Try the fastest way to create flashcards

Related questions with answers

Consider the etch rate data. Test the hypothesis H0:σ12=σ22 against H1:σ12σ22 using α=0.05,H_{0} : \sigma_{1}^{2}=\sigma_{2}^{2} \text { against } H_{1} : \sigma_{1}^{2} \neq \sigma_{2}^{2} \text { using } \alpha=0.05, and draw conclusions. 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 etch solution. The observed etch rates are as follows (in mils/min):

Solutio 1Solution 29.910.610.210.09.410.310.610.29.310.010.710.79.610.310.410.410.210.110.510.3\begin{matrix} \text{Solutio 1} & \quad & \text{Solution 2}\\ \text{9.9} & \text{10.6} & \text{10.2} & \text{10.0}\\ \text{9.4} & \text{10.3} & \text{10.6} & \text{10.2}\\ \text{9.3} & \text{10.0} & \text{10.7} & \text{10.7}\\ \text{9.6} & \text{10.3} & \text{10.4} & \text{10.4}\\ \text{10.2} & \text{10.1} & \text{10.5} & \text{10.3}\\ \end{matrix}

(a) Do the data support the claim that the mean etch rate is the same for both solutions? In reaching your conclusions, use a fixed-level test with α = 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% CI 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.

Question

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):

 Solution 1  Solution 2 9.910.610.210.09.410.310.610.29.310.010.710.79.610.310.410.410.210.110.510.3\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 α=0.05\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

Verified
Step 1
1 of 8

Given:

n1=Sample size=10n2=Sample size=10c=Confidence coefficient=95%=0.95α=Significance level=5%=0.05\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:

x1=9.9+10.6+9.4+...+10.3+10.2+10.1109.97\overline{x}_1=\dfrac{9.9+10.6+9.4+...+10.3+10.2+10.1}{10}\approx 9.97

x2=10.2+10.0+10.6+...+10.4+10.5+10.31010.4\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 n1n-1. The standard deviation is the square root of the variance:

s1=(9.99.97)2+....+(10.19.97)21010.4218s_1=\sqrt{\dfrac{(9.9-9.97)^2+....+(10.1-9.97)^2}{10-1}}\approx 0.4218

s2=(10.210.4)2+....+(10.310.4)21010.2309s_2=\sqrt{\dfrac{(10.2-10.4)^2+....+(10.3-10.4)^2}{10-1}}\approx 0.2309

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Probability and Statistics for Engineers and Scientists 9th Edition by Keying E. Ye, Raymond H. Myers, Ronald E. Walpole, Sharon L. Myers

Probability and Statistics for Engineers and Scientists

9th EditionISBN: 9780321629111 (4 more)Keying E. Ye, Raymond H. Myers, Ronald E. Walpole, Sharon L. Myers
1,204 solutions
Applied Statistics and Probability for Engineers 3rd Edition by Douglas C. Montgomery, George C. Runger

Applied Statistics and Probability for Engineers

3rd EditionISBN: 9780471204541Douglas C. Montgomery, George C. Runger
1,384 solutions
Probability and Statistics for Engineers and Scientists 4th Edition by Anthony J. Hayter

Probability and Statistics for Engineers and Scientists

4th EditionISBN: 9781111827045 (2 more)Anthony J. Hayter
1,341 solutions
Applied Statistics and Probability for Engineers 6th Edition by Douglas C. Montgomery, George C. Runger

Applied Statistics and Probability for Engineers

6th EditionISBN: 9781118539712 (1 more)Douglas C. Montgomery, George C. Runger
2,027 solutions

More related questions

1/4

1/7