# Normal Population Mean Test

When the population mean and variance are both unknown and you want to test whether the mean is $\mu_0$ (in this case $\mu_0 = 0$), you typically use the t-test.

For a single sample t-test:

**Null and alternative hypotheses**: $H_0: \mu = 0$ $H_1: \mu \neq 0$ (This is for a two-sided test. For a one-sided test, the alternative hypothesis would be $\mu > 0$ or $\mu < 0$ depending on the test direction).**Compute the sample mean and standard deviation**: Let $\bar{x}$ be the sample mean and $s$ be the sample standard deviation.**Compute the t-statistic**: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ Where $n$ is the sample size.**Determine the degrees of freedom**: $df = n - 1$**Find the critical value(s)**: Using a t-distribution table (or a statistical software package), find the critical value for the desired significance level (often $\alpha = 0.05$) and the given degrees of freedom. For a two-sided test at $\alpha = 0.05$, you'd typically look up the critical value for $\alpha = 0.025$ in each tail.**Decision**:- For a two-sided test: If $|t| >$ critical value, reject $H_0$.
- For a one-sided test (where $H_1: \mu > 0$): If $t >$ critical value, reject $H_0$.
- For a one-sided test (where $H_1: \mu < 0$): If $t < -$ critical value, reject $H_0$.

**Compute the p-value**(optional but often desired): This is the probability of observing a value as extreme as, or more extreme than, the observed statistic under the null hypothesis. You can find the p-value using t-distribution tables or statistical software.**Conclusion**: Based on the comparison of the t-statistic to the critical value and/or the p-value to your significance level, make a decision to reject or fail to reject the null hypothesis. Report your results.

Remember, failing to reject the null hypothesis doesn't mean the population mean is definitively 0, just that there's not enough evidence from your sample to suggest otherwise. Similarly, rejecting the null hypothesis doesn't prove the population mean isn't 0, but suggests it's unlikely to be 0 based on your sample data.