Final answer:
For the true proportion of students failing comm 215, use the formula CI = proportion ± (Z * sqrt((proportion * (1 - proportion)) / sample size)).
To estimate the sample size required to satisfy specific criteria, use the formulas n = ((Z * standard deviation) / margin of error)^2 and n = ((Z^2 * proportion * (1 - proportion)) / margin of error^2).
Step-by-step explanation:
To find a 95% confidence interval for the true average grade on the comm 215 final exam, we can use the formula:
CI = mean ± (Z * standard deviation)
where CI is the confidence interval, mean is the sample mean, Z is the z-score corresponding to the desired confidence level, and standard deviation is the sample standard deviation.
For part a, we have a sample size of 100, a sample mean of 60%, and a sample standard deviation of 7.5%. The z-score for a 95% confidence level is approximately 1.96. Using the formula, we can calculate the confidence interval as:
CI = 60% ± (1.96 * 7.5%)
CI = 60% ± 14.7%
CI = (45.3%, 74.7%)
Therefore, we can be 95% confident that the true average grade on the comm 215 final exam is between 45.3% and 74.7%.
To find a 95% confidence interval for the true proportion of students failing comm 215 last semester, we can use the formula:
CI = proportion ± (Z * sqrt((proportion * (1 - proportion)) / sample size))
where CI is the confidence interval, proportion is the sample proportion, Z is the z-score corresponding to the desired confidence level, and sample size is the number of observations.
For part b, we have a sample size of 100 and a sample proportion of 7/100 = 0.07. The z-score for a 95% confidence level is approximately 1.96. Using the formula, we can calculate the confidence interval as:
CI = 0.07 ± (1.96 * sqrt((0.07 * (1 - 0.07)) / 100))
CI = 0.07 ± 0.064
CI = (0.006, 0.134)
Therefore, we can be 95% confident that the true proportion of students failing comm 215 last semester is between 0.6% and 13.4%.
For part c, we need to find the sample size required to estimate the true average final exam grade and the true proportion of students not achieving a minimum final exam grade with the desired confidence level and margin of error. Using the formulas:
n = ((Z * standard deviation) / margin of error)^2
n = ((Z^2 * proportion * (1 - proportion)) / margin of error^2)
where Z is the z-score corresponding to the desired confidence level, standard deviation is the sample standard deviation, proportion is the estimated proportion, and margin of error is the desired margin of error.
For the true average final exam grade, we want a margin of error of 5%. The z-score for a 99% confidence level is approximately 2.58. Using the formula, we can solve for the sample size:
n = ((2.58 * 7.5%) / 5%^2)^2
n ≈ 671.95
Therefore, the smallest sample size that can be used to estimate the true average final exam grade to within 5% with 99% confidence is 672.
For the true proportion of students not achieving a minimum final exam grade, we want a margin of error of 0.1. The z-score for a 99% confidence level is approximately 2.58. Using the formula, we can solve for the sample size:
n = ((2.58^2 * 0.07 * (1 - 0.07)) / 0.1^2)
n ≈ 303.58
Therefore, the smallest sample size that can be used to estimate the true proportion of students not achieving a minimum final exam grade to within 0.1 with 99% confidence is 304.