Homework Sourse1 A grower believes that one in five of his citrus trees are
1) A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken if the grower wishes to estimate the proportion of his trees that are infected with citrus red mite to within 0.07 with probability 0.95?
2) A random sample of n = 400 observations from a binomial population produced x = 161 successes. Find a 90% confidence interval for p.
from....to....
3) A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. They expect that about 75% of the student body would respond favorably.
(a) What sample size is required to obtain a 95% confidence interval with an approximate margin of error of 0.046?
(b) Suppose that 55% of the sample responds favorably. Calculate the margin of error for the 95% confidence interval.
Solution
1.
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
E = 0.07
p = 0.2
Thus,
n = 125.4353901
Rounding up,
n = 126 [ANSWER]
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2.
Note that
p^ = point estimate of the population proportion = x / n = 0.4025
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.024520081
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
Margin of error = z(alpha/2)*sp = 0.040331944
lower bound = p^ - z(alpha/2) * sp = 0.362168056
upper bound = p^ + z(alpha/2) * sp = 0.442831944
Thus, the confidence interval is
( 0.362168056 , 0.442831944 ) [ANSWER]
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