Question

In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metallization. The etch rate is an important characteristic of 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. Assume the variances are equal. The etch rates are as follows (in ml per minute):
Solution 1 Solution 2
9.8 10.4 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
(a) Calculate the sample mean for solution 1.

(b) Calculate the sample standard deviation for solution 1.

(c) Calculate the sample mean for solution 2.

(d) Calculate the sample standard deviation for solution 2.

(e) Test the hypothesis
H
0
:
m
1
=
m
2
versus
H
1
:
m
1
â 
m
2
.

Calculate
t
0
.

(f) Calculate a 95% two-sided confidence interval on the difference in mean etch rate.

Answer 1

Answer:

$(a)$ $\bar x_1 = 9.94$

$(b)$ $x_2 = 10.4$

$(c)$ $\sigma_1 = 0.395$

$(d)$ $\sigma_2 = 0.231$

(e) There is sufficient evidence to reject the claim that the mean etch rate is the same for both solutions

$(f)$ $-0.75644 < \mu_1 - \mu_2 < -0.1556$

Explanation:

Given

Solutions 1 and 2

Solving (a): Mean of solution 1

For Solution 1, we have the following data:

$9.8\ 10.4\ 9.4\ 10.3\ 9.3\ 10.0\ 9.6\ 10.3\ 10.2\ 10.1$

The sample mean is calculated as:

$\bar x = \frac{\sum x}{n}$

Where

$n_1 = 10$

So, we have:

$\bar x_1 = \frac{9.8+ 10.4+ 9.4+ 10.3+ 9.3+ 10.0+ 9.6+ 10.3+ 10.2+ 10.1}{10}$

$\bar x_1 = \frac{99.4}{10}$

$\bar x_1 = 9.94$

Solving (b): Mean of Solution 2

For Solution 1, we have the following data:

$10.2\ 10.0\ 10.6\ 10.2\ 10.7\ 10.7\ 10.4\ 10.4\ 10.5\ 10.3$

The sample mean is calculated as:

$\bar x = \frac{\sum x}{n}$

Where

$n_2 = 10$

So, we have:

$x_2 = \frac{10.2+ 10.0+ 10.6+ 10.2+ 10.7+ 10.7+ 10.4+ 10.4+ 10.5+ 10.3}{10}$

$x_2 = \frac{104}{10}$

$x_2 = 10.4$

Solving (c): Sample Standard Deviation of solution 1

This is calculated as:

$\sigma_1 = \sqrt{\frac{\sum (x-\bar x_1)^2}{n_1-1}}$

This gives:

$\sigma_1 = \sqrt{\frac{(9.8 - 9.94)^2+ (10.4- 9.94)^2+ (9.4- 9.94)^2+................+ (10.2- 9.94)^2+ (10.1- 9.94)^2}{10 - 1}}$

$\sigma_1 = \sqrt{\frac{1.404}{9}}$

$\sigma_1 = \sqrt{0.156}$

$\sigma_1 = 0.39496835316$

$\sigma_1 = 0.395$ --- approximated

Solving (d): Sample Standard Deviation of solution 2

This is calculated as:

$\sigma_2 = \sqrt{\frac{\sum (x-\bar x_2)^2}{n_2-1}}$

So, we have:

$\sigma_2 = \sqrt{\frac{(10.2 - 10.4)^2+ (10.0 - 10.4)^2+ (10.6 - 10.4)^2+ (10.2 - 10.4)^2+.......................+ (10.3 - 10.4)^2}{10-1}}$

$\sigma_2 = \sqrt{\frac{0.48}{9}}$

$\sigma_2 = \sqrt{0.05333333333}$

$\sigma_2 = 0.23094010766$

$\sigma_2 = 0.231$

Solving (e): Test the hypothesis

$H_0: \bar x_1 = \bar x_2$

$H_1: \bar x_1 \ne \bar x_2$

Start by calculating pooled standard deviation

$s_p = \sqrt{\frac{(n_1 - 1)*\sigma_1^2 + (n_2 - 1)*\sigma_2^2}{n_1 + n_2 -2}$

$s_p = \sqrt{\frac{(10 - 1)*0.395^2 + (10 - 1)*0.231^2}{10+ 10-2}$

$s_p = \sqrt{\frac{1.884474}{18}$

$s_p = \sqrt{0.104693}$

$s_p = 0.324$

Calculate test statistic

$t = \frac{\bar x_1 - \bar x_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

$t = \frac{9.94 - 10.4}{0.324 * \sqrt{\frac{1}{10} + \frac{1}{10}}}$

$t = \frac{-0.46}{0.324 * \sqrt{0.2}}$

$t = \frac{-0.46}{0.324 * 0.4472}$

$t = \frac{-0.46}{0.1448928}$

$t = -3.175$

Calculate the degree of freedom

$df = n_1 + n_2 - 2$

$df = 10+10 - 2$

$df = 18$

The p value is the in the column title of the student t distribution in row 18

$p = 002621$

The p value is less than the significance level (0.05).

i.e.

$p < 0.05$

So: Reject the null hypothesis

Solving (f): 95% two-sided confidence interval

$c = 95\%$

Calculate the degree of freedom

$df = n_1 + n_2 - 2$

$df = 10+10 - 2$

$df = 18$

Calculate $\alpha$

$\alpha = (1 - c)/2$

$\alpha = (1 - 95\%)/2$

$\alpha = (0.05)/2$

$\alpha = 0.025$

Using student t distribution

$t_{\alpha/2} = 2.101$

Calculate margin of error (E)

$E = t_{\alpha/2} * s_p * \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

$E = 2.101 * 0.324 * \sqrt{\frac{1}{10} + \frac{1}{10}}$

$E = 2.101 * 0.324 * \sqrt{0.2}$

$E = 0.3044$

The boundaries of the confidence are:

$(\bar x_1 - \bar x_2) -E$ $= (9.94- 10.4) - 0.3044 = -0.7644$

$(\bar x_1 - \bar x_2) + E$ $= (9.94- 10.4) + 0.3044 = -0.1556$

The confidence interval is:

$-0.75644 < \mu_1 - \mu_2 < -0.1556$