Question: An experimenter claims that the average yield of extracting protein per 50 kg of seaweed is over 4 kg while using in animal feed. It is known that the yield of extracting protein is normally distributed. To support his claim, a sample of 16 determination of protein extract, each based on 50 kg of seaweed, is analyzed. The results yield a sample mean of 4.2 kg and a sample standard deviation of 0.8 kg.

a. Do the results support the experimenter's claim? Test at 0.1 level of significance.

b. Do we have sufficient evidence to support that the variance of the yield of protein extract per 50 kg of seaweed is less than 1.5? Test at 0.05 level of significance.

Solution: (a) We are interested in testing the following null and alternative hypotheses

\[\begin{aligned}{{H}_{0}}:\mu {\le} {4}\, \\ {{H}_{A}}:\mu {>} {4} \\ \end{aligned}\]

Considering that the population standard deviation \(\sigma\) is not given, we must use a t-test using the following term:

\[t =\frac{\bar{X}-\mu }{s / \sqrt{n}}\]

This corresponds to a right-tailed t-test. The t-statistics is given by the following formula:

\[t=\frac{\bar{X}-\mu }{s /\sqrt{n}}=\frac{{4.2}-4}{0.8/\sqrt{16}}={1}\]

The critical value for \(\alpha = 0.1\) and for \(df = n- 1 = 16 -1 = 15\) degrees of freedom for this right-tailed test is \(t_{c} = 1.34\). The rejection region is given by

\[R = \left\{ t:\,\,\,t>{ 1.34 } \right\}\]

Since we know that \(t = 1 {<} t_c = 1.34\), then we fail to reject the null hypothesis H₀.

Hence, we don't have enough evidence to support the claim that the average yield of extracting protein per 50 kg of seaweed is over 4 kg.