Ask Question
4 June, 22:44

Dr. Patton is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 A. M. section of 40 students, the mean number of misspelled words was 6.05. Construct a 95% confidence interval for the mean number of misspelled words in the population of student essays. (Round your answers to 3 decimal places.)

+2
Answers (1)
  1. 4 June, 23:12
    0
    (5.294, 6.806) Answer

    Explanation:

    SD = 2.44

    Mean = 6.05

    z for 95% confidence interval = 1.96

    n=40

    Confidence Interval:

    (Mean - (z*SD) / sqrt (n), Mean + (z*SD) / sqrt (n))

    = (6.05 - (1.96*2.44) / sqrt (40), 6.05 + (1.96*2.44) / sqrt (40)))

    = (5.294, 6.806) Answer
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Dr. Patton is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the ...” in 📗 Business if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers