Be careful, your changing alpha from 0.01 to later in the calcs to 0.1!
Might want to give it another try.
Reject the null hypothesis if |t calc| > |t crit|
I've read the stickied thread about this topic, but I still have some questions/I want to confirm that I'm doing this correctly.
We have experimental data that maps golf shots to how far away they are from the hole being aimed at. The data has the location of the hole labeled at 0, values shorter than the hole labeled at values from -5 to -1 and values farther than the hole labeled at values from 1 to 5.
So the problem is as follows:
Use the t-distribution to determine if you would reject or fail to reject the null hypothesis (that you shoot the proper distance to the whole). Use a significance value of .01. Then calculate a P-value
So I can calculate the sample standard deviation of the distance along with the sample mean. THen since the null hypothesis is that we hit the hole (origin), would be 0 and
So for my test statistic it would be :
In my case, sample standard deviation was 5.7 and number of tests was 30. The sample mean was .1, so in all I get:
t using .1 (rounded up) and N-1 = 29 gives 1.311
And t using a/2 = .1/2 = .05 and 29 gives 1.699
So since using the significance is less than the value I calculated, I should reject the null, correct? Or should the first equation be a lookup with .961/2 = .481. In the latter case I don't have a table entry for .481, but rounding to .5 would make the point of this calculation moot, so I'm guessing that since the value of the t distribution increases as alpha decreases, my calculated value is >= the value using the significance, so I shouldn't reject H0?
As for the P value, since this is a two-tailed test, I could use ?
Thanks.
Alright, here's a concise version of my calcs. The values I looked up are correct, so as long as the logic is fine, I'm good. I may have made a typo in one of my above posts (I got mixed up between the significance being .01 and the sample mean being .1), but this has all the correct numbers
Level of significance = .01 -> 0.01/2 = 0.005
Sample mean of the range = .1
Sample standard deviation = 5.714
.1 / 1.0406 = .0961
.0961 / 2 = .0481 (two tailed distribution since the ball can go too far or not far enough)
t value in the table is much larger for a = 0.005,29 than .0481,29 so our critical value is bigger and we should not reject the null hypothesis.
As for the t value, df = N-1 = 29 and I rounded up to the t value for 0.05,29 since that's the only value I have in the table.
So the p value for 29 and 1.699 (1.7) is 0.05 for a one-sided and there for .1 for a two-sided test.
Thanks again.
OK up to this point. Now you should recognise that your observed mean is about one tenth of a standard error from the mean for the null hypothesis, this should be a warning that this difference in statistical terms is utterly insignificant and the null hypothesis will not be rejected for any reasonable level of significance.
No you don't multiply by two here, you look the t-score up in a 2-sided table, or double the p-value from a one-sided table..0961 / 2 = .0481 (two tailed distribution since the ball can go too far or not far enough)
CBt value in the table is much larger for a = 0.005,29 than .0481,29 so our critical value is bigger and we should not reject the null hypothesis.
As for the t value, df = N-1 = 29 and I rounded up to the t value for 0.05,29 since that's the only value I have in the table.
So the p value for 29 and 1.699 (1.7) is 0.05 for a one-sided and there for .1 for a two-sided test.
Thanks again.
Alright, so if I use .0961 and round it to .1, I multiply the value in the T table to get almost 2.6ish?
Would I do the same thing for alpha? As in would I look up .1 and multiply it by two instead of rounding? I.e. would I use the value for 0.5 or use .1 * 2? Because if not, my calculated value would be much larger than the critical value, which may or may not be the point.
And then for the P value, would I just take the one-sided P value and multiply it by two?
It'll either be the value for 1.3 (value for t(.1,29)), 29 = .102 * 2 = 0.204
Or 2.6,29 = .007
For the P value. Not sure which it is, because I don't remember us ever multiplying things by 2 and it's throwing me off.
Thanks again guys.
I have no idea what you think you are doing in the above.
Assuming the null hypothesis, the probability of getting a t-value of absolute value greater than 0.1 with 29 degrees of freedom is 0.918 which is the p-value (or 91.8%, so there is no way to reject the null hypothesis at any reasonable level)
CB