Friday, July 31, 2009

Help with statistics Question#2?

A new lab technician read a report that the average number of students using the computer lab per hour was 16. To test this hypothesis, he selected a day at random and kept track of the number of students who used the lab over an 8-hour period. The results were as follows:





20 24 18 16 16 19 21 23





A) At α=0.01 is there enough evidence to support this claim?





B) Find the 99% confidence interval of the true mean.





C) Does the Confidence Interval interpretation agree with the hypothesis test result? Explain.

Help with statistics Question#2?
Small Sample Hypothesis Test for mean:





In order for this test to be valid the data must come from a normal population. If this is not the case then this test is not valid and other methods, such as a randomization test or permutation test should be used.





Assuming the normality assumption is valid to test the null hypothesis





H0: μ ≤ Δ or


H0: μ ≥ Δ or


H0: μ = Δ


Find the test statistic t = (xbar - Δ ) / (sx / √ (n))





where xbar is the sample average


sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.


n is the sample size





and t follows the Student t distribution with n - 1 degrees of freedom. We use the Student t distribution to account for the uncertainty in the estimate of the variance.


As the degrees of freedom approach infinity the Student t converges in probability to the Standard Normal. In most cases the values of the percentiles of the Student t are close enough to the Standard Normal when the degrees of freedom are greater than 30. This is the source of the empirical rule of thumb that samples of size %26gt; 30 have a mean that is normally distributed. Keep that in mind as well, for these hypothesis tests we are assuming the mean is normally distributed. This assumption is easy to verify if the data is normally distributed. The Central Limit Theorem accounts of all other means.





The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.





H1: μ %26gt; Δ; p-value is the area to the right of t


H1: μ %26lt; Δ; p-value is the area to the left of t


H1: μ ≠ Δ; p-value is the area in the tails greater than |t|





If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.





If the p-value is greater than the significance level, i.e., p-value %26gt; α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.





The hypothesis test in this question is:





H0: μ = 16 vs. H1: μ ≠ 16





The test statistic is:


t = ( 19.625 - 16 ) / ( 2.973094 / √ ( 8 ))


t = 3.448613





The p-value = P( | t_ 7 | %26gt; t )


= P( t_ 7 %26lt; -3.448613 ) + P( t_ 7 %26gt; 3.448613 )


= 2 * P( t_ 7 %26lt; -3.448613 )


= 0.01071143





Since the p-value is greater than the significance level of 0.01 we fail to reject the null hypothesis and conclude μ = 16 is plausible.











Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.


In order for the Confidence Interval to be valid you must have data from a normal distribution, at least if you are using the method here. If you do not have normal data then this type of confidence interval is not valid.





To clear up the notation I will use here. "t" is the test statistic and "t_(n-1)" is a Student t random variable with n - 1 degrees of freedom, e.g. a Student t random variable with 18 degrees of freedom is denoted as t_18.For small sample confidence intervals about the mean you have:


xBar ± t * sx / sqrt(n)





where xBar is the sample mean


t is the t - score with n - 1 degrees of freedom such that α% of the data in the tails, i.e., P( |t_(n-1)| %26gt; t) = α sx is the sample standard deviation


n is the sample size





The sample mean xbar = 19.625


The sample standard deviation sx = 2.973094


The sample size n = 8





The t score for a 0.99 confidence interval is the t score such that 0.005 is in each tail.


t = 3.499483


The confidence interval is:





( xbar - t * sx / sqrt( n ) , xbar + t * sx / sqrt( n ) )


( 15.94653 , 23.30347 )








because the value of 16 is in the interval we consider 16 to be a plausible value for the true mean.
Reply:B) Find the 99% confidence interval of the true mean.





CORRECT ANSWER to B)





( xbar - t * sx / sqrt( n ) , xbar + t * sx / sqrt( n ) )


( 16.6 , 22.7) confidence interval of the true mean





C) Confidence Interval does not agree with Null Hypothesis.


No comments:

Post a Comment