91影视

• The F-test is used to compare standard deviations. When comparing one sample to another, or a sample to what we would expect to find given a certain population distribution, we want to determine if the spread or dispersion of the two sets of data is comparable.
• A t-test is a statistical test used to compare the means of two groups that are related in some way.

To utilize distinct t tests to compare the results from Lab C with those from Lab A and Lab B, we must first conduct an F test in each of them to determine which specific case of t test to apply.

(i) Lab C and Lab A

F test

$$\begin{gathered} {\mathrm{F}}_{\mathrm{calc}}=\frac{{\mathrm{s}}_1^2}{{\mathrm{s}}_2^2} \\ =\frac{{\left(0.78\right)}^2}{{\left(0.14\right)}^2} \\ =31 \end{gathered}$$

${\mathrm{F}}_{\mathrm{table}}$= 3.88 (based on Table 4 - 3 degrees of freedom for: ${\mathrm{s}}_1=2\mathrm{and}{\mathrm{s}}_2=12$)

Since ${\mathrm{F}}_{\mathrm{calc}}>{\mathrm{F}}_{\mathrm{table}}$at the 95 % confidence level, there is a considerable discrepancy between the variances and standard deviations of the two lab values.

(i) Lab C and Lab A

t test

Find the case 2 b since it is comparing two findings with standard deviations that are significantly different, i.e., role="math" localid="1663561377979" ${\mathrm{F}}_{\mathrm{calc}}>{\mathrm{F}}_{\mathrm{table}-}$

$$\begin{gathered} {\mathrm{t}}_{\mathrm{calc}}=\frac{\left|{\overline{\mathrm{x}}}_1-{\overline{\mathrm{x}}}_2\right|}{\sqrt{\left({\mathrm{s}}_1^2/{\mathrm{n}}_1\right)+\left({\mathrm{s}}_2^2/{\mathrm{n}}_2\right)}} \\ =\frac{\left|2.68-1.59\right|}{\sqrt{{\displaystyle \frac{{\left(0.78\right)}^2}{3}}+{\displaystyle \frac{{\left(0.14\right)}^2}{13}}}} \\ =2.41 \end{gathered}$$

${\mathrm{t}}_{\mathrm{table}}=4.303$(based on Table 4-4; degrees of freedom -2; 95 % confidence level)

Since ${\mathrm{t}}_{\mathrm{calc}}<{\mathrm{t}}_{\mathrm{table}}$ at the 95 %confidence level, there is no significant difference between the means of the two lab reports.

(ii) Lab C and Lab B

F test

$$\begin{gathered} {\mathrm{F}}_{\mathrm{calc}}=\frac{{\mathrm{s}}_1^2}{{\mathrm{s}}_2^2} \\ =\frac{{\left(0.78\right)}^2}{{\left(0.56\right)}^2} \\ =1.94 \end{gathered}$$

${\mathrm{F}}_{\mathrm{table}}=4.74$ (based on Table 4-3; degrees of freedom for: ${\mathrm{s}}_1=2$and ${\mathrm{s}}_2=7$)

Since ${\mathrm{F}}_{\mathrm{calc}}<{\mathrm{F}}_{\mathrm{table}}$ at the 95 %confidence level, there is no significant difference between the variances and standard deviations of the two lab values.

(ii) Lab C and Lab B

t test

Because it will compare two findings with standard deviations that aren't considerably different and select Case 2 a., i.e., ${\mathrm{F}}_{\mathrm{calc}}<{\mathrm{F}}_{\mathrm{table}}$

First, start by calculating the ${\mathrm{s}}_{\mathrm{pooled}}$as a precondition for solving ${\mathrm{t}}_{\mathrm{calc}}$-

$$\begin{gathered} {\mathrm{S}}_{\mathrm{poonled}}=\sqrt{\frac{{\mathrm{s}}_1^2({\mathrm{n}}_1-1)+{\mathrm{s}}_2^2({\mathrm{n}}_2-1)}{{\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}}}} \\ =\sqrt{\frac{{\left(0.78\right)}^2(3-1)+{\left(0.56\right)}^2(8-1)}{3+8-2}} \\ =0.6157199941 \end{gathered}$$

Now, calculate for ${\mathrm{t}}_{\mathrm{calc}}$and contrast it to ${\mathrm{t}}_{\mathrm{table}}$.

$$\begin{gathered} {\mathrm{t}}_{\mathrm{calc}}=\frac{\left|{\overline{\mathrm{x}}}_1-{\overline{\mathrm{x}}}_2\right|}{{\mathrm{S}}_{\mathrm{pooled}}}\sqrt{\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}} \\ =\frac{\left|2.68-1.65\right|}{0.6157199941}\sqrt{\frac{\left(3\right)\left(8\right)}{3+8}} \\ =2.47 \end{gathered}$$

${\mathrm{t}}_{\mathrm{table}}=2.262$ (Degrees of freedom are based on Table 4-4. localid="1663562198639" ${\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}}=3+8-2=9$; 95% level of assurance)