Answer:
(a) [tex]H_{0}: \mu = 3.00[/tex] vs [tex]H_{1}: \mu > 3.00[/tex] (upper-tail alternative)
(b) We reject the null hypothesis at the significance level of 0.05.
(c) The p-value is 3.167124e-05
Step-by-step explanation:
[tex]H_{0}: \mu = 3.00[/tex] vs [tex]H_{1}: \mu > 3.00[/tex] (upper-tail alternative)
We have [tex]\bar{x} = 3.06[/tex], [tex]\sigma = 0.09[/tex] and n = 36. We have a large sample and our test statistic is
[tex]Z = \frac{\bar{X}-3.00}{\sigma/\sqrt{n}}[/tex] which shoud be normal standard when [tex]H_{0}[/tex] is true. We have observed
[tex]z = \frac{3.06-3.00}{0.09/\sqrt{36}} = 4[/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 RR = {z | z > 1.6449}. Because the observed value 4 is greter than 1.6449, we reject the null hypothesis at the significance level of 0.05.
The p-value is computed as P(Z > 4) = 3.167124e-05
