For this problem, can we simply jump to the conclusion that since the mean of X is 3, and $Y_i=0.17+0.1X_i$, the mean of $Y_i$ is just 0.3?
2026-03-28 13:23:02.1774704182
Null hypothesis testing
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Expectation is a linear function, so we you can conclude that the sample mean of $Y$ is $\bar{Y} = \frac{1}{n}\sum 0.17+0.1X_i=0.17 + 0.1 \frac{1}{n}\sum X_i=0.17 + 0.1\bar{X}=0.17 + 0.1 \cdot 3=0.47$.