A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 405 gram setting. It is believed that the machine is overfilling the bags. A 22 bag sample had a mean of 408 grams with a standard deviation of 13. Assume the population is normally distributed. Is there sufficient evidence at the 0.1 level that the bags are overfilled? Step 2 of 6: Find the value of the test statistic. Round your answer to three decimal places.

Answer:
Given:
The hypothesized mean $(\mu) = 405$
The sample mean $(\overline{x}) = 408$
The sample standard deviation $(s) = 13$
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 $(t):$
$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}$

$= \frac{408 – 405}{\frac{13}{\sqrt{22}}}$

$= 1.082$

The degree of freedom $(df):$
$df = n – 1$
$= 22 – 1$
$= 21$

The p-value:
$\text{The p-value} = P(t > 1.082)$

$= 0.1457$

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:

$H_0: \mu = 405$
$H_a: \mu > 405$

The test statistic $(t) = 1.082$

The p-value $= 0.1457$

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.