Final answer:
To achieve a 90% confidence level and a margin of error of 2.5% for studying group habits of statistics students, a sample size of approximately 907 students is required.
Step-by-step explanation:
To determine the sample size required for a statistics study with specific margin of error and confidence level, we can use the formula for estimating a proportion in a population:
n = (Z^2 * p * (1-p)) / E^2,
where:
- Z is the Z-score associated with the desired level of confidence,
- p is the estimated proportion of the attribute present in the population,
- E is the maximum margin of error tolerated.
In this case, you are looking for a 90% confidence level, which corresponds to a Z-score of approximately 1.645 (since 90% confidence leaves 5% in each tail of the standard normal distribution). The estimated proportion (p) of students studying in groups is 0.30 (30%), and the maximum margin of error (E) is 0.025 (2.5%).
Plugging the values into the formula, we get:
n = (1.645^2 * 0.30 * (1-0.30)) / 0.025^2
n = (2.708025 * 0.30 * 0.70) / 0.000625
n = (0.56677275) / 0.000625
n = 906.8364
Therefore, you would need to sample approximately 907 statistics students to meet the requirements of your study parameters.
Note: This formula assumes a simple random sample and a large enough population so that the formula remains valid.