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:
[tex]1-\alpha/2 = 0.99[/tex]
[tex]\alpha/2 = 0.01[/tex]
[tex]\alpha = 0.005[/tex]
Looking in the z-table, we have z = 2.575.
Now we can find the standard error of the mean:
[tex]\sigma_{\bar{x} }= s_x/\sqrt{n}[/tex]
[tex]\sigma_{\bar{x} }= 14.4/\sqrt{100}[/tex]
[tex]\sigma_{\bar{x} }=1.44[/tex]
Finding the 99% confidence interval, we have:
[tex]99\%\ interval = (\bar{x} - z\sigma_{\bar{x}}, \bar{x} + z\sigma_{\bar{x}})[/tex]
[tex]99\%\ interval = (32.2 - 2.575*1.44, 32.2 + 2.575*1.44)[/tex]
[tex]99\%\ interval = (28.492, 35.908)[/tex]
The mean of 50 mg/liter is not inside the 99% interval, so there is not enough evidence to support their claim.
