The confidence interval is (0.616, 0.686).
To find the confidence interval, we first find p, the proportion of students:
658/1010 = 0.6515
The confidence interval follows the formula
![p\pm z(\sqrt{\frac{p(1-p)}{N}})](https://tex.z-dn.net/?f=p%5Cpm%20z%28%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7BN%7D%7D%29)
To 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 is closer to the z-score 2.33.
Using our information, we have:
![0.6515\pm 2.33(\sqrt{\frac{0.6515(1-0.6515)}{1010}}) \\ \\0.6515\pm 0.0349](https://tex.z-dn.net/?f=0.6515%5Cpm%202.33%28%5Csqrt%7B%5Cfrac%7B0.6515%281-0.6515%29%7D%7B1010%7D%7D%29%0A%5C%5C%0A%5C%5C0.6515%5Cpm%200.0349)
This gives us the interval (0.6515-0.0349, 0.6515+0.0349) or (0.616, 0.686).