Most water treatment facilities monitor the quality of their drinking water on an hourly basis. one variable measured is ph, which measures the degree of alkalinity or acidity in the water. a ph below 7.0 is acidic, one above 7.0 is alkaline, and a ph of 7.0 is neutral. one water-treatment plant has a target ph of 8.5. (most try to maintain a slightly alkaline level) the mean and standard deviation of 1 hour test results based on 49 water samples at this plant are: = 8.42; s = 0.16 does this sample provide sufficient evidence that the mean ph level in the water differs from 8.5?

There is not enough evidence to conclude that the mean pH level is 8.5.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 8.5

Sample mean, = 8.42

Sample size, n = 49

Alpha, α = 0.05

Sample standard deviation, s = 0.16

First, we design the null and the alternate hypothesis

We use two-tailed t test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since, the test statistic does not lies in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

There is not enough evidence to conclude that the mean pH level is 8.5.

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week

they go with it happens or they mite give a chance?

step-by-step explanation: