## Related questions with answers

In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic in this process and known to follow a normal distribution. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. The observed etch rates are as follows (in mils per minute):

$\begin{array}{cccc} \hline {\text { Solution 1 }} & & {\text { Solution 2 }} \\ \hline 9.9 & 10.6 & 10.2 & 10.0 \\ 9.4 & 10.3 & 10.6 & 10.2 \\ 9.3 & 10.0 & 10.7 & 10.7 \\ 9.6 & 10.3 & 10.4 & 10.4 \\ 10.2 & 10.1 & 10.5 & 10.3 \\ \hline \end{array}$

(a) Do the data support the claim that the mean etch rate is the same for both solutions? In reaching your conclusions, use $\alpha=0.05$ and assume that both population variances are equal. (b) Calculate a P-value for the test in part (a). (c) Find a 95% confidence interval on the difference in mean etch rates. (d) Construct normal probability plots for the two samples. Do these plots provide support for the assumptions of normality and equal variances? Write a practical interpretation for these plots.

Solution

VerifiedGiven:

$\begin{align*} n_1&=\text{Sample size}=10 \\ n_2&=\text{Sample size}=10 \\ c&=\text{Confidence coefficient}=95\%=0.95 \\ \alpha&=\text{Significance level}=5\%=0.05 \end{align*}$

The mean is the sum of all values divided by the number of values:

$\overline{x}_1=\dfrac{9.9+10.6+9.4+...+10.3+10.2+10.1}{10}\approx 9.97$

$\overline{x}_2=\dfrac{10.2+10.0+10.6+...+10.4+10.5+10.3}{10}\approx 10.4$

The variance is the sum of squared deviations from the mean divided by $n-1$. The standard deviation is the square root of the variance:

$s_1=\sqrt{\dfrac{(9.9-9.97)^2+....+(10.1-9.97)^2}{10-1}}\approx 0.4218$

$s_2=\sqrt{\dfrac{(10.2-10.4)^2+....+(10.3-10.4)^2}{10-1}}\approx 0.2309$

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