What is the area of this rectangle

Naddika [18.5K]

Wait, people still play Fortnite?

3 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

What is the surface area of the rectangular prism below 7, 12, 14

Rus_ich [418]

What is the surface area of the rectangular prism below 7, 12, 14

Formula to find the surface area of the rectangular prism is:

2wl + 2lh + 2hw,

where w is width, l is length, and h is height of the rectangular prism.

Given dimension of the prism is 7, 12, 14.

So, w=7, l=12 and h= 14.

First step is to plug in the above values of l, w and h in the above formula to get the surface area. So,

Surface area of the rectangular prism =2wl + 2lh + 2hw

=2*7*12 + 2*12*14 + 2*14*7

=168+336+196

= 700

So, the surface area of the rectangular prism is 700 units².

3 0

4 years ago

Work out the area of this shape. ? cm2 4 cm 9 cm 6 cm 12 cm Help!!!

mars1129 [50]

Answer:

84 cm^2

Step-by-step explanation:

Area of triangle + area of rectangle

Area of triangle = 1/2*l*b

= 1/2*8*3= 12 cm^2

Area of rectangle = l*b

= 12*6 = 72 cm^2

Total area= 72+12

= 84 cm^2

7 0

3 years ago

A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs

Makovka662 [10]

Answer:

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be 95% confident that the mean cost, μ, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this interval, but we can be 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be 95% confident that the mean cost, μ,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this interval, but we can be 95% confident that it does.

Hope this Helps!!!

4 0

3 years ago