Homework SourseConfidence Interval for proportions A survey of 50 firsttime
(Confidence Interval for proportions)
A survey of 50 first-time white-water canoers showed that 23 did not want to repeat the experience.
A) find a 90% confidence interval of the true proportion of canoers who did not wish to canoe the rapids a second time. (Hint: critical value can be found in the Z-table by using probability 0.95).
(Confidence Interval for proportions)
A survey of 50 first-time white-water canoers showed that 23 did not want to repeat the experience.
A) find a 90% confidence interval of the true proportion of canoers who did not wish to canoe the rapids a second time. (Hint: critical value can be found in the Z-table by using probability 0.95).
A survey of 50 first-time white-water canoers showed that 23 did not want to repeat the experience.
A) find a 90% confidence interval of the true proportion of canoers who did not wish to canoe the rapids a second time. (Hint: critical value can be found in the Z-table by using probability 0.95).
Solution
Note that
p^ = point estimate of the population proportion = x / n = 0.46
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.070484041
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
lower bound = p^ - z(alpha/2) * sp = 0.34406407
upper bound = p^ + z(alpha/2) * sp = 0.57593593
Thus, the confidence interval is
( 0.34406407 , 0.57593593 ) [answer]
