Question

Polls: Sample Size and Margin of Error In June 2012 the National Post ran an article about Canadians being able to buy private insurance for all forms of medically necessary treatment (such as cancer care and heart surgery), which could then be accessed outside the public health care system. The article cited a June 2012 Ipsos poll about Canadians' opinions of this practice, and included the following Ipsos statement about its survey methods: Results based on a simple random sample of 1101 Canadians would have an estimated margin of error of 3.0 percentage points, 19 times out of 20 (or 95% confidence).

(i) Verify the statement by calculation, and find the margin of error from the sample size. Round your margin of error to the nearest whole percentage.

(ii) Suppose that only 500 Canadians comprised the simple random sample reported in this poll. What would be the survey's margin of error? (Use 95% confidence.)

Answer 1

To verify the statement, we calculate the margin of error using the formula, and find the margin of error from the sample size. If the sample size is 1101, the margin of error is approximately 2.7%. If the sample size is 500, the margin of error is approximately 4.4%.

Step-by-step explanation:

To verify the statement, we can calculate the margin of error using the formula:

Margin of Error = (Z * Standard Deviation) / √(Sample Size)

Where Z is the z-score for the desired confidence level. For a 95% confidence level, Z is approximately 1.96. The standard deviation can be estimated using the formula:

Standard Deviation = √((p(1-p)) / n)

Where p is the estimated proportion and n is the sample size.

Plugging the values into the formulas:

Margin of Error = (1.96 * √((0.5*(1-0.5)) / 1101) = 0.027 or 2.7% (rounded to the nearest whole percentage).

If the sample size is 500, the margin of error can be calculated in the same way:

Margin of Error = (1.96 * √((0.5*(1-0.5)) / 500) = 0.044 or 4.4% (rounded to the nearest whole percentage).