Data from a representative sample were used to estimate that 31% of all computer users in a recent year had tried to get on a Wi-Fi network that was not their own in order to save money. You decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05

The required sample size is 329.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

31% of all computer users in a recent year had tried to get on a Wi-Fi network that was not their own in order to save money.

This means that

Confidence level is not given, so we assume 95%.

95% confidence level

So , z is the value of Z that has a pvalue of , so .

What is the required sample size if you want to estimate this proportion with a margin of error of 0.05?

We need a sample size of n. n is found when M = 0.05. So

Rounding up

The required sample size is 329.

y = 5x + 17

step-by-step explanation:

a perpendicular line will be the opposite reciprocal slope of the given equation.   the slope given is -1/5 so the opposite reciprocal is 5

find b:

7 = 5 (-2) + b

7 = -10 + b

17 = b

write equation

y = 5x + 17

sorry man i accidentally pressed yes my bad

step-by-step explanation: