A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 406 gram setting. It is believed that the machine is overfilling the bags. A 24 bag sample had a mean of 410 grams with a standard deviation of 14. 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 $(\mu) = 406$
The sample mean $(\overline{x}) = 410$
The sample standard deviation $(s) = 14$
The sample size $(n) = 24$

The significance level $(\alpha) = 0.1$

Solution:
The null and alternative hypothesis:

$H_0: \mu = 406$
$H_a: \mu > 406$

The test statistic $(t):$
$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}$

$= \frac{410 – 406}{\frac{14}{\sqrt{24}}}$

$= 1.4$

The degree of freedom $(df):$
$df = n – 1$
$= 24 – 1$
$= 23$

The p-value:
$\text{The p-value} = P(t > 1.4)$

$= 0.0875$

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:

$H_0: \mu = 406$
$H_a: \mu > 406$

The test statistic $(t) = 1.4$

The p-value $= 0.0875$

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.