Corn: In a random sample of 84 ears of corn, farmer Carl finds that 14 of them h

Question

Corn: In a random sample of 84 ears of corn, farmer Carl finds that 14 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 z/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
z/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 26% of Carl's corn has worms?

No, because 0.26 is below the upper limit of the confidence interval.Yes, because 0.26 is above the upper limit of the confidence interval.     Yes, because 0.26 is below the upper limit of the confidence interval.No, because 0.26 is above the upper limit of the confidence interval.

Explanation / Answer

a)

Proportion of positive results = P = x/N = 14/84 =0.167

b)

z/2 =   1.645

c)

E = z/2 *[p*(1-p)/N]

= 1.645 *[0.167*(1-0.167)/84]

=0.067

d)

Lower bound = P - E=0.167 -0.067= 0.100

Upper bound = P + E=0.167 +0.067= 0.234

hence

0.100 < p < 0.234

e)

Yes, because 0.26 is above the upper limit of the confidence interval.

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