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

Answer:

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

$F_{calc} = 8.3457$

Step-by-step 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

      $F_{calc} = \frac{\sigma_1 ^2}{\sigma_2^2}$

=>    $F_{calc} = \frac{1.56^2}{0.54^2}$

=>    $F_{calc} = 8.3457$

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%