Solution :
Given that,
n = 200
x = 80
Point estimate = sample proportion = = x / n = 80/200=0.4
1 - = 0.6
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576 ( Using z table )
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.576 (((0.4*0.6) /200 )
= 0.089
A 99% confidence interval for population proportion p is ,
- E < p < + E
0.4-0.089 < p < 0.4+0.089
0.311< p < 0.489
lower bound 0.311
upper bound 0.489
9.1.17 Question Help Construct a 99% confidence interval of the population proportion using the given information....
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