Answer:
Given:
The hypothesized mean $(\mu) = 405$
The sample mean $(\overline{x}) = 410$
The sample standard deviation $(s) = 12$
The sample size $(n) = 22$
The significance level $(\alpha) = 0.1$
Solution:
The null and alternative hypothesis:
$H_0: \mu = 405$
$H_a: \mu > 405$
The test statistic $(t):$
$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}$
$= \frac{410 – 405}{\frac{12}{\sqrt{22}}}$
$= 1.954$
The degree of freedom $(df):$
$df = n – 1$
$= 22 – 1$
$= 21$
The p-value:
$\text{The p-value} = P(t > 1.954)$
$= 0.0321$
The final conclusion:
The p-value is less than the significance level. Therefore, there is sufficient evidence to conclude that the bags are overfilled at the 0.1 significance level.
Final Answer:
The null and alternative hypothesis:
$H_0: \mu = 405$
$H_a: \mu > 405$
The test statistic $(t) = 1.954$
The p-value $= 0.0321$
The final conclusion:
The p-value is less than the significance level. Therefore, there is sufficient evidence to conclude that the bags are overfilled at the 0.1 significance level.
