Question

The amount of ketchup dispensed into a container follows a normal distribution with a mean of 16 ounces and a standard deviation of 0.2 ounce. A sample of 25 containers revealed a mean fill of 16.04 ounces. Is the sample mean significantly different from 16 ounces? Use the 0.05 significance level.

Answer 1

Answer: Yes, the sample mean is significantly different from 16 ounces.

Step-by-step explanation:

Since we have given that

$H_0:\mu =16\\\\H_1:\mu\neq 16$

Mean = 16 ounces

Standard deviation = 0.2 ounces

$\bar{X}=16.04$

Number of containers = 25

Since there are 25 containers i.e. n = 25

n<30.

so, we will do t-test.

Using the normal distribution, we get that

$z=\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\t=\dfrac{16.04-16}{\dfrac{0.2}{\sqrt{25}}}\\\\t=\dfrac{0.04\times 5}{0.2}\\\\z=1$

Here, degree of freedom v = n-1 = 25-1 = 24

So, $t_{\alpha,v}=t_{0.05,24}=1.711$

Since we can check that

$1<1.711$

So, we will reject the null hypothesis.

Hence, Yes, the sample mean is significantly different from 16 ounces.