Question

A consumer research organization states that the mean caffeine content per 12-ounce bottle of a population of caffeinated soft drinks is 37.8 milligrams. You find a random sample of 48 12-ounce bottles of caffeinated soft drinks that has a mean caffeine content of 41.5 milligrams. Assume the population standard deviation is 12.5 milligrams. At α=0.05, what type of test is this and can you reject the organization’s claim using the test statistic?

Answer 1

Answer:

This test is z test and we reject the organization’s claim .

Step-by-step explanation:

Claim : The mean caffeine content per 12-ounce bottle of a population of caffeinated soft drinks is 37.8 milligrams.

$H_0:\mu \neq 37.8\\H_a:\mu = 37.8$

n = No. of samples = 48

The population standard deviation is 12.5 milligrams.

Since n > 30 and population standard deviation is given .

So, we will use z test

$x = 41.5\\\sigma = 12.5 \\n = 48\\ \mu = 37.8$

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

Substitute the values in the formula :

$z=\frac{41.5-37.8}{\frac{12.5}{\sqrt{48}}}$

$z=2.050$

Refer the z table for p value

p value = 0.9798

α=0.05

p value > α

So, we accept the null hypothesis.

So, we reject the organization’s claim .

Hence this test is z test and we reject the organization’s claim .