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

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 confidence interval for the difference of 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".

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:

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

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 the confidence interval would be given by

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

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

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

$10.862 \leq \mu_A -\mu_B \leq 33.138$