Answer:
Given:
The hypothesized mean $(\mu) = 408$
The sample mean $(\overline{x}) = 413$
The sample standard deviation $(s) = 12$
The sample size $(n) = 23$
The significance level $(\alpha) = 0.1$
Solution:
The null and alternative hypothesis:
$H_0: \mu = 408$
$H_a: \mu > 408$
The test statistic $(t):$
$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}$
$= \frac{413 – 408}{\frac{12}{\sqrt{23}}}$
$= 1.998$
The degree of freedom $(df):$
$df = n – 1$
$= 23 – 1$
$= 22$
The p-value:
$\text{The p-value} = P(t > 1.998)$
$= 0.0291$
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:
$H_0: \mu = 408$
$H_a: \mu > 408$
The test statistic $(t) = 1.998$
The p-value $= 0.0291$
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.
