(a) To construct a confidence interval for the proportion of adult Americans who think lying is never justified, we can use the formula for a confidence interval for a proportion:

CI = p̂ ± z * √( (p̂*(1-p̂))/n )

where:
CI = Confidence Interval
p̂ = Sample Proportion
z = Critical Value (corresponding to the desired confidence level)
n = Sample Size

Given that 56% responded that lying was never justified, p̂ = 0.56. The sample size is 1000, so n = 1000.

First, we need to find the critical value z. Since the confidence level is 90%, we can find the critical value by using the Z-table or a calculator. For a 90% confidence level, the critical value is approximately 1.645.

Now we can plug the values into the formula:

CI = 0.56 ± 1.645 * √( (0.56*(1-0.56))/1000 )

Calculating this, we get:

CI ≈ 0.56 ± 1.645 * 0.014

Simplifying, we get:

CI ≈ 0.56 ± 0.023

So the 90% confidence interval for the proportion of adult Americans who think lying is never justified is approximately (0.537, 0.583).

(b) To construct a confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone's feelings, we can use the same formula as before:

CI = p̂ ± z * √( (p̂*(1-p̂))/n )

Given that 670 responded that this was often or sometimes OK, p̂ = 670/1000 = 0.67. The sample size is still 1000, so n = 1000.

The critical value z for a 90% confidence level is still approximately 1.645.

Now we can plug the values into the formula:

CI = 0.67 ± 1.645 * √( (0.67*(1-0.67))/1000 )

Calculating this, we get:

CI ≈ 0.67 ± 1.645 * 0.012

Simplifying, we get:

CI ≈ 0.67 ± 0.020

So the 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone's feelings is approximately (0.650, 0.690).

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