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2 years ago

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Julia, the head executive of fine woolen sweaters, inc., is a committed christian who strongly adheres to the ten commandments.

one of julia's employees is found to be stealing sweaters and giving them to a local homeless shelter. julia is likely to

1 answer:

ZanzabumX [31]2 years ago

7 0

<span>We can assume that her devotion to Christian ideals also guide her daily decisions. In this case, Julia's employee has violated the commandment "Thou shalt not steal", and so Julia would likely confront him or her for this transgression. Julia will likely punish the employee in some way, perhaps even firing him or her, even though the employee had selfless intentions in committing the sin.</span>

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3 years ago

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Effectus [21]

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.

$p < 0.05$

<em>So: Reject the null hypothesis</em>

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:

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

4 0

2 years ago