Bio statistics Exercises and Solutions

Dear Students in this section, you can find solutions to few problems discussed in theory class

Statistical Table: Standard Normal Distribution $P(Z \leq z)$

z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5000	0.5040	0.5080	0.5120	0.5160	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.5910	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.6480	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.6700	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.6950	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.7190	0.7224
0.6	0.7257	0.7291	0.7324	0.7357	0.7389	0.7422	0.7454	0.7486	0.7517	0.7549
0.7	0.7580	0.7611	0.7642	0.7673	0.7704	0.7734	0.7764	0.7794	0.7823	0.7852
0.8	0.7881	0.7910	0.7939	0.7967	0.7995	0.8023	0.8051	0.8078	0.8106	0.8133
0.9	0.8159	0.8186	0.8212	0.8238	0.8264	0.8289	0.8315	0.8340	0.8365	0.8389
1.0	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531	0.8554	0.8577	0.8599	0.8621
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749	0.8770	0.8790	0.8810	0.8830
1.2	0.8849	0.8869	0.8888	0.8907	0.8925	0.8944	0.8962	0.8980	0.8997	0.9015
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115	0.9131	0.9147	0.9162	0.9177
1.4	0.9192	0.9207	0.9222	0.9236	0.9251	0.9265	0.9279	0.9292	0.9306	0.9319
1.5	0.9332	0.9345	0.9357	0.9370	0.9382	0.9394	0.9406	0.9418	0.9429	0.9441
1.6	0.9452	0.9463	0.9474	0.9484	0.9495	0.9505	0.9515	0.9525	0.9535	0.9545
1.7	0.9554	0.9564	0.9573	0.9582	0.9591	0.9599	0.9608	0.9616	0.9625	0.9633
1.8	0.9641	0.9649	0.9656	0.9664	0.9671	0.9678	0.9686	0.9693	0.9699	0.9706
1.9	0.9713	0.9719	0.9726	0.9732	0.9738	0.9744	0.9750	0.9756	0.9761	0.9767
2.0	0.9772	0.9778	0.9783	0.9788	0.9793	0.9798	0.9803	0.9808	0.9812	0.9817
2.1	0.9821	0.9826	0.9830	0.9834	0.9838	0.9842	0.9846	0.9850	0.9854	0.9857
2.2	0.9861	0.9864	0.9868	0.9871	0.9875	0.9878	0.9881	0.9884	0.9887	0.9890
2.3	0.9893	0.9896	0.9898	0.9901	0.9904	0.9906	0.9909	0.9911	0.9913	0.9916
2.4	0.9918	0.9920	0.9922	0.9925	0.9927	0.9929	0.9931	0.9932	0.9934	0.9936
2.5	0.9938	0.9940	0.9941	0.9943	0.9945	0.9946	0.9948	0.9949	0.9951	0.9952
2.6	0.9953	0.9955	0.9956	0.9957	0.9959	0.9960	0.9961	0.9962	0.9963	0.9964
2.7	0.9965	0.9966	0.9967	0.9968	0.9969	0.9970	0.9971	0.9972	0.9973	0.9974
2.8	0.9974	0.9975	0.9976	0.9977	0.9977	0.9978	0.9979	0.9979	0.9980	0.9981
2.9	0.9981	0.9982	0.9982	0.9983	0.9984	0.9984	0.9985	0.9985	0.9986	0.9986
3.0	0.9987	0.9987	0.9987	0.9988	0.9988	0.9989	0.9989	0.9989	0.9990	0.9990
3.1	0.9990	0.9991	0.9991	0.9991	0.9992	0.9992	0.9992	0.9992	0.9993	0.9993
3.2	0.9993	0.9993	0.9994	0.9994	0.9994	0.9994	0.9994	0.9995	0.9995	0.9995
3.3	0.9995	0.9995	0.9995	0.9996	0.9996	0.9996	0.9996	0.9996	0.9996	0.9997
3.4	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9998
3.5	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998
3.6	0.9998	0.9998	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999
3.7	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999
3.8	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999	0.9999

Example 1:

The display below shows some information for each of 12 tablets taken from a batch of a drug product, and for each of 10 human subjects administered tablets from the same batch.  The "in vitro" figures are the percentages of drug dissolved half an hour after the tablet was introduced into a dissolution medium.

   

5.94	6.22	5.09	6.22	5.37
5.51	4.1	5.51	4.52	4.52

     
Compute arithmetic mean and standard deviation of vitro figures.

Ans:  Let $x$ be the variable takes values on measurement of percentage of drug dissolved half an    hour after the tablet was introduced. The mean of $x$ is $\bar{x}=\dfrac{\sum\limits_{i=1}^{10}{x_{i}}}{10}=\dfrac{53}{10}=5.3 \%$
The standard deviation of $x$ is
 $S_{x}=\sqrt{\dfrac{\sum\limits_{i=1}^{10}{{x_{i}}^2}-\dfrac{({\sum\limits_{i=1}^{10}{x_{i}}})^2}{10}}{9}}$

$ =\sqrt{\dfrac{285.7964 - \dfrac{53^{2}}{10}}{9}}$

$= \sqrt{\dfrac{285.7964-280.9}{9}}$

$ = \sqrt{0.544044444}=0.737593685 \%$

OR

$x_{i}$	$x_{i}-\bar{x}$	$(x_{i}- \bar{x})^{2}$
5.94	0.64	0.4096
6.22	0.92	0.8464
5.09	-0.21	0.0441
6.22	0.92	0.8464
5.37	0.07	0.0049
5.51	0.21	0.0441
4.1	-1.2	1.4400
5.51	0.21	0.0441
4.52	-0.78	0.6084
4.52	-0.78	0.6084
Total	0	4.8964

Therefore

$S_{x}= \sqrt{\dfrac{\sum\limits_{i=1}^{10}{(x_{i}-\overline{x})}^2}{9}}$ =$\sqrt{\dfrac{4.8964}{9}}$ =$\sqrt{0.544044444}=0.737593685\% $

Example 2
In a series of experiments on the determination of tin in foodstuffs, samples were boiled with hydrochloric acid under reflux for different times. Some of the results are shown below:

30

55

57

59

56

Compute $1^{st}$ and $3^{rd}$ quartiles of tin value.
Let $x$ be the variable takes values on measurement of tin $mg kg^{-1}$ for reflux time 30 min.
Arranging the observations in ascending order we obtain

$X_{(1)}=55,X_{(2)}=56,X_{(3)}=56,X_{(4)}=57,X_{(5)}=59,X_{(6)}=59$
Position for First Quartile = $\dfrac{6+1}{4}=1.75$

Quartile 1 = $X_{(1)}+0.75\times (X_{(2)}-X_{(1)}) = 55+0.75\times(56-55)=55.75 mg kg^{-1}$
Position for Third Quartile = $\dfrac{3\times(6+1)}{4}=5.25$

Quartile 3 = $X_{(5)}+0.25\times (X_{(6)}-X_{(5)}) = 59+0.25\times(59-59)=59 mg kg^{-1}$

Example 3
Suppose a random variable X has Normal distribution with mean $\mu = 3$ and standard deviation $\sigma = 2$, compute the following probabilities

1. $P(X < 4)$
2. $P(X > 2)$
3. $P(1 < X < 4)$

Ans:
To Compute $P(X < 4)$
Consider the transformation $z = \dfrac{x-\mu}{\sigma}$. Replacing $x$ by 4 in $z$ we can write
$P(X < 4)=P\left(Z < \dfrac{4-3}{2}\right)$

$=P(Z < 0.5)$
This problem is represented graphically as

Fig 1. Area representing $P(Z < 0.5)$

The area can be read from Statistical Table
we can write
$P(X < 4) = P(Z < 0.5) = 0.6915$










To Compute $P(X > 2)$
Consider the transformation $z = \dfrac{x-\mu}{\sigma}$. Replacing $x$ by 2 in $z$ we can write
$P(X > 2)=P\left(Z > \dfrac{2-3}{2}\right)$

$=P(Z > -0.5)$
This problem is represented graphically as

Fig 2. Area representing $P(Z > -0.5)$

This is not the same Area represented by Statistical Table
Due to symmetric nature of graph we can represent mirror image of the Area as (Fig 3.) This is the same Area represented by Statistical Table. We write

$P(X >2) = P(Z > -0.5) = P(Z  < 0.5) = 0.6915$

Fig 3.Showing Mirror Area representing $P(Z < 0.5)$

To Compute $P(1 < X < 4)$
Consider the transformation $z = \dfrac{x-\mu}{\sigma}$. Replacing $x$ by 1 , 4 in $z$ we can write

$P(1 < X < 4) = P\left( \dfrac{1-3}{2} < Z < \dfrac{4-3}{2} \right)$

$=P(-1 < Z < 0.5)$
This problem is represented graphically as

Fig 4. Area representing $P(-1 < Z < 0.5)$

This is not the same Area represented by Statistical Table







This Area can be divided in to different areas as follows

Fig 5. Area $P(Z  < 0.5)$

Fig 6. Area $P(Z \leq  -1)$

That is we write $P(-1 < Z < 0.5)=P(Z < 0.5) - P( Z \leq -1)$

Using Symmetric property of graph we represent mirror image of graph $P( Z \leq -1)$ as (Fig 7)

Fig 7. Area $P(Z \geq 1)$

Because the total area below the curve is 1, this gives computation of $P(Z \geq 1)$ by following area

Fig 8. $P(Z < 1)$

We can write $P(Z \geq 1)=1 - P(Z < 1)$ (Fig 8)






Therefore
$P(Z \leq -1) = P(Z \geq 1) $

$=1 - P(Z < 1) $

$= 1 - 0.8413 $

$= 0.1587 $

and hence

$P(-1 < Z < 0.5) = P(Z < 0.5)-P( Z \leq -1) $

$=0.6915 - 0.1587 $

$=0.5328 $

We get  $P(1 < X < 4)=0.5328$

Example 4
It is suspected that an acid–base titrimetric method has a significant indicator error and thus tends to give results with a positive systematic error (i.e. positive bias). To test this an exactly 0.1 M solution of acid is used to titrate 25.00 ml of an exactly 0.1 M solution of alkali, with the following results (ml)
Volume of Acid: 25.06 25.18 24.87 25.51 25.34 25.41

1. Determine Mean and Standard deviation of the result obtained.
2. Test Hypothesis of no bias  $H_{0}:\mu = 25ml$ against there is bias  $H_{1}:\mu > 25 ml$ for this experiment at 5$\%$ level of significance.You are given $t_{5,0.05}=2.015,t_{5,0.025}=2.571$

Ans

Let $x$ denotes volume (ml) of test result and $n$ number  of test results.Therefore $x_{1}=25.06,x_{2}=25.18,x_{3}=24.87,x_{4}=25.51,x_{5}=25.34,x_{6}=25.41$ , Therefore

$\bar{x} = \dfrac{\sum\limits_{i=1}^{6}x_{i}}{6}$

$=\dfrac{151.37}{6}$

$=25.22833$

$s^{2} = \dfrac{\sum\limits_{i=1}^{6}x_{i}^{2}-\dfrac{\left( \sum\limits_{i=1}^{6}x_{i}\right)}{6}}{5}$

$=\dfrac{3819.0967-3818.8130}{5}$

$=0.05677 $

$s = \sqrt{0.05677}$

$=0.2383$

Using computations of sample mean and sample standard deviation, the computation of test statistic to test $H_{0}$ is

$t  = \dfrac{\bar{x}-\mu}{s/\sqrt{n}}$

$=\dfrac{25.22833-25}{0.2383/\sqrt{6}}$

$=2.347$

As Computed test statistic $t = 2.347$ exceeds $t_{5,0.05}=2.015$,we reject hypothesis $H_{0}$ in favor  of $H_{1}$ at 5$\%$ level of significance.

Example 5

Two types of chemical solution A and B were tested for their pH (degree of acidity of the solution). Analysis of 6 samples of A showed a mean pH of 7.52 with a standard deviation of 0.024. Analysis of 5 samples of B showed a mean pH of 7.49 with a standard deviation of 0.032. Using a 5$\%$ significance level, Test the hypothesis $H_0:\mu_1 = \mu_2$ against $H_1: \mu_1 \neq \mu_2$. Where $\mu_1$ and $\mu_2$ are true mean pH values of two solutions A and B. You are given 

$t_{9,0.05}=1.833,t_{9,0.025}=2.262$

Ans:

Let $X$ denotes measurement of pH of chemical solution  obtained by Analyst A and $Y$ denotes measurement of pH of chemical solution  obtained by Analyst B.

Also $m  = 6$  and $n = 5$ number of samples used by Analysts A and B respectively. We are given
Analyst A:
Sample Mean $\bar{X} = 7.52$
Sample Standard deviation $S_{1} = 0.024$
Analyst B
Sample Mean $\bar{Y} = 7.49$
Sample Standard deviation $S_{2} = 0.032$

To test the Hypothesis 

$H_0:\mu_1 = \mu_2$ against $H_1: \mu_1 \neq \mu_2$ 

We assume

1. The sample of observations on $X$ and sample of observations on $Y$ are varying according to Normal distribution with means $\mu_1$ , $\mu_2$  and variances $\sigma_{1}^{2}$ ,$\sigma_{2}^{2}$ respectively.
2. Here $\sigma_{1}^{2}$ ,$\sigma_{2}^{2}$ are unknown and $\sigma_{1} = \sigma_{2}$ 

Then

$S_{p}^{2} = \dfrac{(m-1)S_{1}^{2}+(n-1)S_{2}^{2}}{m+n-2}$

$ = \dfrac{(5)0.024^{2}+(4)0.032^{2}}{6+5-2}$

$=\dfrac{0.006976}{9}$

$=0.000775$

$S_{p} = \sqrt{0.000775} =0.027841 $

$S_{p}\sqrt{\dfrac{1}{m}+\dfrac{1}{n}} = 0.027841 \times \sqrt{\dfrac{1}{6}+\dfrac{1}{5}} $

$ =0.027841 \times \sqrt{0.366667}$

$  = 0.016858$

We use $t_{9,0.025}=2.262$,

Therefore $t_{9,0.025} \times 0.016858 =0.038134 $

Criteria For Rejection of $H_0$:
Reject $H_{0}$ if $|\bar{X}-\bar{Y}| > 0.038134$

Here $|\bar{X}-\bar{Y}| = | 7.52 - 7.49| = 0.03$

As $|\bar{X}-\bar{Y}|  \ngtr 0.038134$  Do not Reject $H_{0}$ at 5$\%$ level of significance.