Question

According to Thomson Financial, last year the majority of companies reporting profits had beaten estimates. A sample of 162 companies showed that 110 beat estimates, 29 matched estimates, and 23 fell short. (a) What is the point estimate of the proportion that fell short of estimates? If required, round your answer to four decimal places. pshort = (b) Determine the margin of error and provide a 95% confidence interval for the proportion that beat estimates. If required, round your answer to four decimal places. ME = (c) How large a sample is needed if the desired margin of error is 0.05? If required, round your answer to the next integer. n* =

Answer 1

Answer:

0.1790

0.0738 , 0.5800, 0.7040

354

Step-by-step explanation:

a. Given:  

n = 162

x = 29

c = 95%

The point estimate of the population proportion is the sample proportion. The sample proportion is the number of successes divided by the sample size:  

p =  x/n

  = 29/162

  = 0.1790

b. Given:  

n=162

x = 110

c = 95%

The point estimate of the population proportion is the sample proportion. The sample proportion is the number of successes divided by the sample size:  

p =  x/n

  =110/162

  =0.6790

For confidence level 1 –$\alpha$ = 0.95, determine z_$\alpha$/2 = z_0.025 using table 1 (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):  

z_$\alpha$/2 = 1.96  

The margin of error is then:

E =  z_$\alpha$/2*√p(1-p)/n =1.96* √0.6790(1-0.6790)/162 =0.0738

The boundaries of the confidence interval are then:  

p-z_$\alpha$/2*√p(1-p)/n = 0.6790-1.645√ 0.6790(1- 0.6790)/162 = 0.5800

p+z_$\alpha$/2*√p(1-p)/n = 0.6790+1.645√ 0.6790(1- 0.6790)/162 = 0.7040

c. Given:  

n = 162

x = 110

c = 95%

The point estimate of the population proportion is the sample proportion. The sample proportion is the number of successes divided by the sample size:  

p =x/n

  =0.6790

Formula sample size:

p known: n = [z_$\alpha$/2 ]^2*pq/E^2

                   = [z_$\alpha$/2 ]^2*p(1-p)/E^2

                n = [z_$\alpha$/2 ]^2*0.25/E^2

For confidence level 1 –$\alpha$= 0.95, determine z_$\alpha$/2 = z_o.025 using table 1 (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):  

z_$\alpha$/2  = 1.96

p is known, then the sample size is (round up!):

n =[z_$\alpha$/2 ]^2*p(1-p)/E^2

  = 354