Answer:
We reject the null hypothesis at the significance level of 0.05.
Step-by-step explanation:
[tex]H_{0}: \mu = 4.1[/tex] vs [tex]H_{1}: \mu > 4.1[/tex] (upper-tail alternative)
We have [tex]\bar{x} = 4.3[/tex], [tex]\sigma^{2} = 0.64[/tex] and n = 120. We have a large sample and our test statistic is
[tex]Z = \frac{\bar{X}-4.1}{\sigma/\sqrt{n}}[/tex] which is normal standard approximately. We have observed
[tex]z = \frac{4.3-4.1}{0.8/\sqrt{120}} = 2.7386[/tex].
We should use the significance level [tex]\alpha = 0.05[/tex]. The 95th quantile of the standard normal distribution is [tex]z_{0.95} = 1.6449[/tex] and the rejection region is given by {z > 1.6449}. Because the observed value 2.7386 is greater than 1.6449, we reject the null hypothesis at the significance level of 0.05.
