The principal randomly selected six students to take an aptitude test. their scores were: 81.4 78.6 71.3 77.8 76.9 77.7 determin
1 answer:
The confidence interval is 
.
We first find the mean. Add together all of the data points and divide by 6, the number of data points; the mean is 77.28.
Next we find the standard deviation. Find the difference between each data point and the mean; square it; find the sum; divide by the number of data points; take the square root. The standard deviation is 3.32.
To find the margin of error, we calculate the z-score associated with this level of confidence. 100-90 = 10% = 0.1; 0.1/2 = 0.05; 1-0.05 = 0.95. Using a z-table (http://www.z-table.com) we see that this is between two scores, 1.64 and 1.65; we will use 1.645.
The margin of error is given by
z * (σ/√n) = 1.645*(3.32/√6) = 2.23.
Thus the confidence interval is 77.28 +/ 2.23.
We first find the mean. Add together all of the data points and divide by 6, the number of data points; the mean is 77.28.
Next we find the standard deviation. Find the difference between each data point and the mean; square it; find the sum; divide by the number of data points; take the square root. The standard deviation is 3.32.
To find the margin of error, we calculate the z-score associated with this level of confidence. 100-90 = 10% = 0.1; 0.1/2 = 0.05; 1-0.05 = 0.95. Using a z-table (http://www.z-table.com) we see that this is between two scores, 1.64 and 1.65; we will use 1.645.
The margin of error is given by
z * (σ/√n) = 1.645*(3.32/√6) = 2.23.
Thus the confidence interval is 77.28 +/ 2.23.
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