In a poll, 408 of 1020 randomly selected adults aged 18 or older stated that the

Question

In a poll, 408 of 1020 randomly selected adults aged 18 or older stated that they believe there is too little spending on national defense.

Use this information to complete the following questions.

(c.) Construct a 95% confidence interval for the proportion of adults aged 18 or older who believe there is too little spending on national defense.

answer: 0.370, 0.430.

(e) Use the results of part (c) to construct a 95% confidence interval for the population proportion of adults aged 18 or older who do not believe there is too little spending on national defense. (Round to three decimal places as needed.)

The 95% confidence interval is (_____,_____)

Explanation / Answer

we know that the confidence interval of proportion is given as

p +- Z*SEp

where SEp = sqrt[ p * ( 1 - p ) / n ]

now n = 1020 and p = 408/1020 = 0.4

and SEp = sqrt[ p * ( 1 - p ) / n ]

=sqrt[ 0.4* ( 1 - 0.4 ) / 1020 ] = 0.0153

also from the z table we see that the z for 95% CI is 1.96

so

0.4+- 1.96*0.0153, solving for positive and negative signs we get the answer as

0.37 and 0.43

d)

p in this case would be

1020-408 = 612

so p in this case would be 612/1020 = 0.6

and SEp = sqrt[ p * ( 1 - p ) / n ]

=sqrt[ 0.6* ( 1 - 0.6 ) / 1020 ] = 0.0153

also from the z table we see that the z for 95% CI is 1.96

so

0.6+- 1.96*0.0153, solving for positive and negative signs we get the answer as

0.57 and 0.63

Hope this helps !!

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