Question

an educational researcher is interested in estimating the average gpa of all ucf students. the gpa is collected from 150 random students. a 98% confidence interval for μ is: (2.71, 3.01). the researcher wants to reduce the margin of error to 0.1. how many students should be sampled to be 98% confident that the true mean is within 0.1 of the sample mean? assume the standard deviation is 0.85.

Answer 1

To calculate the number of students needed to be sampled to reduce the margin of error to 0.1, we can use the formula n = ((Z * σ) / E)².

Step-by-step explanation:

To calculate the number of students needed to be sampled in order to have a 98% confidence that the true mean is within 0.1 of the sample mean, we can use the formula: n = ((Z * σ) / E)² where n is the sample size, Z is the z-score for the desired confidence level (2.33 for 98% confidence), σ is the standard deviation, and E is the desired margin of error. In this case, the margin of error is 0.1 and the standard deviation is 0.85.

1. Plug in the values into the formula: n = ((2.33 * 0.85) / 0.1)²
2. Solve for n: n = ((1.9795) / (0.1))^2 = 39.5067² = 1565

Therefore, approximately 1565 students should be sampled in order to be 98% confident that the true mean is within 0.1 of the sample mean.

Answer 2

To estimate the average GPA with a margin of error of 0.1 at a 98% confidence level, given the standard deviation of 0.85, the researcher would need to sample approximately 396 students.

Step-by-step explanation:

An educational researcher is interested in estimating the average GPA of all UCF students. The researcher has a 98% confidence interval for the population mean μ, which is currently (2.71, 3.01) based on a sample of 150 students, and wishes to reduce the margin of error to 0.1. Assuming the standard deviation of the GPAs is 0.85, to achieve the desired margin of error, we must first determine the z-score for a 98% confidence level and then use the formula for the required sample size: n = (z * σ / E)^2, where σ is the standard deviation and E is the desired margin of error.

Step 1: Find the z-score for the 98% confidence level

For a 98% confidence level, the z-score is approximately 2.33 (obtained from a standard normal distribution table or a statistical software).

Step 2: Use the formula to calculate the sample size

n = (2.33 * 0.85 / 0.1)^2. When calculated, this results in n being approximately 395.87. Since we cannot have a fraction of a student, we round up to the nearest whole number. Therefore, a sample of 396 students is required to estimate the average GPA with a margin of error of 0.1 at a 98% confidence level.