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.
![Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cdisplaystyle%5Cfrac%7B%5Ctext%7BSum%20of%20all%20observations%7D%7D%7B%5Ctext%7BTotal%20number%20of%20observation%7D%7D)
![Mean =\displaystyle\frac{10959}{25} =438.36](https://tex.z-dn.net/?f=Mean%20%3D%5Cdisplaystyle%5Cfrac%7B10959%7D%7B25%7D%20%3D438.36)
Sum of squares of differences = 175413.76
![S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49](https://tex.z-dn.net/?f=S.D%20%3D%20%5Csqrt%7B%5Cdfrac%7B175413.76%7D%7B24%7D%7D%20%3D%2085.49)
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
![H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}](https://tex.z-dn.net/?f=H_%7B0%7D%3A%20%5Cmu%20%5Cgeq%20425%5Ctext%7B%20mg%20per%20Litre%7D%5C%5CH_A%3A%20%5Cmu%20%3C%20425%5Ctext%7B%20mg%20per%20Litre%7D)
We use one-tailed t test to perform this hypothesis.
Formula:
![t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }](https://tex.z-dn.net/?f=t_%7Bstat%7D%20%3D%20%5Cdisplaystyle%5Cfrac%7B%5Cbar%7Bx%7D%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%20%7D)
Putting all the values, we have
Now, ![t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108](https://tex.z-dn.net/?f=t_%7Bcritical%7D%20%5Ctext%7B%20at%200.05%20level%20of%20significance%2C%2024%20degree%20of%20freedom%20%7D%20%3D%20-1.7108)
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.