A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 407 gram setting. It is believed that the machine is overfilling the bags. A 19 bag sample had a mean of 412 grams with a standard deviation of 15. 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 $(\mu) = 407$
The sample mean $(\overline{x}) = 412$
The sample standard deviation $(s) = 15$
The sample size $(n) = 19$

The significance level $(\alpha) = 0.1$

Solution:
The null and alternative hypothesis:

$H_0: \mu = 407$
$H_a: \mu > 407$

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

$= \frac{412 – 407}{\frac{15}{\sqrt{19}}}$

$= 1.453$

The degree of freedom $(df):$
$df = n – 1$
$= 19 – 1$
$= 18$

The p-value:
$\text{The p-value} = P(t > 1.453)$

$= 0.0817$

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:

$H_0: \mu = 407$
$H_a: \mu > 407$

The test statistic $(t) = 1.453$

The p-value $= 0.0817$

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.