Answer:
Given:
The Hypothesized Mean $ (\mu) = 425 $
The Sample Mean $ (\bar{x}) = 430 $
The Sample Variance $ (s^2) = 100 $
The Sample Size $ (n) = 11 $
$\therefore$ The Sample Standard Deviation $ (s) = \sqrt{100} = 10 $
The Significance Level $ (\alpha) = 0.02 $
Solution:
The null and alternative hypothesis:
$ H_0: \mu = 425 $
$ H_1: \mu > 425 $
The test statistic $ (t): $
$t = \frac{\bar{x} – \mu}{\frac{s}{\sqrt{n}}} $
$ = \frac{430 – 425}{\frac{10}{\sqrt{11}}} $
$ = 1.658 $
The degree of freedom $ (df): $
$ df = n – 1 $
$ = 11 – 1 $
$ = 10 $
The p-value:
$ \text{p-value} = \text{P}(t_{10} > 1.658) $
$ = 0.0641 $
The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the bags are overfilled.
Final Answer:
The null and alternative hypothesis:
$ H_0: \mu = 425 $
$ H_1: \mu > 425 $
The test statistic $ (t) = 1.658 $
The p-value $ = 0.0641 $
The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the bags are overfilled.