Please help due in 30 min

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 Ro

Studentka2010 [4]

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.

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