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%
[tex]F_{calc} = 8.3457[/tex]
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 [tex]\= x_1 = 24.6\ mmol/L[/tex]
The standard deviation is [tex]\sigma_1 = 1.56\ mmol/L[/tex]
The mean concentration for the new instrument is [tex]\= x_2 = 25.9 mmol/L[/tex]
The standard deviation is [tex]\sigma_2 = 0.54 mmol/L[/tex]
The confidence level is 95%
The level of significance is mathematically represented as [tex]\alpha = (100 - 95)\%[/tex]
[tex]\alpha = 0.05[/tex]
Generally the test statistics is mathematically represented as
[tex]F_{calc} = \frac{\sigma_1 ^2}{\sigma_2^2}[/tex]
=> [tex]F_{calc} = \frac{1.56^2}{0.54^2}[/tex]
=> [tex]F_{calc} = 8.3457[/tex]
Generally the degree of freedom for the old instrument is is mathematically evaluated as
[tex]df = n -1[/tex]
=> [tex]df = 6 -1[/tex]
=> [tex]df = 5[/tex]
Generally the degree of freedom for the new instrument is is mathematically evaluated as
[tex]df_1 = n -1[/tex]
=> [tex]df_1 = 6 -1[/tex]
=> [tex]df_1 = 5[/tex]
For the f distribution table the critical value of [tex]\alpha[/tex] at df and [tex]df_1[/tex] is
[tex]F_{tab} =5.0503[/tex]
Generally given that the [tex]F_{calc} > F_{tab}[/tex] 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%
