Answer:
Given:
The hypothesized mean $(\mu) = 409$
The sample mean $(\overline{x}) = 413$
The sample standard deviation $(s) = 11$
The sample size $(n) = 20$
The significance level $(\alpha) = 0.1$
Solution:
The null and alternative hypothesis:
$H_0: \mu = 409$
$H_a: \mu > 409$
The test statistic $(t):$
$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}$
$= \frac{413 – 409}{\frac{11}{\sqrt{20}}}$
$= 1.626$
The degree of freedom $(df):$
$df = n – 1$
$= 20 – 1$
$= 19$
The p-value:
$\text{The p-value} = P(t > 1.626)$
$= 0.0602$
The final conclusion:
The p-value is greater than the significance level. Therefore, there is insufficient evidence to conclude that the bags are overfilled at the 0.1 significance level.
Final Answer:
The null and alternative hypothesis:
$H_0: \mu = 409$
$H_a: \mu > 409$
The test statistic $(t) = 1.626$
The p-value $= 0.0602$
The final conclusion:
The p-value is greater than the significance level. Therefore, there is insufficient evidence to conclude that the bags are overfilled at the 0.1 significance level.
