Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points. The instr
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Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).
If we had increased the confidence level, the margin of error also would have increased.
<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. Increasing the confidence level, z also increases, hence the margin of error also would have increased.
The sample size and the estimate are given as follows:
.
The lower and the upper bound of the interval are given, respectively, by:
The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).
More can be learned about the z-distribution at brainly.com/question/25890103
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