SAMPLE 1 MEAN 5.61 MEDIAN 5.6 MODE 5.6 STANDARD ERROR 0.39992384 STANDARD DEVIATION 2.1904731 SAMPLE VARIANCE 4.79817241 KUTOSIS -0.07445435 SKEWNESS -0.69709258 RANGE 7.9 MINIMUM 1.1 MAXIMUM 9 SAMPLE COUNT 30 SUM 168.3 SAMPLE 2 MEAN 7.326666667 MEDIAN 7.3 MODE 8.7 STANDARD ERROR 0.248717785 STANDARD DEVIATION 1.362283411 SAMPLE VARIANCE 1.855816092 KUTOSIS -0.118922156 SKEWNESS -0.015286932 RANGE 5.6 MINIMUM 4.6 MAXIMUM 10.2 SAMPLE COUNT 30 SUM 219.8 AND THIS IS MY t-TEST-ASSUMING EQUAL VARIANCE Variable 1 Variable 2 Mean 5.61 7.326666667 Variance 4.798172414 1.855816092 Observations 30 30 Pooled Variance 3.326994253 Hypothesized Mean Difference 0 df 58 t Stat -3.645067529 P(T<=t) one-tail 0.000286392 t Critical one-tail 1.671552762 P(T<=t) two-tail 0.000572783 t Critical two-tail 2.001717484 IS IT RIGHT THAT MY T-STAT CAME OUT AS A NEGATIVE, AND ARE MY CALCULATIONS CORRECT?
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
SAMPLE 1
| 5.61 | |
| 5.6 | |
| 5.6 | |
| STANDARD ERROR | 0.39992384 |
| STANDARD DEVIATION | 2.1904731 |
| SAMPLE VARIANCE | 4.79817241 |
| KUTOSIS | -0.07445435 |
| SKEWNESS | -0.69709258 |
| 7.9 | |
| MINIMUM | 1.1 |
| MAXIMUM | 9 |
| SAMPLE COUNT | 30 |
| SUM | 168.3 |
SAMPLE 2
| MEAN | 7.326666667 |
| MEDIAN | 7.3 |
| MODE | 8.7 |
| STANDARD ERROR | 0.248717785 |
| STANDARD DEVIATION | 1.362283411 |
| SAMPLE VARIANCE | 1.855816092 |
| KUTOSIS | -0.118922156 |
| SKEWNESS | -0.015286932 |
| RANGE | 5.6 |
| MINIMUM | 4.6 |
| MAXIMUM | 10.2 |
| SAMPLE COUNT | 30 |
| SUM | 219.8 |
AND THIS IS MY t-TEST-ASSUMING EQUAL VARIANCE
| Variable 1 | Variable 2 | |
| Mean | 5.61 | 7.326666667 |
| Variance | 4.798172414 | 1.855816092 |
| Observations | 30 | 30 |
| Pooled Variance | 3.326994253 | |
| Hypothesized Mean Difference | 0 | |
| df | 58 | |
| t Stat | -3.645067529 | |
| P(T<=t) one-tail | 0.000286392 | |
| t Critical one-tail | 1.671552762 | |
| P(T<=t) two-tail | 0.000572783 | |
| t Critical two-tail | 2.001717484 |
IS IT RIGHT THAT MY T-STAT CAME OUT AS A NEGATIVE, AND ARE MY CALCULATIONS CORRECT?
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