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The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

8 0

4 years ago

What is the difference between 4Σn=1, 2n+1 and 4Σi=1, (2i+1)?

alina1380 [7]

Answer:

0

Step-by-step explanation:

Each expression is a way to write the sum ...

3 + 5 + 7 + 9

That sum in each case is 24, so the difference is 24-24 = 0.

6 0

3 years ago

Find the midpoint of (1-k,-4) (2,k+1)

satela [25.4K]

The midpoint is k-3/ 1-k

8 0

4 years ago

Which sequence of transformations creates a similar but not congruent triangle?

Hitman42 [59]

Answer:

dilation and translation

Step-by-step explanation:

5 0

3 years ago

Please answer right away!!!

liraira [26]

Answer:

22.9m

Step-by-step explanation:

Using Pythagorean theorem, we can get two equations using the angles.

From Point A:

∠A = 20°

AB = 20m

From Point B:

∠B = 29°

BD = x

What we are looking for is the opposite side of each right triangle, each person makes because we have one adjacent side. We also know that both opposite sides will be equal.

So we use this formula for both point of views:

$Tan\theta=\dfrac{Opposite}{adjacent}$

Where:

Opposite = height of the building

Adjacent = distance from the building

We are looking for the opposite side so we can tweak our formula to get an equation for the height

$height=(Tan\theta)(distance)$

Using our given, we can solve for the distance of point B to D:

$(Tan20)(20+x) = (Tan29)(x)\\\\(0.3640)(20+x) = (0.5543)(x)\\\\\dfrac{(7.28+0.3640x)}{0.5543}=x\\\\13.1337 + 0.6567x = x\\\\13.1337 = x - 0.6567x\\\\13.1337 = 0.3433x\\\\\dfrac{13.1337}{0.3433}=x\\\\38.2572 = x$

The distance of point B to D is 38.2572 m.

Now that we know the distance of BD we can solve for the height of the building using only the given from point B.

$height=(Tan\theta)(distance)$

$height=(Tan29)(38.2572m)$

$height=(0.5543)(38.2572m)$

$height=21.21m$

But this is only the height from the line of sight. To get the height of the building from the ground, we just add the height of the viewer, which is 1.7m.

21.21m + 1.7m = 22.91m

The closest answer is: 22.91 m

8 0

4 years ago