Question

Fizzy Waters promotes their alkaline water product for everyone on the basis that alkaline water is good for health as it neutralizes acids produced in the body. They boast having a mean alkalinity level of 50 mg/liter. Alkaline water has a higher pH level than regular drinking water and Fizzy Waters claims that its higher Hydrogen content provides better hydration than regular water. To test their claim, you contact Fizzy Waters and they allow you to collect samples from their manufacturing plant to test for yourself. You collect 100 random samples of their alkaline water and find that the mean and standard deviation are y = 32.2mg/liter and 14.4mg/liter. With 99% confidence, is there enough evidence to support their claim that the population mean exceeds 50 mg/liter?

Answer 1

Answer:

The mean of 50 mg/liter is not inside the 99% interval, so there is not enough evidence to support their claim.

Step-by-step explanation:

First we need to find the z-value for a confidence of 99%

The value of alpha for a 99% confidence is:

$1-\alpha/2 = 0.99$

$\alpha/2 = 0.01$

$\alpha = 0.005$

Looking in the z-table, we have z = 2.575.

Now we can find the standard error of the mean:

$\sigma_{\bar{x} }= s_x/\sqrt{n}$

$\sigma_{\bar{x} }= 14.4/\sqrt{100}$

$\sigma_{\bar{x} }=1.44$

Finding the 99% confidence interval, we have:

$99\%\ interval = (\bar{x} - z\sigma_{\bar{x}}, \bar{x} + z\sigma_{\bar{x}})$

$99\%\ interval = (32.2 - 2.575*1.44, 32.2 + 2.575*1.44)$

$99\%\ interval = (28.492, 35.908)$

The mean of 50 mg/liter is not inside the 99% interval, so there is not enough evidence to support their claim.