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)

 Steps for hypothesis test

For the first part we want to prove the claim that mean life of time of battery exceeds 400 hours, so Ho would be that the mean is 400 and Ha is that the mean is greater than 400
step 1: Ho: Mu = 400
Ha: Mu > 400
for the second step, recall that since sample size is small and standard deviation is not known, have to use the t-distribution, so test statistic is t
t = (sample mean - population mean)/(sample standard deviation/square root (n))
x-bar = 473.46
s = 210.77 as done on the calculator
n = 13
t = (473.46 - 400)/(210.77/square root(13))
t = 1.26
For the next step,we know using the p-value approach that we reject if p-value is less than alpha level of the test. So we reject Ho if p-value < .025
For the next step, I obtained the p-value from this site..
http://www.socscistatistics.com/pvalues/tdistribution.aspx
Putting in the values for t-statistic and one-tailed test, p-value is .11581
Last step, since .115811 > .025, we do not reject Ho since there is not significant evidence to conclude mean battery life is more than 400 hours.

A random sample of size 49 is taken from a population with mean 31and standard deviation of 12.
What are the expected value and the standard deviation of the sample mean ?
Describe the probability distribution of .
What is the probability that the sample mean is greater than 32?
What is the probability that the sample mean falls between 27 and 29?
What is the probability that the sample mean will be within ±3 of the population mean?

The distribution is the normal distribution which is symmetrical about the mean. Symmetric means it looks the same on both sides of the mean.
P(sample mean > 32)
We need to find a Z-score, which shows how many standard deviations away from the mean a value is.
Z = (sample mean - population mean)/(standard deviation/square root(n))
Z = (32 - 31)/(12/square root (49))
Z = 1/(12/7)
Z = 0.58

Now find Z(0.58) on the standard normal distribution chart. Since the chart shows probabilities less than, we need to take 1- Z(0.58)
Here is the chat I used for all these. http://www.regentsprep.org/regents/math/algtrig/ats7/zchart.htm
Z(0.58) = .7190
1- .7190 = .2010
The probability it falls between 27 and 29 ... P(27 < X < 29)
We need 2 Z-scores now, one for 27 and one for 29
Z = (27 -31)/(12/sqrt(49))
Z = -4/(12/7)
Z = -2.33
and
Z = (29 - 31)/(12/sqrt(49))
Z = -2/(12/7)
Z = -1.17
Now we get Z(-1.17) and Z(-2.33) and subtract them
.1210 - .0099 = .1111
Probability sample mean is within +/- 3standard deviations we simply find Z(3) and Z(-3) and subtract
So we get .9987 - .0013 = .9974

he 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .26.
a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)? Use 95% confidence.

b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population (to 4 decimals)?

c. What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

The margin of error is the 1.96(sqrt(pq/n) and that has to equal .02. We know p =.26, so q = .74

So we have 1.96(sqrt(.26*.74/n) = .02

3.8416(.1924)/n = .0004

0.73912384/n = .0004

n = 1848

b) point estimate is 524/1848 = .2835

c) .26 +/- .02 = (.24, .28)