Question

Corn: In a random sample of 92 ears of corn, farmer Carl finds that 16 of them have worms. He wants to find the 95% 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 z_α/2) for a 95% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
z_α/2 =

(c) What is the margin of error (E) for a 95% confidence interval? Round your answer to 3 decimal places.
E =

(d) Construct the 95% confidence interval for the proportion of all of Carl's corn that has worms. Round your answers to 3 decimal places.
< p <

(e) Based on your answer to part (d), are you 95% confident that less than 28% of Carl's corn has worms?

Yes, because 0.28 is below the upper limit of the confidence interval.No, because 0.28 is above the upper limit of the confidence interval.     No, because 0.28 is below the upper limit of the confidence interval.Yes, because 0.28 is above the upper limit of the confidence interval.

Answer 1

a)

Point estimate of proportion = = 16 / 92 = 0.174

b)

Z critical value at 0.05 significance level = 1.96 (From Z table)

c)

Margin of error E = Z/2 * sqrt [ ( 1 - ) / n ]

= 1.96 * Sqrt [ 0.174 ( 1 - 0.174) / 92 ]

= 0.077

d)

95% confidence interval for p is

- E < p < + E

0.174 - 0.077 < p < 0.174 + 0.077

0.097 < p < 0.251

e)

We are 95% confident that less than 28% of Carl's corn has worms.

Yes, because 0.28 is above the upper limit of the confidence interval.