Question

# Corn: In a random sample of 88 ears of corn, farmer Carl finds that 14 of...

Corn: In a random sample of 88 ears of corn, farmer Carl finds that 14 of them have worms. He wants to find the 99% confidence interval for the proportion of all his corn that has worms.
(a) What is the point estimate for the proportion of all of Carl's corn that has worms? Round your answer to 3 decimal places.

(b) What is the critical value of z (denoted zα/2) for a 99% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
zα/2 =

(c) What is the margin of error (E) for a 99% confidence interval? Round your answer to 3 decimal places.

(d) Construct the 99% confidence interval for the proportion of all of Carl's corn that has worms. Round your answers to 3 decimal places.

(e) Based on your answer to part (d), are you 99% confident that less than 29% of Carl's corn has worms?
No, because 0.29 is above the upper limit of the confidence interval.
Yes, because 0.29 is above the upper limit of the confidence interval.
Yes, because 0.29 is below the upper limit of the confidence interval.
No, because 0.29 is below the upper limit of the confidence interval.

Solution(a)

Point estimate for the proportion of all of the Carl's corn = X/N = 14/88 = 0.159

Solution(b)

Alpha=0.01, alpha/2 = 0.005

Zalpha/2 = 2.575

Solution(c)

Margin of error for 99% confidence interval is

Zalpha/2*sqrt(p*(1-p)/n))

2.575*sqrt(0.159*0.841/88) = 2.575*0.0387 = 0.1

Solution(d)

99% confidence interval is

Point estimate +/- margin of error

0.159 +/-0.1

So 99% confidence interval is

0.059 to 0.259 and 5.9% to 25.9%

Solution(e)

Its answer is A. I.e. No, because 0.29 is above upper limit of the confidence interval.

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