Let X equal the thickness of spearmint gum manufactured for vending machines. Assume that the distribution of X is N(mu, sigma^2). The target thickness is 7.5 hundreds of an inch. We shall test the null hypothesis H_0: mu=7.5 against a two-sided alternative hypothesis, using 10 observations. Let m denote the sample mean and S^2 denote the sample variance.
Define the test statistic in terms of m and s.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean for the population is 7.5 or no, the system of hypothesis would be:

Null hypothesis:[tex]\mu =7.5[/tex]

Alternative hypothesis:[tex]\mu \neq 7.5[/tex]

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=(\sqrt{10})\frac{m-7.5}{s}[/tex]

and we have our statistic in terms of m (mean) and the sample standard deviation s.