Anyone know what 1.27 repeating is as a fraction??

A random sample of 1014 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a l

arlik [135]

Answer:

a) $\hat p=\frac{538}{1014}=0.531$ estimated proportion of people indicated that televisions are a luxury they could do without

b) Verify that the requirements for constructing a confidence interval about p are satisfied. The sample is a simple random sample, the value of np(1-p)=1014*0.531*(1-0.531)=252.53 is, which is > 10, and the less than or equal to 5% of the population.

c) The 95% confidence interval would be given by (0.500;0.562)

B. We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.500 and 0.562

d) We can say that is possible that more than 60% of adult Americans believe that television is a luxury they could do without because we need to remember that the confidence interval is just a point of estimation and iits probable that the true proportion would be not contained on the interval. But we can say that is not that likely that the true proportion is greater than 60%

e) The 95% confidence interval would be given by (0.438;0.500)

Step-by-step explanation:

Notation and definitions

$X=538$ number of people indicated that televisions are a luxury they could do without

$n=1014$ random sample taken

$\hat p=\frac{538}{1014}=0.531$ estimated proportion of people indicated that televisions are a luxury they could do without

$p$ true population proportion of people indicated that televisions are a luxury they could do without

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

a) $\hat p=\frac{538}{1014}=0.531$ estimated proportion of people indicated that televisions are a luxury they could do without.

b) Verify that the requirements for constructing a confidence interval about p are satisfied. The sample is a simple random sample, the value of np(1-p)=1014*0.531*(1-0.531)=252.53 is, which is > 10, and the less than or equal to 5% of the population.

c) In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.531 - 1.96\sqrt{\frac{0.531(1-0.531)}{1014}}=0.500$

$0.531 + 1.96\sqrt{\frac{0.531(1-0.531)}{1014}}=0.562$

The 95% confidence interval would be given by (0.500;0.562)

B. We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.500 and 0.562

d) We can say that is possible that more than 60% of adult Americans believe that television is a luxury they could do without because we need to remember that the confidence interval is just a point of estimation and iits probable that the true proportion would be not contained on the interval. But we can say that is not that likely that the true proportion is greater than 60%

e) The new propotion on this case would be

$\hat p =1-0.531=0.469$

$0.469 - 1.96\sqrt{\frac{0.469(1-0.469)}{1014}}=0.438$

$0.469 + 1.96\sqrt{\frac{0.469(1-0.469)}{1014}}=0.500$

The 95% confidence interval would be given by (0.438;0.500)

4 0

4 years ago

Which line is the best model for the data in the scatter plot? A scatterplot with an x axis labeled, variable A from twenty to s

Firlakuza [10]

The line which is the best model for the data in a scatter plot is: D. A scatterplot with an x axis labeled, variable A from twenty to sixty in increments of ten and the y axis labeled, variable B from twenty to forty in increments of five with a negative trend line through the middle of eight points.

<h3>What is a scatter plot?</h3>

A scatter plot can be defined as a graph which uses Cartesian coordinates (x-axis and y-axis) to display the values of two (2) variables, with the resulting points revealing any correlation between the data.

<h3>What is a trend line?</h3>

A trend line refers to an statistical (analytical) tool that is used in conjunction with a scatter plot, so as to determine whether or not there's any correlation between a data.

For example, a scatterplot with an x-axis with its points ranging from twenty (20) to sixty (60) in increments of ten (10) while the y-axis has its points ranging from twenty (20) to forty (40) in increments of five (5) with a negative trend line through the middle of eight (8) points i.e above and below the trend line.

Read more on scatterplot here: brainly.com/question/6592115

7 0

2 years ago

What is the value of the ratio of the surface area to the volume of the cylinder? 0.38 2.66 0.94 0.7

yKpoI14uk [10]

Answer:

the correct answer is 0.38, I used the other persons answer and got it wrong.

Step-by-step explanation:

3 0

4 years ago

The equation for the circle below is x2 + y2 = 100. what is the length of the circle's radius?

stealth61 [152]

X²+y² =(Radius)²

x²+y² =10². So the radius is 10

4 0

3 years ago

Read 2 more answers

Eduardo is building a sandbox that has an area of 84 square feet. What are the possible whole number measurements for the length

kicyunya [14]

Try this option:
1. note, that the number 84 is divided by numbers: 1;2;3;4;6;7;12;14;21;28;42 and 84. It means, that measurements of the length may be: 1;2;3;4;6;7;12;14;21;28;42;84 and for the width: 84;42;28;21;14;12;7;6;4;3;2;1.
2. according the numbers calculated in item 1, the possible number measurements for the length and width of the sandbox are (length; width):
1;84
2;42
3;28
4;21
6;14
7;12
12;7
14;6
21;4
28;3
42;2
84;1

4 0

4 years ago

Read 2 more answers