Answer:
The mean water hardness of lakes in Kansas is 425 mg/L or greater.
Step-by-step explanation:
We are given the following data set:
346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496
Formula:
where
are data points,
is the mean and n is the number of observations.


Sum of squares of differences = 175413.76

Population mean, μ = 425 mg/L
Sample mean,
= 438.36
Sample size, n = 25
Alpha, α = 0.05
Sample standard deviation, s = 85.49
First, we design the null and the alternate hypothesis

We use one-tailed t test to perform this hypothesis.
Formula:

Putting all the values, we have
Now, 
Since,
The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.
Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.