900 high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average was 2.7 , and the standard deviation was 0.4 . what is the margin of error, assuming a 95% confidence level?
Answers
Answer:
0.026
Step-by-step explanation:
0.025.
Given,
n=900
the mean grade-point average =2.7
standard deviation=0.4
To Find,
what is the margin of error.
Solution:
- Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.95 = 0.05
- Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.05/2 = 0.975
- Find the degrees of freedom (df): df = n - 1 = 900 -1 = 899
the crucial value. Since we are unsure about the population standard deviation, we will use a t statistic to indicate the crucial value. For this issue, the t statistic with 899 degrees of freedom and a cumulative probability of 0.975 will be used. The crucial number, according to the t Distribution Calculator, is 1.96.
Next, we find the standard error of the mean: 0.4 / sqrt( 900 ) = 0.4 / 30 = 0.013
- Lastly, we determine the margin of error (ME).
ME = Critical value x Standard error = 1.96 * 0.013 = 0.025
Since there is a 0.025 margin of error, we can say with 95% certainty that the population's mean grade point average is 2.7 with or without a 0.025 error.
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