A survey of 50 first time whitewater canoers showed that 23

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: the critical value can be found in the z-table by using probability .95)

B.) assume before collecting data, you want to control the margin of error to be smaller than .2 in this case, at least how many Samples you need to collect?

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: the critical value can be found in the z-table by using probability .95)

B.) assume before collecting data, you want to control the margin of error to be smaller than .2 in this case, at least how many Samples you need to collect?

A) find a 90% confidence interval of the true proportion of canoers who did not wish to canoe the rapids a second time. (Hint: the critical value can be found in the z-table by using probability .95)

B.) assume before collecting data, you want to control the margin of error to be smaller than .2 in this case, at least how many Samples you need to collect?

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]

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Note that

n = z(alpha/2)^2 p (1 - p) / E^2

where

alpha/2 =    0.05


Using a table/technology,

z(alpha/2) =    1.644853627

Also,

E =    0.2
p =    0.46

Thus,

n =    16.80142485

Rounding up,

n =    17   [ANSWER]

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Hi! n = 17 seems to be too small for part b. Are you sure the margin of error is 0.2, or 0.02? If there\'s a typo here, please resubmit the question so we can help you. Thanks!

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

A survey of 50 first time whitewater canoers showed that 23

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