In: Statistics and Probability
| College Name | State | Public (1)/ Private (2) | in-state tuition |
| Berry College | GA | 2 | 8050 |
| University of Massachusetts at Dartmouth | MA | 1 | 1836 |
| Olivet College | MI | 2 | 10500 |
| Saint Olaf College | MN | 2 | 14350 |
| Francis Marion University | SC | 1 | 2920 |
| Huron University | SD | 2 | 7260 |
| Tennessee Technological University | TN | 1 | 1740 |
| Abilene Christian University | TX | 2 | 7440 |
| Southwestern Adventist College | TX | 2 | 7536 |
| Radford University | VA | 1 | 2924 |
The attached Excel file has a sample of tuition rates at 10 colleges in the US. Based on this sample consider the hypothesis that average tuition is $7500. Ho: µ=7500. Calculate the t value for this test. Enter your answer to 3 decimal places.
The attached Excel file has a sample of tuition rates at 10 colleges in the US. Based on this sample consider the hypothesis that average tuition is at least $8000. Ho: µ≥8000. Calculate the t value for this test. Enter your answer to 3 decimal places.
Ans a ) using excel
| t Test for Hypothesis of the Mean | |
| Data | |
| Null Hypothesis m= | 7500 |
| Level of Significance | 0.05 |
| Sample Size | 9 |
| Sample Mean | 6278.444444 |
| Sample Standard Deviation | 4326.226506 |
| Intermediate Calculations | |
| Standard Error of the Mean | 1442.0755 |
| Degrees of Freedom | 8 |
| t Test Statistic | -0.8471 |
| Two-Tail Test | |
| Lower Critical Value | -2.3060 |
| Upper Critical Value | 2.3060 |
| p-Value | 0.4216 |
| Do not reject the null hypothesis |
the t value for this test = -0.847
Ans b ) using excel
| t Test for Hypothesis of the Mean | |
| Data | |
| Null Hypothesis m= | 8000 |
| Level of Significance | 0.05 |
| Sample Size | 9 |
| Sample Mean | 6278.444444 |
| Sample Standard Deviation | 4326.226506 |
| Intermediate Calculations | |
| Standard Error of the Mean | 1442.0755 |
| Degrees of Freedom | 8 |
| t Test Statistic | -1.1938 |
| Lower-Tail Test | |
| Lower Critical Value | -1.8595 |
| p-Value | 0.1334 |
| Do not reject the null hypothesis |
the t value is -1.194