487 + 310 ≈ 500 + 300 ≈ 800
Margin of error, e = Z*SD/Sqrt (N), where N = Sample population
Assuming a 95% confidence interval and substituting all the values;
At 95% confidence, Z = 1.96
Therefore,
0.23 = 1.96*1.9/Sqrt (N)
Sqrt (N) = 1.96*1.9/0.23
N = (1.96*1.9/0.23)^2 = 262.16 ≈ 263
Minimum sample size required is 263 students.
Hello!
Let's multiply each thing here.
4(4)=16
3(5)=15
15=16=31
50-31=29
Therefore, her change is $29.
I hope this helps!
