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 m

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 confidence interval for the difference of 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".

Data given and notation

`\bar X_(A)=242` represent the mean for the sample A

`\bar X_(B)=220` represent the mean for the sample B

`s_(A)=20` represent the sample standard deviation for the sample A

`s_(B)=31` represent the sample standard deviation for the sample B

`n_(A)=47` sample size selected A

`n_(B)=42` sample size selected B

`\alpha=0.05` represent the significance level

Confidence =0.95 or 95%

Solution to the problem

For this case the confidence interval for the difference of means
`\mu_A -\mu_B` 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 the confidence interval would be given by

`(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 confidence interval for the difference of means is given by:

`10.862 \leq \mu_A -\mu_B \leq 33.138`