# 1. and the model would indeed perfectly fit the

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

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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