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3 years ago

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What 2 things did Adams hold strong views about

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer: Adams believed that the danger to American society in 1800 came not from excessive authority but from conflict and anarchy and Adams asserted that a good government is a government of laws,

Explanation:

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Use the midpoint method to calculate the price elasticity of demand for potato chips that increased in price from $2.00 to $3.00

fiasKO [112]

Answer:

$e_p$ ≅ -1.7.

Explanation:

Let's calculate the change in quantity:

$\frac{Q_2 - Q_1}{\frac{Q_2+Q_1}{2} } = \frac{50-100}{\frac{50+100}{2} } = (approx.) -0.67.$

Let's now calculate the change in price:

$\frac{P_2 - QP_1}{\frac{P_2+P_1}{2} } = \frac{3.00-2.00}{\frac{3.00+2.00}{2} } = 0.40.$

Finally, let's calculate the price elasticity of demand:

$e_p = \frac{-0.67}{0.40} = (approx) -1.7.$

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2 years ago

Which of the following people came up with the term “hypnosis”?

Over [174]

James Braid.
He got the term "hypnosis" from Hypnos the Greek God of sleep.

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The last hieroglyphs was found at Tikal in what year

vladimir1956 [14]

Answer:

The last hieroglyphs was found at Tikal in

Year 869.

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3 years ago

musickatia [10]

I think the answer is D

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In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalliza

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

3 years ago