n an EKOS survey on the 2008 Canadian federal vote intentions, we find the following note at the bottom of the table presenting the results of the survey (the estimated proportions of votes for each political party):

"The field dates for this survey are September 2 to September 4, 2008. In total, a random sample of 2000 Canadians aged 18 and over responded to the survey. A sample of this size provides a margin of error of +/- 3.4 percentage points, 19 times out of 20".

Explain the last sentence of this quote. Write your answer directly in the box.

What is the standard error? Your answer should be in percentage points and have two digits after the period. Your answer should be in percentage points and have two digits after the period. So, if your answer is 50.023 percentage points, enter 50.02.

What would be the margin of error for the 86% C.I., if the sample size was 1500 instead of 2000? Your answer should be in percentage points and have two digits after the period. So, if your answer is 50.023 percentage points, enter 50.02.

The survey has a margin of error of +/- 3.4 percentage points, and for an 86% confidence interval with a sample size of 1500, the margin of error is about 2.6 percentage points.

The last sentence of the quote means that the margin of error for the survey is +/- 3.4 percentage points. This indicates that the estimated proportions of votes for each political party may vary by up to 3.4 percentage points from the actual proportions in the population.

To calculate the standard error, we use the formula:

Standard Error = √(P * (1 - P) / n)

where:

P = estimated proportion (sample proportion)

n = sample size

Given that the estimated proportion (P) is not provided in the quote, we cannot calculate the standard error without additional information.

To find the margin of error for an 86% confidence interval (C.I.) with a sample size of 1500, we can use the formula:

Margin of Error = Z * √(P * (1 - P) / n)

where:

Z = Z-score corresponding to the desired confidence level (86% confidence level in this case)

P = estimated proportion (sample proportion)

n = sample size

For an 86% confidence level, the Z-score is approximately 1.0803 (corresponding to the 43rd percentile of the standard normal distribution).

Assuming P = 0.5 (which gives the maximum margin of error for a given sample size),

Margin of Error = 1.0803 * √(0.5 * (1 - 0.5) / 1500)

Margin of Error ≈ 0.026 or 2.6 percentage points (rounded to two decimal places)

So, the margin of error for the 86% confidence interval with a sample size of 1500 would be approximately 2.6 percentage points.

Learn more about margin of error here:

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