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An Economist/YouGov national poll conducted from July 27-30, 2019, found that 41% of 1,500 randomly sampled Americans approve of the way Donald Trump is handling his job as President.

Task:
1. You are asked to write a newspaper article about the percentage of Americans who approve of the way Donald Trump is handling his job as President. Report the findings with a margin of error at the 95% confidence level.

2. Your coworker proposes the following summary statement for the article:
"An Economist/YouGov national poll was conducted from July 27-30, 2019, to see what proportion of Americans approve of the way Donald Trump is handling his job as President. The poll was conducted online. The margin of sampling error for overall results is plus or minus 2.5 percentage points."
Identify the two pieces of information missing in this statement for approval.

3. Suppose you want a confidence interval at the 99.9% confidence level. Predict if this interval will be wider or narrower than the 95% confidence interval, explain your reasoning, and then calculate a 99.9% interval to find out.

Answer :

The missing information in the proposed summary statement is the actual point estimate of the percentage of Americans who approve of the way Donald Trump is handling his job as President.

The summary statement provides the margin of sampling error but doesn't mention the specific proportion that the margin of error is applied to.

When increasing the confidence level from 95% to 99.9%, the confidence interval will become wider. This is because higher confidence requires more certainty in capturing the true population parameter, and thus a wider interval is needed to accommodate that increased certainty. The trade-off is between confidence and precision.

To calculate the 99.9% confidence interval, you can use the following formula: Confidence Interval = Point Estimate ± Z * (Standard Error)

Where: Point Estimate = 41% (given)

Z = Z-score corresponding to the desired confidence level (99.9%)

Standard Error = sqrt[(Point Estimate * (1 - Point Estimate)) / Sample Size]

For a 99.9% confidence level, the Z-score is approximately 3.29 (looked up from a standard normal distribution table or calculated using statistical software).

Given:Point Estimate = 41%, Sample Size = 1500, Z = 3.29

Calculate Standard Error: Standard Error = sqrt[(0.41 * 0.59) / 1500] ≈ 0.01425

Calculate Margin of Error: Margin of Error = Z * Standard Error ≈ 3.29 * 0.01425 ≈ 0.04689

Calculate Confidence Interval:

Confidence Interval = 41% ± 0.04689 = (36.11%, 45.89%)

So, the 99.9% confidence interval is (36.11%, 45.89%), which is wider than the 95% confidence interval, reflecting the increased confidence at the expense of narrower precision.

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