Question

1. a survey of 4213 people found that they exercise on average for 1.4 hours each week. the standard deviation was 0.3 hours. what is the margin of error in the sample mean? based upon the sample mean, a realistic expectation of the mean is within what interval?

1. a survey of 56 random people found that they go to the movies 2.5 times/month. the standard deviation is 0.7. what is the margin of sampling error? it is reasonable to think

Explanation:

Problem 1

Step1: Recall Margin of Error Formula

For a sample mean, the margin of error (ME) for a 95% confidence interval (a common choice) is approximately \( ME = 1.96\times\frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation and \( n \) is the sample size. Here, \( \sigma = 0.3 \), \( n = 4213 \).
First, calculate \( \frac{\sigma}{\sqrt{n}}=\frac{0.3}{\sqrt{4213}}\approx\frac{0.3}{64.91}\approx0.0046 \).
Then, \( ME\approx1.96\times0.0046\approx0.0090 \).

Step2: Find Confidence Interval

The confidence interval is \( \bar{x}\pm ME \), where \( \bar{x} = 1.4 \).
Lower bound: \( 1.4 - 0.0090 = 1.391 \)
Upper bound: \( 1.4 + 0.0090 = 1.409 \)

Step1: Apply Margin of Error Formula

Using \( ME = 1.96\times\frac{\sigma}{\sqrt{n}} \), with \( \sigma = 0.7 \), \( n = 56 \).
Calculate \( \frac{\sigma}{\sqrt{n}}=\frac{0.7}{\sqrt{56}}\approx\frac{0.7}{7.483}\approx0.0935 \).
Then, \( ME\approx1.96\times0.0935\approx0.183 \).

Step2: Determine Confidence Interval

The sample mean \( \bar{x} = 2.5 \).
Lower bound: \( 2.5 - 0.183 = 2.317 \)
Upper bound: \( 2.5 + 0.183 = 2.683 \)

Answer:

Margin of error ≈ 0.009, Interval: (1.391, 1.409) (assuming 95% confidence)

Problem 2