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A recent Gallup Poll asked 4045 randomly chosen American adults to rate the ethical standards of 45 occupations. Nurses were at the top, with 73% of the poll respondents rating them very high. Identify the following

(a) What is the population of interest for this study?
(b) What is the sample for this study?
(c) Is the number 73% considered a parameter or a statistic?
(d) How many of the 4045 American adults in the Gallup poll rated nurses with very high ethical standards?
(e) Estimate the margin of error for this poll using 95% confidence. Round to two decimal places as necessary. (f) Give the 95% confidence interval for this poll. (g) Make a confidence statement about this poll in context.

User Mik
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Final answer:

The population of interest for this study is the entire American adult population. The sample for this study is the 4045 randomly chosen American adults who participated in the Gallup poll. The number 73% is considered a statistic because it represents a percentage derived from the sample data.

Step-by-step explanation:

a) The population of interest for this study is the entire American adult population.

b) The sample for this study is the 4045 randomly chosen American adults who participated in the Gallup poll.

c) The number 73% is considered a statistic because it represents a percentage derived from the sample data.

d) To find out how many of the 4045 American adults in the Gallup poll rated nurses with very high ethical standards, you would multiply 4045 by 0.73 (73%). This gives us an estimate of approximately 2954 adults.

e) To estimate the margin of error for this poll using 95% confidence, you can use the formula:

M = Z * sqrt((p * (1 - p)) / n)

where Z is the z-score for the desired confidence level, p is the proportion, and n is the sample size. In this case, Z is approximately 1.96 (for 95% confidence), p is 0.73, and n is 4045. Plugging these values into the formula, we get:

M = 1.96 * sqrt((0.73 * (1 - 0.73)) / 4045) ≈ 0.015 or 1.5%

So, the estimated margin of error for this poll is approximately 1.5%.

f) The 95% confidence interval for this poll can be calculated as:

p ± M

where p is the proportion (0.73) and M is the margin of error (0.015). Plugging in the values, we get:

0.73 ± 0.015

So, the 95% confidence interval for this poll is approximately (0.715, 0.745).

g) A confidence statement about this poll in context would be: With 95% confidence, we can say that between 71.5% and 74.5% of American adults rate nurses' ethical standards as very high.

User Davidnr
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