Assume that a sample is used to estimate a population proportion
p. Find the 99% confidence interval for a sample of size 356 with
320 successes. Enter your answer as an open-interval (i.e.,
parentheses) using decimals (not percents) accurate to three
decimal places.
99% C.I. =
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
Solution :
Given that,
n = 356
x = 320
Point estimate = sample proportion = = x / n = 320/356=0.899
1 - = 1-0.899=0.101
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 (((0.899*0.101) / 356)
= 0.041
A 99% confidence interval for population proportion p is ,
- E < p < + E
0.899-0.041< p < 0.899+0.041
0.858< p < 0.940
(0.858 , 0.940)
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