Statistics - Computing for critical values

Question

Estimate the value of the tcritical if i wish to run the test of acceptance analysis with an alpha and a beta level of 5% each, when detecting moderate to large effects in a right tailed hypothesis. report up to two decimal digits.

Answer 1

A weird question. Cohen defined strong and moderate as .8 and .5 standard deviations. If we knew the standard deviation, then z crit would simply be 1.645, the ordinary value for alpha = .05 and n being calculated as the value necessary for 95% power when the effect size is .5. However, as the problem talks of t crit, we have to find the minimum n necessary, taking the fact that t changes as n changes.

As we have an effect size of .5, and alpha and beta are equal, the cutoff will be at .25 standard deviations. (.25 + .25 = .5)

Then, we need .25 sqrt(n) = t crit, or n = 16 t crit^2, where we have n - 1 degrees of freedom.

When n = 45, 16 t crit^2 = 45.17; When n = 46, 16 t crit ^2 = 45.17.

Thus, n = 46 is sufficient.

For this n, t crit = , in Excel, =T.INV(0.95, 45) = 1.68

This would be our answer.

1.68/sqrt(46) = .247618 (no roundoffs actually used).

Then, to get beta, t.dist((.247618 - .5)*sqrt(46), 45) = .0469, slightly less than the .05 required (that is because n is an integer).

Answer 2

If your alpha level (this is the level of significance) is 5% in a right tailed hypothesis, then you are looking for the t-value that corresponds to the 95 percentile for a t-distribution with n-1 degrees of freedom, where n is the sample size. Since your problem does not give a sample size, you can estimate the t critical value by using a normal distribution, but this will not be a good estimate if the sample size is small. You can use t-tables or the function invNorm on a TI calculator to find this value. The 95 percentile for the standard normal curve is 1.645 or 1.65 rounded to 2 decimal places.

Let me know if you have any questions.