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 m

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".

Data given and notation

[tex]\bar X_{A}=242[/tex] represent the mean for the sample A

[tex]\bar X_{B}=220[/tex] represent the mean for the sample B

[tex]s_{A}=20[/tex] represent the sample standard deviation for the sample A

[tex]s_{B}=31[/tex] represent the sample standard deviation for the sample B

[tex]n_{A}=47[/tex] sample size selected A

[tex]n_{B}=42[/tex] sample size selected B

[tex]\alpha=0.05[/tex] represent the significance level

Confidence =0.95 or 95%

Solution to the problem

For this case the confidence interval for the difference of means [tex]\mu_A -\mu_B [/tex] is given by:

[tex] (\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}} [/tex]

The degrees of freedom are given by:

[tex] df = n_A +n_B -2= 47+42-2=87[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]t_{\alpha/2}=1.988[/tex]

And the confidence interval would be given by

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

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

And the confidence interval for the difference of means is given by:

[tex] 10.862 \leq \mu_A -\mu_B \leq 33.138[/tex]