Question

Corn: In a random sample of 84 ears of corn, farmer Carl finds that 7 of them have worms. He wants to find the 90% 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 90% 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 90% confidence interval? Round your answer to 3 decimal places. E = ____

(d) Construct the 90% 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 90% confident that less than 16% of Carl's corn has worms?

Yes, because 0.16 is below the upper limit of the confidence interval.

Yes, because 0.16 is above the upper limit of the confidence interval.

No, because 0.16 is above the upper limit of the confidence interval.

No, because 0.16 is below the upper limit of the confidence interval.

Answer 1

Solution:

Given , n = 84 and x = 7

a)

point estimate for the proportion

= Sample proportion

= x/n

= 7/84

= 0.083

b)

What is the critical value of z (denoted zα/2) for a 90% confidence interval?

Here , c = 90% = 0.90

= 1 - c = 1 - 0.90 = 0.10

/2 = 0.05

= 1.65

(c) What is the margin of error (E) for a 90% confidence interval?

E = /2 *  

= 1.65 * [ 0.083 *(1 - 0.083)/84]

= 0.050

E = 0.050

d)

( - E)   ( + E)

(0.083 - 0.050)   (0.083 + 0.050)

0.033 0.133

e)

Based on your answer to part (d), are you 90% confident that less than 16% of Carl's corn has worms?

Yes, because 0.16 is above the upper limit of the confidence interval.