Question

In a random sample of 84 ears of corn, farmer Carl finds that 9 of them have worms. He wants to find the 99% confidence interval for the proportion of all his corn that has worms.

(c) What is the margin of error (E) for a 99% confidence interval?
E =  

(d) Construct the 99% confidence interval for the proportion of all of Carl's corn that has worms.

___ < p < ____

(e) Based on your answer to part (d), are you 99% confident that less than 22% of Carl's corn has worms?

Yes, because 0.22 is below the upper limit of the confidence interval.

No, because 0.22 is below the upper limit of the confidence interval.   

  Yes, because 0.22 is above the upper limit of the confidence interval.

No, because 0.22 is above the upper limit of the confidence interval.

Answer 1

Solution :

Given that,

n = 84

x = 9

Point estimate = sample proportion = = x / n = 9 /84=0.107

1 - = 1 -0.107 =0.893

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576 (((0.107*0.893) /84 )

E = 0.0869

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.107 - 0.0869< p <0.107+ 0.0869

0.0201< p < 0.1939

The 99% confidence interval for the population proportion p is : 0.0201< p < 0.1939