In a recent poll, 370 people were asked if they liked dogs, and 7% said they did. find the margin of error of this poll, at the 90% confidence level. give your answer to three decimals.

Step-by-step explanation:

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".  

represent the real population proportion of interest

represent the estimated proportion for the sample

n=370 is the sample size required (variable of interest)

represent the critical value for the margin of error

The population proportion have the following distribution  

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 90% of confidence, our significance level would be given by and . And the critical value would be given by:  

The margin of error for the proportion interval is given by this formula:  

(a)  

And replacing into formula (a) the values provided we got:

The margin of error on this case would be ME=0.0218

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

1696 hundred thousand

step-by-step explanation: