One question that typically hunts students of basic statistics when attempting to solve a hypothesis testing question, be it from a homework or a test, is how to assess what type of tail a hypothesis test has.

The problem of determining the type of tail is simply reduced to the correct specification of the null and alternative hypothesis. One has correctly determined the hypotheses for a test, the problem of knowing what type of tail is the correct one (right-tailed, left-tailed or two-tailed) is simple.

In order to see the type of tail, we need to look at the alternative hypothesis. If the sign in the alternative hypothesis is "<", then we have a left-tailed test. Or if the sign in the alternative hypothesis is ">", then we have a right-tailed test. Or, on the other hand, if sign in the alternative hypothesis is "≠", then we have a two-tailed test.

__LET US CONSIDER THE FOLLOWING EXAMPLE__:

Assume that a simple random sample of the weights of 19 green M&Ms has a mean of 0.8635 grams, and also assume that the population standard deviation \(\sigma\) is known to be 0.0565 g. Let us use a 0.05 significance level to test the claim that the mean weight of all green M&Ms is equal to 0.8535 g, which is the mean weight required so that M&Ms have the weight printed on the package label. Do green M&Ms appear to have weights consistent with the package label?

__This is how we solve it__

We want to test the following null and alternative hypotheses

\[\begin{align}{{H}_{0}}:\mu {=} {0.8535}\, \\ {{H}_{A}}:\mu {\ne} {0.8535} \\ \end{align}\]

Given that the population standard deviation is known, with \(\sigma = 0.0565\) we use the normal distribution. The z-statistic is calculated as

\[z =\frac{\bar{X}-\mu }{\sigma / \sqrt{n}}\]

We know that this is two-tailed z-test (since the sign in the alternative hypothesis is "≠").

The z-statistics is computed by the following formula:

\[z =\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}=\frac{{0.8635}-0.8535}{0.0565 /\sqrt{19}}={0.7715}\]

The critical value for \(\alpha = 0.05\) for this two-tailed test found to be \(z_{c} = {1.96}\). The rejection region is corresponds to

\[R=\left\{ z:\,\,\,|z|>{1.96} \right\}\]

Since \(|z| = 0.7715 {<} z_c = 1.96\), then we fail to reject the null hypothesis H_{0}.

*Thus, we don’t have enough evidence to reject the claim green M&Ms appear to have weights consistent with the package label.*

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