Question

One important test for kidney disease involves measuring the levels of bicarbonate (HCO−3) in blood. Normal bicarbonate levels for a person ranging in age from 18 to 59 years old is 23–29 mmol/L. A lab purchased a new instrument to measure bicarbonate levels in blood and needs to certify it against their current instrument. The bicarbonate levels in the blood of a 38 year old woman was measured using the old and new instrument. The blood was tested 6 times using each instrument. The mean concentration of bicarbonate using the old instrument was found to be 24.6 mmol/L with a standard deviation of 1.56 mmol/L. The new instrument yielded a mean concentration of 25.9 mmol/L with a standard deviation of 0.54 mmol/L. Determine if there is a significant difference in the standard deviations of the two sets of measurements made by the two instruments at the 95% confidence level. Determine the value of Fcalc.

Answer 1

Yes there is a significant difference in the standard deviations of the two sets of measurements made by the two instruments at the 95%

`F_(calc) = 8.3457`

Explanation:

From the question we are told that

The normal bicarbonate level is k = 23-29 mmol/L

The number of times the blood was tested is n = 6

The mean concentration for old instrument is
`\= x_1 = 24.6\ mmol/L`

The standard deviation is
`\sigma_1 = 1.56\ mmol/L`

The mean concentration for the new instrument is
`\= x_2 = 25.9 mmol/L`

The standard deviation is
`\sigma_2 = 0.54 mmol/L`

The confidence level is 95%

The level of significance is mathematically represented as
`\alpha = (100 - 95)\%`

`\alpha = 0.05`

Generally the test statistics is mathematically represented as

`F_(calc) = (\sigma_1 ^2)/(\sigma_2^2)`

=>
`F_(calc) = (1.56^2)/(0.54^2)`

=>
`F_(calc) = 8.3457`

Generally the degree of freedom for the old instrument is is mathematically evaluated as

`df = n -1`

=>
`df = 6 -1`

=>
`df = 5`

Generally the degree of freedom for the new instrument is is mathematically evaluated as

`df_1 = n -1`

=>
`df_1 = 6 -1`

=>
`df_1 = 5`

For the f distribution table the critical value of
`\alpha` at df and
`df_1` is

`F_(tab) =5.0503`

Generally given that the
`F_(calc) > F_(tab)` it means that there is a significant difference in the standard deviations of the two sets of measurements made by the two instruments at the 95%