Respuesta :
Answer:
[tex](a)[/tex] [tex]\bar x_1 = 9.94[/tex]
[tex](b)[/tex] [tex]x_2 = 10.4[/tex]
[tex](c)[/tex] [tex]\sigma_1 = 0.395[/tex]
[tex](d)[/tex] [tex]\sigma_2 = 0.231[/tex]
(e) There is sufficient evidence to reject the claim that the mean etch rate is the same for both solutions
[tex](f)[/tex] [tex]-0.75644 < \mu_1 - \mu_2 < -0.1556[/tex]
Explanation:
Given
Solutions 1 and 2
Solving (a): Mean of solution 1
For Solution 1, we have the following data:
[tex]9.8\ 10.4\ 9.4\ 10.3\ 9.3\ 10.0\ 9.6\ 10.3\ 10.2\ 10.1[/tex]
The sample mean is calculated as:
[tex]\bar x = \frac{\sum x}{n}[/tex]
Where
[tex]n_1 = 10[/tex]
So, we have:
[tex]\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}[/tex]
[tex]\bar x_1 = \frac{99.4}{10}[/tex]
[tex]\bar x_1 = 9.94[/tex]
Solving (b): Mean of Solution 2
For Solution 1, we have the following data:
[tex]10.2\ 10.0\ 10.6\ 10.2\ 10.7\ 10.7\ 10.4\ 10.4\ 10.5\ 10.3[/tex]
The sample mean is calculated as:
[tex]\bar x = \frac{\sum x}{n}[/tex]
Where
[tex]n_2 = 10[/tex]
So, we have:
[tex]x_2 = \frac{10.2+ 10.0+ 10.6+ 10.2+ 10.7+ 10.7+ 10.4+ 10.4+ 10.5+ 10.3}{10}[/tex]
[tex]x_2 = \frac{104}{10}[/tex]
[tex]x_2 = 10.4[/tex]
Solving (c): Sample Standard Deviation of solution 1
This is calculated as:
[tex]\sigma_1 = \sqrt{\frac{\sum (x-\bar x_1)^2}{n_1-1}}[/tex]
This gives:
[tex]\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}}[/tex]
[tex]\sigma_1 = \sqrt{\frac{1.404}{9}}[/tex]
[tex]\sigma_1 = \sqrt{0.156}[/tex]
[tex]\sigma_1 = 0.39496835316[/tex]
[tex]\sigma_1 = 0.395[/tex] --- approximated
Solving (d): Sample Standard Deviation of solution 2
This is calculated as:
[tex]\sigma_2 = \sqrt{\frac{\sum (x-\bar x_2)^2}{n_2-1}}[/tex]
So, we have:
[tex]\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}}[/tex]
[tex]\sigma_2 = \sqrt{\frac{0.48}{9}}[/tex]
[tex]\sigma_2 = \sqrt{0.05333333333}[/tex]
[tex]\sigma_2 = 0.23094010766[/tex]
[tex]\sigma_2 = 0.231[/tex]
Solving (e): Test the hypothesis
[tex]H_0: \bar x_1 = \bar x_2[/tex]
[tex]H_1: \bar x_1 \ne \bar x_2[/tex]
Start by calculating pooled standard deviation
[tex]s_p = \sqrt{\frac{(n_1 - 1)*\sigma_1^2 + (n_2 - 1)*\sigma_2^2}{n_1 + n_2 -2}[/tex]
[tex]s_p = \sqrt{\frac{(10 - 1)*0.395^2 + (10 - 1)*0.231^2}{10+ 10-2}[/tex]
[tex]s_p = \sqrt{\frac{1.884474}{18}[/tex]
[tex]s_p = \sqrt{0.104693}[/tex]
[tex]s_p = 0.324[/tex]
Calculate test statistic
[tex]t = \frac{\bar x_1 - \bar x_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}[/tex]
[tex]t = \frac{9.94 - 10.4}{0.324 * \sqrt{\frac{1}{10} + \frac{1}{10}}}[/tex]
[tex]t = \frac{-0.46}{0.324 * \sqrt{0.2}}[/tex]
[tex]t = \frac{-0.46}{0.324 * 0.4472}[/tex]
[tex]t = \frac{-0.46}{0.1448928}[/tex]
[tex]t = -3.175[/tex]
Calculate the degree of freedom
[tex]df = n_1 + n_2 - 2[/tex]
[tex]df = 10+10 - 2[/tex]
[tex]df = 18[/tex]
The p value is the in the column title of the student t distribution in row 18
[tex]p = 002621[/tex]
The p value is less than the significance level (0.05).
i.e.
[tex]p < 0.05[/tex]
So: Reject the null hypothesis
Solving (f): 95% two-sided confidence interval
[tex]c = 95\%[/tex]
Calculate the degree of freedom
[tex]df = n_1 + n_2 - 2[/tex]
[tex]df = 10+10 - 2[/tex]
[tex]df = 18[/tex]
Calculate [tex]\alpha[/tex]
[tex]\alpha = (1 - c)/2[/tex]
[tex]\alpha = (1 - 95\%)/2[/tex]
[tex]\alpha = (0.05)/2[/tex]
[tex]\alpha = 0.025[/tex]
Using student t distribution
[tex]t_{\alpha/2} = 2.101[/tex]
Calculate margin of error (E)
[tex]E = t_{\alpha/2} * s_p * \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}[/tex]
[tex]E = 2.101 * 0.324 * \sqrt{\frac{1}{10} + \frac{1}{10}}[/tex]
[tex]E = 2.101 * 0.324 * \sqrt{0.2}[/tex]
[tex]E = 0.3044[/tex]
The boundaries of the confidence are:
[tex](\bar x_1 - \bar x_2) -E[/tex] [tex]= (9.94- 10.4) - 0.3044 = -0.7644[/tex]
[tex](\bar x_1 - \bar x_2) + E[/tex] [tex]= (9.94- 10.4) + 0.3044 = -0.1556[/tex]
The confidence interval is:
[tex]-0.75644 < \mu_1 - \mu_2 < -0.1556[/tex]
