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