Confidence Interval A simple random sample of 56 students who picked up all tests they completed was selected, and the mean score on final exam for this sample of students was 83 with a standard deviation of 10.4. An independent simple random sample of 51 students who had one or more tests that were not picked up was selected, and the mean score on the final exam for this sample of students was 67 with a standard deviation of 24.2. Both distributions are skewed heavily to the left [note: this is the same information provided in question above). If appropriate, use this information to calculate, and interpret a 99% confidence interval for the difference in the mean score on the final exam for all students who picked up all their tests and the mean score on the final exam for all students who had one or more tests that were not picked up. 16 4 points Are the assumptions met? No, because the distributions are heavily skewed to the left. Yes, because both samples are larger than 40 and there are 2 simple random samples Yes, because any sample size will work and there are 2 simple random samples Yes, because only sample sizes of 15 are needed and there are 2 simple random samples, 17 6 points What are the confidence interval values aning 99% confide (11.615, 26.385) (6.2876, 25.712) (13.506, 20.494) (6.568, 25.434) (6.6985, 25.302)