The water content of a soil volume is measured four times gravimetrically by oven drying. The mean value of the water content is 23.2%. If the standard deviation of the four measurements is 1.0%, the 99% confidence interval for the soil water content is most nearly:. A. (19.5%, 26.9%)
B. (18.6%, 27.8%)
C. (20.6%, 25.8%)
D. (20.9%, 25.5%)

99% confidence interval for soil water content is  approx : A (19.5%, 26.9%)

Given : Mean x' = 23.2% = 0.232 , Standard Deviation s = 1% = 0.01 , Confidence Limit = 99%

Confidence Internal = x' + t ( s / √n)

t at 99% confidence limit, ie 1% ( 0.01 ) critical value  

t at 0.01/ 2, for two sided = t 0.005 = 7.4534

0.232 +  7.4534 ( 0.01 / √4 )

= 0.232 + 0.03736

= ( 0.232 + 0.03736 , 0.232 - 0.03736 )

= 0.19474 , 0.26926

= 19.5% , 26.9%

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99% confidence interval for the soil water content is determined by

(0.19474, 0.26926)

Step-by-step explanation:

Explanation:-

Given mean of the sample = 23.2% = 0.232

Given standard deviation of the sample = 1.0% =0.01

Given sample size 'n' =4

Degrees of freedom = n-1 =4-1 =3

critical value = 0.01

99% confidence interval for the soil water content is determined by

( 0.232 -0.03726 , 0.232 +0.03726)

(0.19474, 0.26926)

Final -

99% confidence interval for the soil water content is determined by

(0.19474, 0.26926)

i dont get the question

step-by-step explanation: