Answer:
<u>The correct answer is B. between 56.45% and 57.55%
</u>
Complete statement and question:
The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?
between 55.89% and 58.11%
between 56.45% and 57.55%
between 56.54% and 57.46%
between 56.07% and 57.93%
Source: brainly.com/question/1068489
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
Mean percentage of a population of people eating out at least once a week = 57%
Standard deviation = 3.5%
Sample size = 40
Confidence level = 68%
2. In which interval will 68% of the sample means occur?
For answering this question, we should find out the standard deviation of the sample, using this formula:
Standard deviation of the sample = Standard deviation of the population/√Sample size
Standard deviation of the sample = 3.5/√40
Standard deviation of the sample = 3.5/6.32
Standard deviation of the sample = 0.55
Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean + 1 time the standard deviation of the sample. Thus:
57 - 1 * 0.55 = 57 - 0.55 = 56.45
57 + 1 * 0.55 = 57 + 0.55 = 57.55
<u>The correct answer is B. between 56.45% and 57.55%
</u>
