Remember that removing an outlier can greatly affect the correlation between two variables.

Suppose x,y are as follows

x  1  1 1  2 2  2  3 3  3  1
y  8 9 10 8 9 10 8 9 10 10

This has a relatively strong negative correlation, but if you remove the data point (1,10), all you have left is a block of data with correlation coefficient of 0.

When finding correlation coefficient r by hand, as well as regression slope and intercept, you will see many calculations involving terminology you are unfamiliar with.

This is what is all means

sum(x) = all x's added together
sum(y) = all y's added together
sum(x^2) = square each x, then sum the total
sum(y^2) = square each y, then sum the total
sum(xy) = muitiply each xy value , then sum the total
(sumx)^2 = sum of the x's then square
(sumy)^2 = sum of the y's, then square
r^2 is just taking r and squaring it
y-bar is the average of the y's, which is sum(y)/n
x-bar is the average of the x's, which is sum(x)/n

Hypothesis test procedure

1. state the Ho and Ha
2. find test statistic z or t, for proportions we can use z if np > 5 and n(1-p) or nq > 5
3. find critical value for z or t using the appropriate chart
4. make the decision on the test, if using p-value, if p-value > significance level then do not reject. If p-value < significance level of the test, reject.
5. state the conclusion
Type I error is when you reject Ho when you should not reject Ho
Type II error is when you do not reject Ho when you should reject Ho
The power of the test is the probability of not committing a type two error, which is basically the probability of making the correct decision of rejecting Ho.

When using critical value approach for decision making with a hypothesis test, proceed as follows:


left tailed test (if test statistic < critical value, then reject Ho), if not, then do not reject Ho
right tailed test (if test statistic >= critical value, then reject Ho), if not, then do not reject Ho
two-tailed test, (if test statistic falls in between the critical values, do not reject Ho, if not then reject Ho


for p-value if p-value is > = alpha level of the test, then reject Ho, if not then do not reject Ho

Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours.
a.
x -bar =____
 Sx=
n=___
n-1 = ____
b. Define the random variables X and X (with a line over top of it)
c. Which distribution should use you for this problem?
d. Construct a 95% confidence interval for the population mean time wasted. State the confidence interval


x-bar is the sample mean which is 8
Sx is the standard deviation of x which is 4
n = sample size of 81
n-1 is the degrees of freedom which is 80
part b, x is the time an individual waited to be called for jury duty and x-bar is the sample mean, so that is the mean waiting time
c) this is t-distribution since population standard deviation is not known
part d, 95% CI, for 80 df, t value is 1.99
8 +/- 1.99(4/sqrt(81))
8 +/- 0.88 = (7.12, 8.88)
the error bound is also known as the margin of error which is the value added and subtract from the mean in the interval which is 0.8

R-square is the percent of variability explained by the model and 64% then is not explained by the model, explained by other factors and possibly due to chance. This is important because the higher the r-squared the better the data fits the model. So with a low r-squared the data isn't the best fit for the model.

First we need the hypotheses:

Ho: Mu = 10
Ha: Mu > 10

now get the test statistic t, since sample size is small and population standard deviation is not known.

t = (x-bar - Mu)/(standard deviation/square root(n))

t = (9.5 - 10)/(2.5/square root(16))

t = -0.8

We get the critical value for the test,

look up t at n-1 df for one tailed area of .05

t, 15df, .05 = 1.753

Since -0.8 < 1.753, we do not reject Ho. There is not enough evidence to support the claim that mean is greater than 10

For part b, the CI is x-bar+/- t(standard deviation/square root(n))

t for 95% interval, 15 df is 2.131

CI = 9.5 +/- 2.131(2.5/sqrt(16)) = 9.5 +/- 1.332

(8.168, 10.832)