Algebra II question, urgent!!
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The confidence interval is
We first find p, our sample proportion. 118/200 = 0.59.
Next we find the z-score associated with this level of confidence:
Convert 98% to a decimal: 98% = 98/100 = 0.98
Subtract from 1: 1-0.98 = 0.02
Divide by 2: 0.02/2 = 0.01
Subtract from 1: 1-0.01 = 0.99
Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 2.33.
The margin of error (ME) is given by
This gives us the confidence interval
We first find p, our sample proportion. 118/200 = 0.59.
Next we find the z-score associated with this level of confidence:
Convert 98% to a decimal: 98% = 98/100 = 0.98
Subtract from 1: 1-0.98 = 0.02
Divide by 2: 0.02/2 = 0.01
Subtract from 1: 1-0.01 = 0.99
Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 2.33.
The margin of error (ME) is given by
This gives us the confidence interval