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

## Friday, November 27, 2015

## Saturday, November 21, 2015

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

## Monday, November 16, 2015

## Wednesday, November 11, 2015

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)

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