where σ is the standard deviation of the sample data and n is the sample size. to estimate the number of…

where σ is the standard deviation of the sample data and n is the sample size. to estimate the number of times cell phone users check their phone each day, a national polling company surveys a random sample of 2000 cell phone users. the sample has a mean of 52 checks per day and a standard deviation of 11 checks per day. a. find the margin of error. round your answer to the nearest thousandth. about ± 1 0.492 correct answers: 1 0.492 interpret the margin of error. 0 / 10000 word limit b. give an interval that is likely to contain the actual population mean. round your answers to the nearest thousandth. the interval is between 1 and 2 51.508 and 52.492 correct answers: 1 51.508 2 52.492

where σ is the standard deviation of the sample data and n is the sample size. to estimate the number of times cell phone users check their phone each day, a national polling company surveys a random sample of 2000 cell phone users. the sample has a mean of 52 checks per day and a standard deviation of 11 checks per day. a. find the margin of error. round your answer to the nearest thousandth. about ± 1 0.492 correct answers: 1 0.492 interpret the margin of error. 0 / 10000 word limit b. give an interval that is likely to contain the actual population mean. round your answers to the nearest thousandth. the interval is between 1 and 2 51.508 and 52.492 correct answers: 1 51.508 2 52.492

Answer

Explanation:

Step1: Identify the formula for margin of error

For a large - sample (when $n\geq30$), the margin of error $E$ is given by $E = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the sample standard deviation and $n$ is the sample size. Here, $\sigma = 11$ and $n=2000$.

Step2: Calculate the margin of error

Substitute the values into the formula: $E=\frac{11}{\sqrt{2000}}$. First, calculate $\sqrt{2000}\approx44.721$. Then, $E=\frac{11}{44.721}\approx0.492$.

Step3: Calculate the confidence interval

The confidence interval for the population mean $\mu$ is given by $\bar{x}-E<\mu <\bar{x} + E$, where $\bar{x}$ is the sample mean. Given $\bar{x}=52$ and $E = 0.492$, we have $52 - 0.492=51.508$ and $52+0.492 = 52.492$.

Answer:

a. $0.492$ b. $51.508$ $52.492$