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4 years ago

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Sample size: A researcher is trying to decide how many people to survey. Which of the following sample sizes will result in a co

nfidence interval with the largest width? 300 700 1000 The width depends on the sample proportion.

1 answer:

oksian1 [2.3K]4 years ago

4 0

Answer:

Step-by-step explanation:

Given that a researcher is trying to decide how many people to survey.

We have confidence intervals are intervals with middle value as the mean and on either side margin of error.

Confidence interval = Mean ± Margin of error

Thus confidence interval width depends on margin of error.

Margin of error = $Critical value *\frac{\sigma}{\sqrt{n} }$

Thus for the same confidence level and std deviation we find margin of error is inversely proportional to square root of sample size.

Hence for small n we get wide intervals.

So if sample size = 300, the researcher will get wider confidence interval

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David invested $340 in an account paying an interest rate of 2\tfrac{1}{8}2 8 1 % compounded continuously. Natalie invested $3

Nutka1998 [239]

Answer:

$53.83

Step-by-step explanation:

For David

David invested $340 in an account paying an interest rate of 2\tfrac{1}{8}2 8 1 % compounded continuously.

r = 2 1/8% = 17/8% = 2.125% = 0.02125

t = 17 years

P = $340

For Compounded continuously, the formula =

A = Pe^rt

A = Amount Invested after time t

P = Principal

r = interest rate

t = time

A = $340 × e^0.02125 × 17

A = $ 487.94

For Natalie

Natalie invested $340 in an account paying an interest rate of 2\tfrac{3}{4}2 4 3 % compounded quarterly.

r = 2 3/4 % = 11/4% = 2.75% = 0.0275

t = 17 years

P = $340

n = compounded quarterly = 4 times

Hence,

Compound Interest formula =

A = P(1 + r/n)^nt

A = Amount Invested after time t

P = Principal

r = interest rate

n = compounding frequency

t = time

A = $340 (1 + 0.0275/4) ^17 × 4

A = $ 541.77

After 17 years, how much more money would Natalie have in her account than David, to the nearest dollar?

This is calculated as

$541.77 - $ 487.94

= $53.83

Hence, Natalie would have in her account, $53.83 than David

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0 yards remaining if she used one yard

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A line passes through the point (10,-8) and has a slope of negative 3/2Write an equation in slope-intercept form for this line.

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

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3 years ago