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3 years ago

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Among 20 golden hamster litters recorded, there was a sample mean of =7.72 baby hamsters, with a sample standard deviation of s=

2.5 hamsters per liter. Create a 98% confidence interval for the mean number of baby hamsters per liter.

1 answer:

Ivan3 years ago

7 0

Answer:

$7.72 -2.539 \frac{2.5}{\sqrt{20}} =6.30$

$7.72 +2.539 \frac{2.5}{\sqrt{20}} =9.14$

Step-by-step explanation:

For this case we have the following info given:

$\bar X= 7.72$ represent the sample mean

$s= 2.5$ represent the sample deviation

$n=20$ represent the sample size

The confidence interval is given by:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The confidence interval is 98% and the significance level is $\alpha=0.02$ the degrees of freedom are given by:

$df= n-1 = 20-1=19$

And the critical value would be:

$t_{\alpha/2}= 2.539$

And replacing we got:

$7.72 -2.539 \frac{2.5}{\sqrt{20}} =6.30$

$7.72 +2.539 \frac{2.5}{\sqrt{20}} =9.14$

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Step-by-step explanation:

<em>To solve polynomial long division problems like these, it's helpful to build a long division table. Getting used to building these can make problems like this much simpler to solve.</em>

Begin by looking at the first term of the cubic polynomial.

What would we have to multiply $4x + 5$ by to get an expression containing $16x^3$ ? The answer is $4x^2$ , since $(4x + 5) \times 4x^2 = 16x^2 + 20x$ .

This is the first step of our long division, and we write out the start of our long division table like this:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,16x^3 + 20x^2\\$

On the left is the divisor. On top is $4x^2$ . In the middle is the polynomial we are dividing, and on the bottom is the result of multiplying our divisor by

The next step is to subtract the bottom expression from the middle one, like so:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,0x^3 - 32x^2\\$

We are left with $-32x^2$ . The next thing to do is to add the next term of the polynomial we are dividing to the bottom line, like this:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\$

Now we return to the beginning of the instructions, and repeat the process: namely, what would we have to multiply $4x + 5$ by to get an expression containing $-32x^2$ ? The answer is $-8x$ , and we fill out our long division table like so:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2 - \,\,\,\,8x\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 - 40x\\$

Once again, we subtract the bottom expression from the one above it, and include the next term of the divisor, like so:

${ }\qquad{ }\qquad{ }\quad{ }4x^2 - \,\,\,\,8x \,+ 15\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{- 32x^2 - 40x}\\{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\,\,\,\,\,60x - 3\\$

And repeat. What do we multiply $4x + 5$ by to get an expression containing $60x$ ? The answer is 15. Our completed long division table looks like this: ${ }\qquad{ }\qquad{ }\quad{ }4x^2 - \,\,\,\,8x \,+15\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{- 32x^2 - 40x}\\{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\,\,\,\,\,60x - 3\\{ }\qquad{ }\hspace{3cm}\,\,\underline{60x + 75}\\{ }\hspace{4.3cm}\,\,-78$

Now, the expression at the top,

$4x^2 - 8x + 20x + 15$

is our quotient, and the last number, $-78$ , is our remainder.

Hence we arrive at the solution of

$\frac{16x^3-12x^2+20x-3}{4x+5} =4x^2 - 8x + 15 - \frac{78}{4x+5}.$

6 0

3 years ago