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

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

`(242-220) +1.988 \sqrt{(20^2)/(47) +(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`

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{(s^2_A)/(n_A) +(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{(20^2)/(47) +(31^2)/(42)}= 10.862`

`(242-220) +1.988 \sqrt{(20^2)/(47) +(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`