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Assume we have a group of 10 rats daily injected with 50 µg Pb/kg b. wt. At the end of experiment, the Pb concentrations in the liver and kidney were measured and tabulated as mean ± standard error in the following table:Pb content Mean Standard error

liver (µg/g dry tissue)kidney (µg/g dry tissue)

6080

0.20.8

Is there any significant difference between the liver and kidneys in the levels of accumulated Pb at confidence level 95%?

Revision

2

So we want to test the null hypothesis H0: σ22 = σ1

2 against the alternate hypothesis HA: σ2

2 ≠ σ12 (2-tailed)

Solution:

∵𝑭 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅=𝒔𝟏𝟐

𝒔𝟐𝟐¿𝟐 .𝟓𝟑❑

𝟐

𝟎 .𝟔❑𝟐 ¿𝟏𝟕 .𝟕𝟖

F0.025(9,9) = 4.03

d.f.= 10 – 1 = 9

In this case, Fcalc (17.78) > Ftabulated (4.03), so we reject H0 that the two standard deviations are unequal, so P < 0.05

Does the type of diet significantly affected the body weight of mice at confidence levels of 95% and 99 %?

Three diets (I, II, III) for mice were tested for differences in body weight (in grams) after a specified period of time. The results are recorded in the following table:

Diet Body weight (g)

I 20 30 20 30

II 10 20 30 10

III 50 40 50 30

Compare between group I and II.

𝑯𝟎 :𝝁𝟏=𝝁𝟐=……….=𝝁𝒌

𝑯 𝑨 :𝒂𝒕 𝒍𝒆𝒂𝒔𝒕 𝒐𝒏𝒆𝒑𝒂𝒊𝒓 𝒐𝒇 𝝁′ 𝒔𝒓𝒆𝒏𝒐𝒕𝒆𝒒𝒖𝒂𝒍 .

Solution

Diet Body weight (g) mean size S

I 20 30 20 30 25 4 5

II 10 20 30 10 17.5 4 8.29

III 50 40 50 30 42.5 4 7.98

For I, σ2=25

For II, σ2=68.75

Source SS df MS F P

Between (Factor)

SSB dfB MSB

MSB/MSW

Within (Error)

SSW dfW MSW

Total SST

𝐗𝐠=𝟒 (𝟐𝟓 )+𝟒 (𝟏𝟕 .𝟓 )+𝟒 (𝟒𝟐 .𝟓 )

𝟒+𝟒+𝟒=𝟐𝟖 .𝟑𝟑

𝐒𝐒𝑩=𝐧𝟏 (𝐗𝟏−𝐗𝐠 )𝟐+𝐧𝟐 (𝐗𝟐−𝐗𝐠 )𝟐+𝐧𝟑 (𝐗𝟑−𝐗𝐠 )𝟐

= 1316.66dfB= h-1= 3-1= 2

MSB= SSB/dfB= 658.33

𝑺𝐒 (𝐖 )=𝒅𝒇 𝟏∗𝑺𝟏𝟐+𝒅𝒇 𝟐∗𝑺𝟐

𝟐+…+𝒅𝒇 𝒌∗𝑺𝒌𝟐

= 472.5dfW= N-h= 12-3= 9

MSW= SSW/dfW= 52.5

Source SS df MS FCal P

Between (Factor)

1316.66 2 658.33 12.54 <0.01

Within (Error)

472.5 9 52.5

Total 1789.16 11

F0.05 (2, 9)= 4.26F0.01 (2, 9)= 8.02

f-distribution Table

Tukey’s HSD (Honestly significance difference) Post-hoc test

𝑸=𝒒(𝒈 ,𝑵 −𝒈 ,∝ ) √𝑴𝑺𝑾𝒏

The critical value for comparison between two averages

Sample size /group

Number of groups = number of meansTotal number of Samples

Critical q value(tabulated)

Tukey’s HSD (Honestly significance difference) Post-hoc test

𝑸=𝒒(𝟑 ,𝟏𝟐−𝟑,𝟎 .𝟎𝟓 )√𝟓𝟐 .𝟓𝟒Q= 3.95 (3.62)=14.31

= 17.5, = 25, = 42.5

-= 7.5 <Q (14.31) insignificant

N (total sample size)- g (number of groups) g (3)12 – 3= 9

The data below represent the levels of blood glucose before and after injection with a certain herbal extract.

Experimental conditions Blood glucose levels (mg/dl)

Before 2, 3, 3, 2, 4, 2after 9, 8, 9, 8, 8, 7

Did the herbal extract cause increased blood glucose levels??

After (X2)

0

0

3

20

0

0

Before (X1) After (X2)

2 9

3 8

3 9

2 8

4 8

2 7

D

-7

-5

-6

-6

-4

-5

D2

49

25

36

36

16

25

𝒕=∑ 𝑫

√𝒏∑ 𝑫𝟐− (∑ 𝑫 )𝟐

𝒏−𝟏

𝑫=𝑿𝟐−𝑿𝟏 d.f. = n - 1

-33 187

=12.84

ttabulated at d.f. 5 = 2.015

𝑯𝟎 :𝝁=𝝁𝟎

𝑯𝟏 :𝝁<𝝁𝟎 ,

tCal (12.84) > ttabulated (2.015)Significant, P<0.05

In an experiment to study the effect of pH value on the hepatic Cd content, the data below were recorded. Test the claim that Cd content at pH 8 is significantly higher than at pH 5?

pH Hepatic Cd content (mg /kg b. wt.)5 4, 8, 8, 30, 10, 12, 128 4, 7, 10, 11, 14, 16,16, 21, 23, 25, 26

𝑯𝟎 :𝝁=𝝁𝟎

𝑯𝟏 :𝝁>𝝁𝟎 ,

pH5 (X1) pH8 (X2)

4 4

8 7

8 10

30 11

10 14

12 16

12 16

21

23

25

26

𝐒𝐒𝟏=∑𝐗𝟏𝟐−

(𝐗𝟏 )𝟐

𝐧𝟏

(X1)2

16

64

64

900

100

144

144

(X2)2

16

49

100

121

196

256

256

441

529

625

676

84 173 1432 3265

𝐒𝐒𝟏=𝟏𝟒𝟑𝟐−(𝟖𝟒)𝟐

𝟕=𝟒𝟐𝟒

𝐒𝐒𝟐=𝟑𝟐𝟔𝟓−(𝟏𝟕𝟑)𝟐

𝟏𝟏=𝟓𝟒𝟒 .𝟐

=12 =15.73

𝒕𝐂𝐚𝐥=𝐗𝟏−𝐗𝟐

√( 𝐒𝐒𝟏+𝐒𝐒𝟐

(𝐧𝟏+𝐧𝟐)−𝟐 )( 𝟏𝐧𝟏

+ 𝟏𝐧𝟐

)

17

d.f. = n1 + n2 - 2 = 7+11 -2= 16

=

= 1.001 ttabulated at d.f. (7+11-2) = 1.746

tCal (1.001) < ttabulated (1.746)

Insignificant, P>0.05

t-distribution Table