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: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.[tex]H_0:\mu \neq 37.8\\H_a:\mu = 37.8[/tex]n = No. of samples = 48The population standard deviation is 12.5 milligrams.Since n > 30 and population standard deviation is given .So, we will use z test[tex]x = 41.5\\\sigma = 12.5 \\n = 48\\ \mu = 37.8[/tex][tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]Substitute the values in the formula :[tex]z=\frac{41.5-37.8}{\frac{12.5}{\sqrt{48}}}[/tex][tex]z=2.050[/tex]Refer the z table for p value p value = 0.9798α=0.05p value > αSo, 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 .