Questions on statistics Essay

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Answer 1:

Standard Deviation ( SD ) ( denoted by) :The standard divergence ( SD ) is a procedure/measure that is used to show the sum of fluctuation of a set of informations values. Discrepancy, in this context, measures how far a set of Numberss is dispersed out. The standard divergence is the square root of discrepancy.

where

N is the size of population

? is the mean

Note: If the given information is merely a sample of the whole population ( N ) , so the expression for Sample Standard Deviation is:

where

N is the size of population ; ( N-1 is the Bessel ‘s rectification )

xI„ is the sample mean

s is the sample criterion divergence

Standard Error of the Mean ( SEM ) :Put compactly, SEM is the standard divergence ( SD ) of the sample mean ‘s approximate of the true mean. It quantifies the preciseness of the mean.

where

s is the sample criterion divergence

N is the sample size

Relationship between Standard Deviation ( SD ) and Standard Error of the Mean ( SEM ) :

It is of import to retrieve that SEM gives the preciseness of the mean, where as SD gives the variableness of a set of informations values. These two are related. Therefore, their relation is given by

SE = ? / vn

where

? is the standard divergence

N is the sample size

Answer 2:

Use of descriptive statistics is really common in the field of biomedical research. As stated earlier, standard divergence ( SD ) quantifies data variableness and standard mistake of the average quantifies the preciseness of the mean. Unlike standard divergence, SEM is non a descriptive statistic and should be merely used to bespeak the preciseness of the mean.

In reply to the inquiry “ Given the experimental design and intent, which one is more appropriate as a sum-up of the variableness in this tabular array, and why? “ ,standard divergence ( SD )is more appropriate thanSEMas a sum-up of the variableness in the given tabular array for the undermentioned grounds.

  • The SD ( s ) represent ( s ) the variableness within the sample ( s ) .
  • The SEM is non a descriptive statistic to sum up the variableness in the given tabular array.
  • Standard divergence gives an thought of thespread( of values in the population ) between ten independent groups. It quantifies spread.
  • Using SDs as a sum-up of variableness along with groups ‘ agencies convey an thought of the enormousness of the difference between the several means as it shows how large the consequence size is.
  • In instance of statistical significance, demoing p values of the attendant statistic is more accurate than demoing SEM

Answer 3:

Beginning

d.f.

amount of squares

Average squares

F ratio

P-value

Between Columns

9

299.35

33.26

1.95

0.065

Within Columns

50

851.51

17.03

Entire

59

1150.86

Between Columns:It represents variableness among group agencies

Within Columns:Variability within the groups. It is besides calledresiduary fluctuation.

Degrees of freedom:Basically, it denotes how many independent things are present. It is a figure of values that are independent to change. They are calculated as follows:

dfbetween = a -1

dfWithin = N – a

dfTotal = N – 1

where

a = Number of groups ( degrees )

N = population size

Sum of squares:It determines the scattering of informations values ( points ) . It is a step of variableness of a set of values around a particular value which is normally average. There are calculated as follows:

Entire SS = SS ( B ) + SS ( W )

where SS ( B ) is Sum of squares between groups and SS ( W ) is Sum of squares within groups

Average squares:It refers to an estimation of the population discrepancy ( ? ) . It is computed as follows:

MS ( B ) = SS ( B ) /dfbetween

MS ( W ) = SS ( W ) /dfWithin

Where MS ( W ) is the Mean square within groups and MS ( B ) is the Mean square between groups

F ratio:This trial statistic decides whether to accept or reject void hypothesis ( H0 ) . For case, if F is greater than 3.1599, so reject Null Hypothesis. It is calculated as follows:

F = MS ( B ) /MS ( W )

P-value:It non merely indicates the chance that one or more column ( group ) means differ ( s ) from the others but besides of rejecting the void hypothesis when it is accurate and right. It is determined from the F ratio and grades of freedom.

Answer 5:

A t-test is used to find if two groups are significantly different from each other. It examines the difference between agencies of two groups ( for case, control group vs. Drug 1 ) . However, when there are more than two groups, t-test must non be used as the familywise mistake rate adds up ; and that consequences in happening an consequence that truly is n’t present. Familywise mistake rate ( FWER ) is the chance of doing type I errors ( false positive ) .

The best illustration to understand the hazard of promoting false positive consequences is the ‘probability of obtaining “ caputs ” in a coin somersault. ‘ In one somersault, the chance is 0.5. But for four coin somersaults, the chance is 1 – ( 0.50 ) 4. = 0.9475 ; that is 94 % of sets of four coin somersaults result in one or more caputs.

We can accommodate the above illustration to understand how the usage of nine t-tests elevates the hazard of false positive consequences above the nominal degree of ? = 0.05. Assume that if the H0 ( void hypothesis ) is true, so the chance of obtaining an undistinguished consequence is 0.95 ( 1 – 0.05 ) . So, the chance ofnon obtainingone or more important consequences in 9 trials is1 – ( 0.95 )9= 0.3698. In other words, the chance ofobtainingone or more important consequences in 9 trials is1 – 0.3698 = 0.6302

For argument interest, assume the research worker performed merely 6 t-tests. Therefore, the chance ofnon obtainingone or more important consequences in 6 trials is1 – ( 0.95 )6= 0.2649. So, the chance ofobtainingone or more important consequences in 6 trials is1 – 0.2649 = 0.7351.

In decision, it is apparent that the usage of 9 t-tests elevates the hazard of falsepositive consequences above the nominal degree of ? = 0.05 where P & A ; lt ; ? is used as a threshold for make up one’s minding a consequence is ‘statistically significant’ .

Answer 6:

95 % Assurance Intervals for control ( i.e no drug ) :

Calculation:

Mean, xI„Control = 20.25

sample size, n = 10

Confidence degree = ( 1 – ? ) 100 %

( 1 – ? ) 100 % = 95 ( since, assurance degree = 95 % )

( 1 – ? ) = 0.95

? = 0.05

Degrees of freedom = n – 1 = 10 – 1 = 9

t0.05 = 2.2621 ( two tailed trial and sample size & A ; lt ; 30 )

Standard Deviation, s = 6.0555

SEM = 1.9149

Assurance Time interval: (xI„ ± T0.05*SEM )

Therefore, Confidence Time intervals: ( 15.9183, 24.5817 ) ie. ,( ±4.3317 )

Interpretation:Given the mean of 20.25, a SD of 6.0555, and a assurance degree of 95 % for control, the corresponding assurance interval would be ± 4.3317. That is to state that there is 95 % certainty that the true population mean ( ? ) of control ( i.e no drug ) falls within the scope of 15.9183 to 24.5817.

95 % Assurance Intervals for drug 3:

Calculation:

Mean, xI„Drug 3 = 17.18

sample size, n = 10

Confidence degree = ( 1 – ? ) 100 %

( 1 – ? ) 100 % = 95 ( since, assurance degree = 95 % )

( 1 – ? ) = 0.95

? = 0.05

Degrees of freedom = n – 1 = 10 – 1 = 9

t0.05 = 2.2621 ( two tailed trial and sample size & A ; lt ; 30 )

Standard Deviation, s = 4.0879

SEM = 1.2927

Assurance interval: (xI„ ± T0.05*SEM )

Therefore, Confidence Time intervals: ( 14.2560, 20.1040 ) ie. , (±2.9240 )

Interpretation:Given the mean of 17.18, a SD of 4.0879, and a assurance degree of 95 % for drug 3, the corresponding assurance interval would be ± 2.9240. That is to state that there is 95 % certainty that the true population mean ( ? ) of drug 3 falls within the scope of 14.2560 to 20.1040.

Answer 7:

95 % Assurance Intervals for difference of agencies:

Calculation:

Difference of Means, xI„Difference = | xI„Control – xI„Drug 3 |

xI„Difference = | 20.25 – 17.18 | = 3.07

sample sizes, nControl = nDrug 3 = 10

Confidence degree = ( 1 – ? ) 100 %

( 1 – ? ) 100 % = 95 ( since, assurance degree = 95 % )

( 1 – ? ) = 0.95

? = 0.05

Degrees of freedom = ( nControl – 1 ) + ( nDrug 3 – 1 ) = 18

t0.05 = 2.1009 ( two tailed trial )

Standard Deviation, sControl – Drug 3 = 2.3104 ( Since, Sdifference = v ( 2*MSE/n ) )

Assurance Time interval: (xI„Difference± T0.05*sControl – Drug 3)

Therefore, Confidence Time intervals: ( -1.7839, 7.9239 ) ie. ,( ±4.8539 )

-1.7839 & A ; lt ;?Control– ?Drug 3& A ; lt ; 7.9239

Interpretation:The significance of the above consequence is ‘there is 95 % opportunity ( 0.95 chance ) that true mean difference lies between the bounds -1.7839 and 7.9239. ‘ It is sensible for a research worker to give precedence to the difference between agencies [ as they are used to find the difference in population agencies ] than the specific average values themselves.

Answer 8:

The undermentioned disagreements are observed between the two sets of drug 3 consequences:

The First Drug 3 ExperimentThe Second Drug 3 Experiment

Sample Size: 6 Sample Size: 10

Mean: 13.6 Mean: 17.18

Standard Deviation: 1.9 Standard Deviation: 4.0879

SEM: 0.8 SEM: 1.2927

  1. Standard Deviation:As the sample size ( n ) additions, the SD of the sample means lessenings because the variableness becomes less. Normally, the standard divergence of the sample average reduces to the square root of the sample size.

– The standard divergence of the ‘first drug 3 experiment ‘ with sample size [ n = 6 ] is [ SD1= 1.9 ]

but the SD of the ‘second Drug 3 experiment ‘ with sample size [ n = 10 ] is [ SD2= 4.0879 ] which is a disagreement.

  1. Standard Error of the Mean:As the sample size ( n ) additions, the SEM decreases. It is reasonable that holding more informations values yield more preciseness. SEM lessenings because N is reciprocally relative as shown in the expression below ( in other words, it present in the denominator of the expression ) .

formula

– The standard mistake of the mean of the ‘first drug 3 experiment ‘ with sample size [ n = 6 ] is[ SEM1= 0.8 ] but the SEM of the ‘second Drug 3 experiment ‘ with sample size [ n = 10 ] is[ SEM2= 1.2927 ] which can be viewed as a disagreement.

The ground for above disagreements could be ‘the adding of values ( or observations ) that are far off from the general bunch around xI„1 [ mean of ‘first drug 3 experiment ‘ ] .

Answer 9:

Assuming that it was decided that two more drugs from table 1 were to be selected for farther testing, I would takeDrug 4andDrug 8. The consequence of the whole experiment is to cognize that any of nine drugs are worth following up. Following is a brief drumhead of Drug 4 and Drug 8:

Drug 4: Average = 16.4, Standard Deviation = 1.7, SEM = 0.7

Drug 8: Average = 19.0, Standard Deviation = 2.1, SEM = 0.9

The sample mean is an indifferent calculator for the population mean. SEM is the standard divergence of the sample mean, s. It quantifies the preciseness of the sample mean. Therefore, smaller the SEM, the more is the preciseness of the mean. For case, SEM = 1.9 indicates that a sample mean (xI„) is a more accurate estimation of population mean than if SEM = 3.5.

The Standard Error of the Mean closer to zero ( 0 ) indicate that the observations for xI„ are clustered around the sample mean.

The purpose of this experiment is non to lose assuring drugs as stated by the foreman ‘ … .For this undertaking it’s more of import non to lose assuring drugs than it is to protect against fanciful ‘family-wise’ mistakes or false finds… ‘ .The SEMs of drugs 4 and 8 are smaller than the remainder which implies that they could be the promising drugs.Therefore, maintaining the purpose of the experiment in head, I would re-test drugs 4 and 8 to guarantee that they are the promising drugs that are deserving following up.

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