Answer:
![\bar x_1 = 9.94](https://tex.z-dn.net/?f=%5Cbar%20x_1%20%3D%209.94)
![x_2 = 10.4](https://tex.z-dn.net/?f=x_2%20%3D%2010.4)
![\sigma_1 = 0.395](https://tex.z-dn.net/?f=%5Csigma_1%20%3D%200.395)
![\sigma_2 = 0.231](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%200.231)
(e) There is sufficient evidence to reject the claim that the mean etch rate is the same for both solutions
![-0.75644 < \mu_1 - \mu_2 < -0.1556](https://tex.z-dn.net/?f=-0.75644%20%3C%20%5Cmu_1%20-%20%5Cmu_2%20%3C%20-0.1556)
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](https://tex.z-dn.net/?f=9.8%5C%2010.4%5C%209.4%5C%2010.3%5C%209.3%5C%2010.0%5C%209.6%5C%2010.3%5C%2010.2%5C%2010.1)
The sample mean is calculated as:
![\bar x = \frac{\sum x}{n}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
Where
![n_1 = 10](https://tex.z-dn.net/?f=n_1%20%3D%2010)
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}](https://tex.z-dn.net/?f=%5Cbar%20x_1%20%3D%20%5Cfrac%7B9.8%2B%2010.4%2B%209.4%2B%2010.3%2B%209.3%2B%2010.0%2B%209.6%2B%2010.3%2B%2010.2%2B%2010.1%7D%7B10%7D)
![\bar x_1 = \frac{99.4}{10}](https://tex.z-dn.net/?f=%5Cbar%20x_1%20%3D%20%5Cfrac%7B99.4%7D%7B10%7D)
![\bar x_1 = 9.94](https://tex.z-dn.net/?f=%5Cbar%20x_1%20%3D%209.94)
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](https://tex.z-dn.net/?f=10.2%5C%2010.0%5C%2010.6%5C%2010.2%5C%2010.7%5C%2010.7%5C%2010.4%5C%2010.4%5C%2010.5%5C%2010.3)
The sample mean is calculated as:
![\bar x = \frac{\sum x}{n}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
Where
![n_2 = 10](https://tex.z-dn.net/?f=n_2%20%3D%2010)
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}](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B10.2%2B%2010.0%2B%2010.6%2B%2010.2%2B%2010.7%2B%2010.7%2B%2010.4%2B%2010.4%2B%2010.5%2B%2010.3%7D%7B10%7D)
![x_2 = \frac{104}{10}](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B104%7D%7B10%7D)
![x_2 = 10.4](https://tex.z-dn.net/?f=x_2%20%3D%2010.4)
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}}](https://tex.z-dn.net/?f=%5Csigma_1%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum%20%28x-%5Cbar%20x_1%29%5E2%7D%7Bn_1-1%7D%7D)
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}}](https://tex.z-dn.net/?f=%5Csigma_1%20%3D%20%5Csqrt%7B%5Cfrac%7B%289.8%20-%209.94%29%5E2%2B%20%2810.4-%209.94%29%5E2%2B%20%289.4-%209.94%29%5E2%2B................%2B%20%2810.2-%209.94%29%5E2%2B%20%2810.1-%209.94%29%5E2%7D%7B10%20-%201%7D%7D)
![\sigma_1 = \sqrt{\frac{1.404}{9}}](https://tex.z-dn.net/?f=%5Csigma_1%20%3D%20%5Csqrt%7B%5Cfrac%7B1.404%7D%7B9%7D%7D)
![\sigma_1 = \sqrt{0.156}](https://tex.z-dn.net/?f=%5Csigma_1%20%3D%20%5Csqrt%7B0.156%7D)
![\sigma_1 = 0.39496835316](https://tex.z-dn.net/?f=%5Csigma_1%20%3D%200.39496835316)
--- 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}}](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum%20%28x-%5Cbar%20x_2%29%5E2%7D%7Bn_2-1%7D%7D)
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}}](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%20%5Csqrt%7B%5Cfrac%7B%2810.2%20-%2010.4%29%5E2%2B%20%2810.0%20-%2010.4%29%5E2%2B%20%2810.6%20-%2010.4%29%5E2%2B%20%2810.2%20-%2010.4%29%5E2%2B.......................%2B%20%2810.3%20-%2010.4%29%5E2%7D%7B10-1%7D%7D)
![\sigma_2 = \sqrt{\frac{0.48}{9}}](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%20%5Csqrt%7B%5Cfrac%7B0.48%7D%7B9%7D%7D)
![\sigma_2 = \sqrt{0.05333333333}](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%20%5Csqrt%7B0.05333333333%7D)
![\sigma_2 = 0.23094010766](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%200.23094010766)
![\sigma_2 = 0.231](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%200.231)
Solving (e): Test the hypothesis
![H_1: \bar x_1 \ne \bar x_2](https://tex.z-dn.net/?f=H_1%3A%20%5Cbar%20x_1%20%5Cne%20%5Cbar%20x_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}](https://tex.z-dn.net/?f=s_p%20%3D%20%5Csqrt%7B%5Cfrac%7B%28n_1%20-%201%29%2A%5Csigma_1%5E2%20%2B%20%28n_2%20-%201%29%2A%5Csigma_2%5E2%7D%7Bn_1%20%2B%20n_2%20-2%7D)
![s_p = \sqrt{\frac{(10 - 1)*0.395^2 + (10 - 1)*0.231^2}{10+ 10-2}](https://tex.z-dn.net/?f=s_p%20%3D%20%5Csqrt%7B%5Cfrac%7B%2810%20-%201%29%2A0.395%5E2%20%2B%20%2810%20-%201%29%2A0.231%5E2%7D%7B10%2B%2010-2%7D)
![s_p = \sqrt{\frac{1.884474}{18}](https://tex.z-dn.net/?f=s_p%20%3D%20%5Csqrt%7B%5Cfrac%7B1.884474%7D%7B18%7D)
![s_p = \sqrt{0.104693}](https://tex.z-dn.net/?f=s_p%20%3D%20%5Csqrt%7B0.104693%7D)
![s_p = 0.324](https://tex.z-dn.net/?f=s_p%20%3D%200.324)
Calculate test statistic
![t = \frac{\bar x_1 - \bar x_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Cbar%20x_1%20-%20%5Cbar%20x_2%7D%7Bs_p%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bn_2%7D%7D%7D)
![t = \frac{9.94 - 10.4}{0.324 * \sqrt{\frac{1}{10} + \frac{1}{10}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B9.94%20-%2010.4%7D%7B0.324%20%2A%20%5Csqrt%7B%5Cfrac%7B1%7D%7B10%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D%7D%7D)
![t = \frac{-0.46}{0.324 * \sqrt{0.2}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-0.46%7D%7B0.324%20%2A%20%5Csqrt%7B0.2%7D%7D)
![t = \frac{-0.46}{0.324 * 0.4472}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-0.46%7D%7B0.324%20%2A%200.4472%7D)
![t = \frac{-0.46}{0.1448928}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-0.46%7D%7B0.1448928%7D)
![t = -3.175](https://tex.z-dn.net/?f=t%20%3D%20-3.175)
Calculate the degree of freedom
![df = n_1 + n_2 - 2](https://tex.z-dn.net/?f=df%20%3D%20n_1%20%2B%20n_2%20-%202)
![df = 10+10 - 2](https://tex.z-dn.net/?f=df%20%3D%2010%2B10%20-%202)
![df = 18](https://tex.z-dn.net/?f=df%20%3D%2018)
The p value is the in the column title of the student t distribution in row 18
![p = 002621](https://tex.z-dn.net/?f=p%20%3D%20002621)
The p value is less than the significance level (0.05).
i.e.
![p < 0.05](https://tex.z-dn.net/?f=p%20%3C%200.05)
<em>So: Reject the null hypothesis</em>
Solving (f): 95% two-sided confidence interval
![c = 95\%](https://tex.z-dn.net/?f=c%20%3D%2095%5C%25)
Calculate the degree of freedom
![df = n_1 + n_2 - 2](https://tex.z-dn.net/?f=df%20%3D%20n_1%20%2B%20n_2%20-%202)
![df = 10+10 - 2](https://tex.z-dn.net/?f=df%20%3D%2010%2B10%20-%202)
![df = 18](https://tex.z-dn.net/?f=df%20%3D%2018)
Calculate ![\alpha](https://tex.z-dn.net/?f=%5Calpha)
![\alpha = (1 - c)/2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%281%20-%20c%29%2F2)
![\alpha = (1 - 95\%)/2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%281%20-%2095%5C%25%29%2F2)
![\alpha = (0.05)/2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%280.05%29%2F2)
![\alpha = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.025)
Using student t distribution
![t_{\alpha/2} = 2.101](https://tex.z-dn.net/?f=t_%7B%5Calpha%2F2%7D%20%3D%202.101)
Calculate margin of error (E)
![E = t_{\alpha/2} * s_p * \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}](https://tex.z-dn.net/?f=E%20%3D%20t_%7B%5Calpha%2F2%7D%20%2A%20s_p%20%2A%20%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bn_2%7D%7D)
![E = 2.101 * 0.324 * \sqrt{\frac{1}{10} + \frac{1}{10}}](https://tex.z-dn.net/?f=E%20%3D%202.101%20%2A%200.324%20%2A%20%5Csqrt%7B%5Cfrac%7B1%7D%7B10%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D%7D)
![E = 2.101 * 0.324 * \sqrt{0.2}](https://tex.z-dn.net/?f=E%20%3D%202.101%20%2A%200.324%20%2A%20%5Csqrt%7B0.2%7D)
![E = 0.3044](https://tex.z-dn.net/?f=E%20%3D%200.3044)
The boundaries of the confidence are:
The confidence interval is:
![-0.75644 < \mu_1 - \mu_2 < -0.1556](https://tex.z-dn.net/?f=-0.75644%20%3C%20%5Cmu_1%20-%20%5Cmu_2%20%3C%20-0.1556)