select the correct answer from each drop - down menu. the mean percentage of a population of people eating…

select the correct answer from each drop - down menu. the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5%. assume that a sample size of 40 people was surveyed from the population an infinite number of times. 95% of the sample mean occurs between % and %.

select the correct answer from each drop - down menu. the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5%. assume that a sample size of 40 people was surveyed from the population an infinite number of times. 95% of the sample mean occurs between % and %.

Answer

Explanation:

Step1: Recall the formula for confidence - interval of sample mean

For a 95% confidence - interval of the sample mean when the population standard deviation $\sigma$ is known, the formula is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the population mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size. The $z$ - score for a 95% confidence interval is $z_{\alpha/2}=1.96$.

Step2: Identify the given values

We are given that $\bar{x} = 57%$, $\sigma=3.5%$, and $n = 40$.

Step3: Calculate the margin of error

The margin of error $E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Substitute the values: $\frac{\sigma}{\sqrt{n}}=\frac{3.5}{\sqrt{40}}\approx\frac{3.5}{6.3246}\approx0.5534$. Then $E = 1.96\times0.5534\approx1.0847$.

Step4: Calculate the lower and upper bounds of the confidence - interval

The lower bound is $\bar{x}-E=57 - 1.0847=55.9153\approx55.92%$. The upper bound is $\bar{x}+E=57 + 1.0847=58.0847\approx58.08%$.

Answer:

The lower bound is 55.92% and the upper bound is 58.08%.