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2 The following functions are equivalent.

uysha [10]

2x2 − 12x + 21. Axis of symmetry. 2.

8 0

3 years ago

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

IRINA_888 [86]

Answer:

a)

[0.5235, 0.5765]

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

b) 1,843 American adults

Step-by-step explanation:

The 99% confidence interval is given by

$\bf p\pm z^*\sqrt{p(1-p)/n}$

where

p = the proportion of American adults surveyed who said they have watched digitally streamed TV programming on some type of device = 55% = 0.55

$\bf z^*$ the z-score for a 99% confidence level associated with the Normal distribution N(0,1). We can do this given that the sample size (2,341) is big enough

n = sample size = 2,341

We can find the $\bf z^*$ value either with a table or with a spreadsheet.

In Excel use NORM.INV(0.995,0,1)

In OpenOffice Calc use NORMINV(0.995;0;1)

We get a value of $\bf z*$ = 2.576

and our 99% confidence interval is

$\bf 0.55\pm 2.576\sqrt{0.55*0.45/2341}=0.55\pm 2.576*0.0103=0.55\pm 0.265 = [0.5235, 0.5765]$

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

We are 99% confident that this interval contains the true population proportion.

(b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)

The sample size n in a simple random sampling is given by

$\bf n=\frac{(z^*)^2p(1-p)}{e^2}$

where

e is the error proportion = 0.03

hence

$\bf n=\frac{(2.576)^2p(1-p)}{(0.03)^2}=7373.0844p(1-p)=7373.0844p-7373.044p^2$

taking the derivative with respect to p, we get

n'(p)=7373.0844-2*7373.0844p

and

n'(p) = 0 when p=0.5

By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of n

This means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.03

Replacing p with 0.5 in the formula for the sample size we get

$\bf n=7373.0844*0.5-7373.044(0.5)^2=1,844$

rounded up to the nearest integer.

6 0

4 years ago

Read 2 more answers

How to solve inequaities?

bulgar [2K]

Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.
But these things will change direction of the inequality:
Don't multiply or divide by a variable (unless you know it is always positive or always negative)

8 0

4 years ago

If angle knm=(8x-5) and angle mnj=(4x-19) find the measure of angle knm?

vitfil [10]

Answer: knm = 131

Step-by-step explanation:

5 0

3 years ago

Amplitude of 16 units, a period of 5 units, a vertical displacement of 3 units up and a phase shift of 2.5 units left.

allochka39001 [22]

The question is incomplete. The complete question is :

A sine function that has an amplitude of 16 units, a period of 5 units,a vertical displacement of 3 units up and a phase shift of 2.5 units left. Graph this function and show each step.

Solution :

The standard form of the sine function is

y = a sin (bx - c) + d

Here given :

Amplitude, a = 16

Period : $\frac{2 \pi}{|b|} = 5$