Cr 914

OK. 20 cardholders randomly slected CR were interviewed and the average amount they charged during the last twelve months has been found at $ 2,017, while the standard deviation was $ 867. Using a level 2.5 importance of assessing the application that the average amount charged by all holders of credit was over $ 1,750, which would be the null hypothesis (Ho) and alternate (Hi) hypothesis? I have a couple of diff. assumptions, but it is a bit confusing ...

Let us assume that a trial against the null hypothesis. The data are evidence against average. You assume the mean is true and try to prove that this is not true. After finding the test statistic and p-value, if the p-value less than or equal to the level of significance of the test, we reject the null hypothesis and conclude the alternative hypothesis is true. If the p-value is greater than a level of significance, then we can not reject the null hypothesis and conclude it is plausible. Note that we can not conclude the null hypothesis is true, just that it is plausible.

If statement question asks whether there is a difference between statistics and a value, then you have a two-tailed test, the null hypothesis, for example, would be μ = D vs the alternative hypothesis μ a‰  d

if the question ask to test an inequality you make sure your results will be worth it. example. say you have a steel bar that will be used in a construction project. If the bar can support a load of 100,000 pounds per square inch, then you'll use the bar, if it does, so you may not use the bar.

if the null hypothesis is μ a‰¥ 100,000 vs the alternate μ <100,000 then will be a test of significance. In this case, if you reject the null hypothesis, you conclude that the alternative hypothesis is true and the average load of the bar can support is less than 100,000 pounds per square inch and you will not be able to use bar. However, if you fail to reject the null then you conclude it is plausible the mean is greater than or equal to 100,000. We can never conclude that the null hypothesis is true. Therefore, you should not use the bar because you have no proof that the average strength is quite high.

if the null hypothesis is μ a‰¤ 100,000 against the alternate μ> 100,000 and you reject the null then you conclude the alternate is true and the bar is strong enough, if you are unable to reject it is plausible the bar is not strong enough, so you do not use it. In this case, you have a significant result.

Whenever you set the hypothesis test you need to consider whether or not the results will be significant.


Small sample hypothesis test to say:

For this test to be valid, the data must come from a normal population. If this is not the case, 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 for testing the null hypothesis

H0: μ a‰¤ Δ or
H0: μ a‰¥ Δ or
H0: μ = Δ
Find the test statistic = (Xbar - Δ) / (SX / aˆš (n))

where Xbar is the sample mean
SX is the standard deviation, if you know the standard deviation of the population, σ, then replace SX with σ in the equation of the test statistic.
n is the sample size

and follows the Student t distribution with n - 1 degrees of freedom. We use the t distribution to account for uncertainty in the estimation of the variance.
As the degrees of freedom approach infinity Student converges in probability to the standard normal. In most cases, the percentile values of the t-statistics are fairly close to normal when the type of degrees of freedom are greater than 30. It is the source of the empirical rule of thumb that the sample size> 30 have an average that is normally distributed. Keep this in mind as well, for these hypotheses.