In: Math
Q: Suppose a new treatment for a certain disease is given to a sample of 200 patients. The treatment was successful for 164 of the patients. Assume that these patients are representative of the population of individuals who have this disease. Calculate a 98% confidence interval for the proportion successfully treated. (Round the answers to three decimal places.)
A: ___ to ___
Solution :
Given that,
n = 200
x = 164
Point estimate = sample proportion = = x / n = 164 / 200 =
0.82
1 - = 1 - 0.82 = 0.18
At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
Z/2
= Z0.01 = 2.326
Margin of error = E = Z / 2 *
((
* (1 -
)) / n)
= 1.96 * (((0.82 * 0.18) /200 )
= 0.063
A 98% confidence interval for population proportion p is ,
- E < p <
+ E
0.82 - 0.063 < p < 0.82 + 0.063
0.757< p < 0.883
The 98% confidence interval for the population proportion p is : 0.757 to 0.883