Question

# For a simple random sample of 15 items from a population that is approximately normally distributed, $\bar{x}=82.0$ and $s=20.5$. At the $0.05$ level of significance, test $H_0: \mu \geq 90.0$ versus $H_1: \mu<90.0$.

Solution

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For this exercise, apply the $t$-test formula to describe the sampling distribution of the given mean because the population standard deviation is unknown and the value of sample standard deviation ($s$) is given. Use $s$ to estimate $\sigma$. The $t$-test formula for a sample mean:

\begin{aligned} t = \dfrac{\overline{x} - \mu_0}{s / \sqrt{n}}, \end{aligned}

where $\overline{x}$ represents the sample mean, $\mu_0$ represents the hypothesized population mean, $s$ represents the sample standard deviation and $n$ represents the sample size.

It is given that the sample size is $n=15$, the sample standard deviation is $s = 20.5$, the sample mean is $\overline{x} = 82.0$, and the level of significance is $\alpha = 0.05$.