1.

A standard deviation is a measure used to

quantify the amount of variation or dispersion of a set of data values in a population.

It is found using the formula

1

It is useful because it allows us to see by

how much the members of a population differ from the mean value for the

population, meaning we can compare data sets more effectively than by just

looking at the average values for data sets.

Standard deviation differs from

standard error as that is the standard deviation of the sample mean, measuring how

accurately a sample represents a population as opposed to standard deviation

which measures dispersion.

2.

The linear regression given is an attempt to

model the relationship between a dependent variable (?) and an independent

variable (T). It is in the form y = a+Tx, 2.04 is the “a” value known as the Y

intercept (so graphically it would cross the y-axis at (0, 2.04)) and 4.22 is

the coefficient of T.

The coefficient of determination is the

explained variation divided by total variation, therefore telling us how well

the model fits the data. In this case it is 0.98 (r^2 =0.98) and so we see that

the model almost fits the data perfectly, since, if the coefficient of

determination was equal to 1, it would mean all variation is accounted for and

the model would indeed perfectly fit the data. So the conclusion that can be

drawn is that the regression line approximates real data very accurately.

The two values, (2.02) and (0.20) are the

result of the standard error test, an indication of the reliability of the

mean, which determines if the parameters (2.04 and 4.22) are significant,

testing the null hypothesis (that T has no relationship to ?). The parameters

are significant as the standard errors are low and the null hypothesis can be

rejected.

3.

My T value was calculated using the following formula using

the difference between means of the two data sets and dividing that by the

square root of the sum of the two variances divided by the amount of values in

each data set.

2

This gave me a value of

0.35883/0.123 = 2.82409

The critical value was found by determining the degrees of

freedom (n1 + n2 -2 =32) and finding the value for p=0.05 and it was 2.037

Since 2.82409>2.037 the T value was higher than the

critical value, the null hypothesis, the hypothesis that there is no

statistically significant difference between the samples is rejected.

I then carried out the T test on excel and got the value of

p=0.006511

ii) A Z-test, a way to compare sample and population means

to determine if there is a significant difference between them, may be used

when the standard variation is known or also when there is a large sample size

because fluctuations may happen in the T –test that will not exist in the

Z-test.

1 https://www.quora.com/Is-there-any-difference-between-the-standard-deviation-and-the-sigma-level-If-yes-how-are-they-both-calculated-What-is-the-relation-between-them

2 https://ncalculators.com/math-worksheets/how-to-calculate-t-test.htm