Question

3. A sample of 95 results in 57 successes. Use Table 1. Please do not post if you are unsure. THANK YOU!

a. Calculate the point estimate for the population proportion of successes. (Do not round intermediate calculations. Round your answer to 3 decimal places.)

POINT ESTIMATE

0.600 (CORRECT)

b. Construct 99% and 90% confidence intervals for the population proportion. (Round intermediate calculations to 4 decimal places. Round "z-value" and final answers to 3 decimal places.)

Confidence Level	Confidence Interval		Confidence Interval
99%	0.590 (WRONG)	to	0.610 (WRONG)
90%	0.510 (WRONG)	to	0.680 (CORRECT)

c. Can we conclude at 99% confidence that the population proportion differs from 0.680?

A. No, since the confidence interval does not contain the value 0.680. (wrong)

B. No, since the confidence interval contains the value 0.680.

C. Yes, since the confidence interval does not contain the value 0.680.

D. Yes, since the confidence interval contains the value 0.680.

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d. Can we conclude at 90% confidence that the population proportion differs from 0.680?

A. No, since the confidence interval contains the value 0.680.

B. No, since the confidence interval does not contain the value 0.680.

C. Yes, since the confidence interval contains the value 0.680. (wrong)

D. Yes, since the confidence interval does not contain the value 0.680.

Answer 1

Given that, sample size ( n ) = 95 and x = 57

sample proportion = 57/95 = 0.60

A 90% confidence level has significance level = 0.10 and critical value is,

A 99% confidence level has significance level = 0.01 and critical value is,

b) The 90% confidence interval is,

a) The 99% confidence interval is,

Therefore,

Confidence Level	Confidence Interval		Confidence Interval
99%	0.471	to	0.729
90%	0.517	to	0.683

c) since, 0.680 lies in 99% confidence interval.

Answer: B) No, since the confidence interval contains the value 0.680.

d) Since, 0.680 lies in 90% confidence interval.

Answer: A) No, since the confidence interval contains the value 0.680.