A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a standard deviation of 6 hours, what is the required sample size if the error is to be less than ½ hour with a 95% level of confidence?
First find the allowed standard deviation for the sample average. Then find the number of members of the group to needed reduce the standard deviation to that value.
What is the standard deviation of the following data? If necessary, round your answer to two decimal places.13, 11, 4, 5, 6, 9, 10, 12, 15, 16
393
To determine the required sample size for a given level of confidence and desired margin of error, we can use the following formula:
n = (Z * σ / E)²
Where:
- n is the required sample size
- Z is the z-score corresponding to the desired level of confidence
- σ is the standard deviation
- E is the desired margin of error
In this case, we want a 95% level of confidence, so the z-score would be the critical value that corresponds to a 95% confidence level, which is approximately 1.96. The standard deviation (σ) is given as 6 hours, and the desired margin of error (E) is 0.5 hours.
Now we can plug in these values into the formula:
n = (1.96 * 6 / 0.5)²
n = 23.52²
n ≈ 553.36
Since we can't have a fraction of a student, you would need a sample size of at least 554 students to achieve a margin of error less than 0.5 hours with a 95% level of confidence.