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In a sample of 144 Canadians, proportion of people who

had income less than $20000 is 0.24. What is 95% condence interval

for the proportion of Canadians that have income less than

$20000.

Answer :

With 95% confidence that the proportion of Canadians with income less than $20000 is between 0.1702 and 0.3098.

We have been given a problem to solve which is all about finding out a 95% confidence interval for the proportion of a whole population, specifically Canadians, which have an income of less than $20000.

Here is a step by step guide on how you can solve the problem:

Step1: Identify the given information from the problem:

a) the sample size. In this case, we know that the size of the sample is 144.

b) the proportion of the sample population who has an income below $20000. This is given as 0.24.

c) The value of the z-score provided for a 95% confidence interval, which is 1.96.

Step2: Discover the standard error of the sample proportion.

The standard error (SE) is a measure of how spread out the proportions are in the sample.

It is obtained from the formula:

SE = sqrt [p(1-p) / n]

Where:

p = sample proportion
n = sample size

Using the provided values,

p = 0.24 and

n = 144, substitute these into the formula for the standard error calculation.

Calculating these values, the standard error comes out to be approximately 0.0356

Step3: The next step is to calculate the confidence interval.

The 95% confidence interval can be calculated using the formula:

CI = p ± (Z*SE)

Where:

Z = Z-score for the desired confidence level
SE = standard error

Substituting the given and calculated values:

Z = 1.96 (from Step 1)
SE = 0.0356 (from Step 2)
p = 0.24 (from Step 1)

Therefore, the 95% confidence interval is 0.24 ± (1.96*0.0356)

This means that the confidence interval falls between 0.1702 and 0.3098.

To know more about standard error visit:

https://brainly.com/question/34299603

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