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Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 95% confidence level, hence, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:
Hence the bounds of the interval are found as follows:
The 95% confidence interval is (0.2316, 0.3112).
More can be learned about the z-distribution at brainly.com/question/25890103
First multiply 3 and the 1
Then simplify b^2-4b+3=0
Then divide both numbers by 3 (b-3/3)(b-1/3)=0
If it doesn’t go in even like the second term then bring it before the B
Then equal both terms to 0 and solve
Then simplify b^2-4b+3=0
Then divide both numbers by 3 (b-3/3)(b-1/3)=0
If it doesn’t go in even like the second term then bring it before the B
Then equal both terms to 0 and solve