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(a) A poll of 2,298 likely voters was conducted on the president’s performance. Approximately what margin of error would the approval rating estimate have if the confidence level is 95%? (Round your answer to 4 decimal places.)  Margin of error  _______            (b) The poll showed that 49 percent approved the president’s performance. Construct a 90 percent confidence interval for the true proportion. (Round your answers to 4 decimal places.)  The 90% confidence interval is from ______  to  _______   (c) Would you agree that the percentage of all voters opposed is likely to be 50 percent?  No, the confidence interval does not contain .50.Yes, the confidence interval contains .50.

Question

(a) A poll of 2,298 likely voters was conducted on the president’s performance. Approximately what margin of error would the approval rating estimate have if the confidence level is 95%? (Round your answer to 4 decimal places.)
  
Margin of error  _______           
 
(b) The poll showed that 49 percent approved the president’s performance. Construct a 90 percent confidence interval for the true proportion. (Round your answers to 4 decimal places.)
  
The 90% confidence interval is from ______  to  _______  
 
(c) Would you agree that the percentage of all voters opposed is likely to be 50 percent?
  

  • No, the confidence interval does not contain .50.
  • Yes, the confidence interval contains .50.
check_circleAnswer
Step 1

(a)

 

The margin of error with confidence level 95% is 0.0204 and it is obtained below:

 

From the given information, there are 2,298 voters. Let us assume that p=0.5.

 

From the standard normal table, the critical value for 0.05 level of significance is 1.96.

Margin of Error = zP1-p)
n
=1.96x0.5(1- 0.5)
2,298
= 1.96x 0.0104
0.0204
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Margin of Error = zP1-p) n =1.96x0.5(1- 0.5) 2,298 = 1.96x 0.0104 0.0204

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Step 2

(b)

 

From the given information, the 49% approved the president’s performance. That is, the p value is 0.49.

 

The 90% confidence interval is from 0.4729 to 0.5071. And it i...

p(1-p)
Confidence Interval = p±Z,
0.49 1.645x0.49(1-49)
2, 298
-0.49 +[1.645 x 0.0104]
-0.49 t0.0171
(0.4729,0.5071)
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Image Transcriptionclose

p(1-p) Confidence Interval = p±Z, 0.49 1.645x0.49(1-49) 2, 298 -0.49 +[1.645 x 0.0104] -0.49 t0.0171 (0.4729,0.5071)

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