Statistic(Population variance unknown)
In the below dataset , in the task three solution, how is t statistic calculated as 3.36... Please can you explain. I did not get...
Dataset | ||||
$ 78,000 | Task 1: | Mean | $ 92,533 | |
$ 90,000 | St. deviation | $ 13,932 | ||
$ 75,000 | Standard error | $ 4,644 | ||
$ 117,000 | ||||
$ 105,000 | Task 2: | Population variance is unknown | ||
$ 96,000 | We have a small sample | |||
$ 89,500 | We assume that the population is normally distributed | |||
$ 102,300 | The appropriate statistic to use is the t-statistic | |||
$ 80,000 | ||||
Task 3: | 99% CI, t8,0.005 | 3.36 | ||
Sample size is 9 => the degrees of freedom are 8. |
1 answers ( 0 marked as helpful)
Hello Smriti!
Here you have a sample size of 9 observations which means the degrees of freedom would be equal to n-1 or 8.
You use the t-table:
d.f. / α | 0.1 | 0.05 | 0.025 | 0.01 | 0.005 |
1 | 3.078 | 6.314 | 12.706 | 31.821 | 63.657 |
2 | 1.886 | 2.920 | 4.303 | 6.965 | 9.925 |
3 | 1.638 | 2.353 | 3.182 | 4.541 | 5.841 |
4 | 1.533 | 2.132 | 2.776 | 3.747 | 4.604 |
5 | 1.476 | 2.015 | 2.571 | 3.365 | 4.032 |
6 | 1.440 | 1.943 | 2.447 | 3.143 | 3.707 |
7 | 1.415 | 1.895 | 2.365 | 2.998 | 3.499 |
8 | 1.397 | 1.860 | 2.306 | 2.896 | 3.355 |
9 | 1.383 | 1.833 | 2.262 | 2.821 | 3.250 |
And examine the case for 99% confidence level. This means that half of alpha would be equal to 0.005 (two-sided test).
You check in the table to see that the corresponding value is equal to 3.36.
Best,
The 365 Team