Question

In a study of the effect of cooling rate on the hardness of welded joints, 50 welds cooled at a rate of 10°C/s had an average Rockwell (B) hardness of 91.1 and a standard deviation of 6.55, and 40 welds cooled at a rate of 30°C/s had an average hardness of 90.7 and a standard deviation of 4.32.
A. Find a 95% confidence interval for the difference in hardness between welds cooled at the different rates.

B. Someone says that the cooling rate has no effect on the hardness. Do these data contradict this claim? Explain

Answer 1

Answer:

a) The 95% confidence interval would be given by $-1.887 \leq \mu_1 -\mu_2 \leq 2.687$  

b) No contradict the result obtained since the confidence interval contains the value 0, so we don't have enough evidence to conclude that we have a significant effect in the colling rate and the hardness.

Step-by-step explanation:

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".  

$\bar X_1 =91.1$ represent the sample mean 1

$\bar X_2 =90.7$ represent the sample mean 2

n1=50 represent the sample 1 size  

n2=40 represent the sample 2 size  

$s_1 =6.55$ sample standard deviation for sample 1

$s_2 =4.32$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Part a

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =91.1-90.7=0.4$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n_1 +n_2 -1=50+40-2=88$  

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(0.025,88)".And we see that $t_{\alpha/2}=1.987$  

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=1.325$

Now we have everything in order to replace into formula (1):  

$0.4-1.987\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=-1.887$  

$0.4+1.987\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=2.687$  

So on this case the 95% confidence interval would be given by $-1.887 \leq \mu_1 -\mu_2 \leq 2.687$  

Part b

No contradict the result obtained since the confidence interval contains the value 0, so we don't have enough evidence to conclude that we have a significant effect in the colling rate and the hardness.