Question

One thousand randomly selected adult Americans participated in a survey. When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" 52% responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 610 responded that this was often or sometimes okay.

(a)Construct a 90% confidence interval for the proportion of adult Americans who think lying is never justified. (Round your answers to three decimal places.)

( _ , _ )

(b)Construct a 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. (Round your answers to three decimal places.)

( _ , _ )

Answer 1

SOLUTION:

From given data,

One thousand randomly selected adult Americans participated in a survey. When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" 52% responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 610 responded that this was often or sometimes okay.

= 1000

(a)Construct a 90% confidence interval for the proportion of adult Americans who think lying is never justified. (Round your answers to three decimal places.)

= 52% = 52/100 = 0.52

90% confidence interval

Confidence interval is 90%

90% = 90/100 = 0.90

= 1 - Confidence interval = 1-0.90 = 0.10

/2 = 0.10 / 2

= 0.05

z critical value

Z_/2 = Z_0.05 = 1.645

Formula to compute a large-sample confidence interval , is shown below

Z_/2 *

0.52   1.645 *

0.52   1.645 * 0.01579873​​​​​​​

0.52   0.0259889

[ (0.52 -0.0259889) , (0.52+0.0259889) ]

[ 0.4940111 , 0.5459889 ]

[ 0.494 , 0.546 ]

interpret:

Based on the sample , we can be 90% confident that P , the proportion of adult Americans who think lying is never justified is between 0.494 and  0.546

(b) Construct a 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. (Round your answers to three decimal places.)

= 610/1000 = 0.61

90% confidence interval

Confidence interval is 90%

90% = 90/100 = 0.90

= 1 - Confidence interval = 1-0.90 = 0.10

/2 = 0.10 / 2

= 0.05

z critical value

Z_/2 = Z_0.05 = 1.645

Formula to compute a large-sample confidence interval , is shown below

Z_/2 *

0.61   1.645 * ​​​​​​​

0.61   1.645 * 0.0154240072​​​​​​​

0.61   0.02537249

[ (0.61 -0.02537249) , (0.61+0.02537249) ]

[ 0.58462751 , 0.63537249 ]

[ 0.584 , 0.635 ]

interpret:

Based on the sample , we can be 90% confident that P , the proportion of adult Americans who think that it is often or sometimes okay to lie to avoid hunting someone's feelings is between 0.584 and  0.635