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:
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