In a random sample of 200 people, 74% of them said that they do go to Truecar.com before they go shopping for a vehicle.  Step 1 of 1: Determine a 90% confidence interval for the true proportion of people who do consult Truecar.com before shopping for a vehicle. A) 0.688 to 0.792 (to three decimal places) B) 0.689 to 0.791 (to three decimal places)C) 0.690 to 0.790 (to three decimal places)D) 0.691 to 0.789  (to three decimal places) Step 2 of 2: Determine a 99% confidence interval for the true proportion of people who do consult Truecar.com before shopping for a vehicle.A) (0.662, 0.822)B) (0.661, 0.821)C) (0.663, 0.823)D) (0.660, 0.820)

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Asked Nov 5, 2019
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In a random sample of 200 people, 74% of them said that they do go to Truecar.com before they go shopping for a vehicle. 
 
Step 1 of 1: 
Determine a 90% confidence interval for the true proportion of people who do consult Truecar.com before shopping for a vehicle.
 
A) 0.688 to 0.792 (to three decimal places) 
B) 0.689 to 0.791 (to three decimal places)
C) 0.690 to 0.790 (to three decimal places)
D) 0.691 to 0.789  (to three decimal places)
 
Step 2 of 2: Determine a 99% confidence interval for the true proportion of people who do consult Truecar.com before shopping for a vehicle.
A) (0.662, 0.822)
B) (0.661, 0.821)
C) (0.663, 0.823)
D) (0.660, 0.820)
 
 
 
 
 
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Expert Answer

Step 1

It is given that 74% out of 200 people said that they do go to Truecar.com before they go shopping for a vehicle.

The sample proportion is 0.74.

The sample size is 200.

Step 2

a.

Since the required confidence interval is 90%, the two-tailed z value at 0.05 (=0.10/2) level of significance is 1.645.

The 90% confidence interval is calculated as follows:

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(1-) CI ptz 0.74 (1-0.74) 0.74+(1.645), 200 = 0.74+0.051 = (0.689to 0.791)

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Step 3

Therefore, the correct answer is option B).

b.

Since the required confidence interval is 99%, the two-tailed z value at 0.005 (=0.01/2) ...

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0.74(1-0.74) CI 0.74t(258) 200 =0.74+0.08 (0.660 to 0.820)

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