It is known that monthly spending by individuals in a community is normally distributed. To estimate the average monthly spending, a statistician asked 4 people randomly about their monthly spending. The following are the number of dollars these for spent in April: 480, 510, 501, 505. With 98% confidence, over what interval doesthe mean of people's monthly spending lie?
2026-03-22 14:52:39.1774191159
Confidence intervals review
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1
Use a $t$-test. Compute a solution $c$ of the equation $$F(c) = \frac{1}{2}(1 + 0.98)$$ where $F(c)$ is the distribution function of the Student's t-distribution (use a table to solve this equation), and compute the sample std. deviation $s$ and the sample mean ${x}$. The required confidence interval is simply $$\left( x - \frac{sc}{\sqrt{n}}, x + \frac{sc}{\sqrt{n}} \right)$$ where $n$ is the number of samples; here $n=4$.