The given equation is a linear equation (all variables to the first power) relating the variables x and y. There are an infinite number of values of x and y that will satisfy this equation.

When graphed on an x-y plane, those solution values will fall on a straight line with a slope of 2. It will cross the y-axis at y=32, and the x-axis at x=-16.

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3 years ago

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

IRINA_888 [86]

Answer:

[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

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