$ sample standard deviation for inhibitor B

$n_B = 42$ sample size for A

Solution to the problem

For this case the confidence interval for the difference of means is given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}$

The degrees of freedom are given by:

$df = n_A +n_B -2 = 47+42-2= 87$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,87)".And we see that $t_{\alpha/2}=1.988$

And replacing we got:

$(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862$

$(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138$

And the 95% confidence interval for the difference in the means is given by: $10.862 \leq \mu_A -\mu_B \leq 33.138$