Module 1: development of practical skills Flashcards
Define standard deviation
How spread out repeats are from the mean
Benefits of standard deviation
Shows the spread of data around the mean, whereas range shows difference between highest and lowest value
Reduces the effect of anomalies
Can be used to indicate whether a difference between results is significant
Why is standard deviation useful for a conclusion
Add or subtract the standard deviation from mean results.
No overlap between the grouped results - likely to be a significant difference.
Overlap between grouped results - unlikely to be a significant difference
Conclusions using S.D.
A- Mean 11, S.D 3.2
B- Mean 8, S.D 2.8
C - Mean 1, S.D 1.3
Evaluate the conclusion that there is no significant difference between the three groups. Using evidence from the results above.
- Calculate the spread of data for each group by adding and subtracting the S.D. from the mean
- Compare spread of data for any overlap
- State evidence for and against the conclusion
- Add context of the question (e.g. refer to the specific investigation)
The standard deviations for group A and B overlap - there is likely no significant difference between the groups.
The standard deviation for group C does not overlap with either A or B - there is likely to be a significant difference. It is significantly lower.
What is a P-value (also significance level)
The probability that a result is due to chance
How to answer statistic question
- Justify use of statistic
- Write a null hypothesis
- State a conclusion
Write a null hypothesis based on an investigation between two variables for a correlation coefficient statistic
There is NO SIGNIFICANT CORRELATION between the two variables
When to use correlation coefficient
Continuous data
Investigating an association between two measurements
Spearman’s rank correlation coefficient (rs) = 1 - 6 ∑ (d²n) divided by (n² - 1)
D - Difference in rank of two measurements
N - number of pairs of data
Spearman’s rank correlation coefficient value
A number between -1 and +1
-1 is perfect negative correlation
+1 is perfect positive correlation
0 is no correlation
This value is then compared to a critical values table to conclude whether or not to accept the null hypothesis
Steps to calculate spearman’s rank correlation
- Rank reach set of data (1 being the smallest)
- Find the difference in rank between the two variables
- Square the difference
- Substitute appropriate values into equation given to you
How to compare the spearman’s rank correlation coefficient to a critical value table.
- For the value of n, find the critical value at the 0.05 significance level/p-value
- If the spearman’s rank correlation coefficient is lower than the critical value - accept the null hypothesis
- If the spearman’s rank correlation coefficient is higher than the critical value - reject the null hypothesis
If the critical value for the investigation between temperature and rate of reaction is 0.6786 and the correlation coefficient is 0.616. State the conclusion.
0.616 is lower than the critical value of 0.6786 at the 5% significance level.
Insufficient evidence to suggest a correlation between temperature and rate of reaction.
We accept the null hypothesis.
When to use Student T-Test
Continuous data
Investigating the difference between two means
Write a null hypothesis based on an investigation between two variables for a T-test statistic
There is no significant difference between the means of the two variables
Student T-test equation
t = (X̄₁ - X̄₂) / √[(s₁² / n₁) + (s₂² / n₂)]
X̄₁ = mean of group 1
X̄₂ = mean of group 2
s₁² = s.d. of group 1
s₂² = s.d. of group 2
n₁ = sample size of group 1
n₂ = sample size of group 2
Paired T-test
what type of data is used
How is degrees of freedom calculated
Two sets of data are from:
- from the same individual
- Both groups have the same sample size
degrees of freedom = n-1
Unpaired T-test
What type of data is used
How is degrees of freedom calculated
Groups have different sample sizes
Degrees of freedom = n₁ +
n₂ - 2
How to compare a calculated Student T-test value to a critical value
- Calculate degrees of freedom
- Find critical value for degree of freedom value at 0.05 p value
- T test value is greater than the critical value, reject the null hypothesis
- T test value is lower than the critical value, accept the null hypothesis
When to use Chi-squared?
Frequency data - counting individuals in a category
Investigating a difference between frequencies
Chi squared equation
χ² = ∑((O - E)² divided by E)
O = observed frequency
𝐸 = expected frequency
observed - expected squared
divided by the expected frequency
calculate for individual groups
Sum of these values
How to draw conclusion with chi-squared value.
- Find critical value at p-value of 0.05/5% significance level. at corresponding degree of freedom (Df = n-1)
- if chi squared value is greater than critical value, reject null hypothesis
- if chi squared value is smaller than critical value, accept null hypothesis
Write a null hypothesis based on an investigation for a chi squared statistic
There is no significant difference between the observed and expected value for a variable.
How to reduce uncertainty of a measurement
Increase the resolution of the measurement instrument
Take repeated readings
Increasing the value of what is being measured
How to find uncertainty for a reading (single judgement) e.g. a thermometer, mass balance?
the uncertainty of a reading is half of the smallest division on the measuring instrument - plus or minus
How to find uncertainty for a measurement (two judgements) e.g. a ruler, stopwatch?
the uncertainty of a measurement is equal to the smallest division on the measuring instrument - plus or minus
How to find uncertainty of a given value?
The uncertainty is plus or minus the last significant digit of the given value e.g. 1730g is plus or minus 10g uncertainty
Equation to calculate percentage uncertainty
Uncertainty divided by measured value/measured mean value
Multiplied by 100
