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.

answer: the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

step-by-step explanation:

the given rate of interest i= 85%

in calculating the monthly payment for a loan with an interest rate compounded monthly, the value of the interest rate per period we use in the formula must be in decimal form.

to convert it in decimal, we divide it by 100, we get

hence, the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

3; 9 is 8 to -4

step-by-step explanation: