Answer:
We reject the null hypothesis at the significance level of 0.05.
Step-by-step explanation:
[tex]H_{0}: \mu = 6.5[/tex] vs [tex]H_{1}: \mu < 6.5[/tex] (lower-tail alternative)
We have [tex]\bar{x} = 5.89[/tex], [tex]\sigma = 3.26[/tex] and n = 100. We have a large sample and our test statistic is
[tex]Z = \frac{\bar{X}-6.5}{\sigma/\sqrt{n}}[/tex] which is normal standard approximately. We have observed
[tex]z = \frac{5.89-6.5}{3.26/\sqrt{100}} = -1.8712[/tex].
We should use the significance level [tex]\alpha = 0.05[/tex]. The 5th quantile of the standard normal distribution is [tex]z_{0.05} = -1.6449[/tex] and the rejection region is given by RR = {z | z < -1.6449}. Because the observed value -1.8712 is less than -1.6449, we reject the null hypothesis at the significance level of 0.05.
