According to a 2021 survey of student loan debt incurred to finance undergraduate education, the average amount of debt for a 4-year college degree with $37,172 with a standard deviation of $9,250. The
Inputs |
$ |
Sample |
$37,172 |
Presumed Population Standard Deviation, σ |
$9,250 |
Please Note: The t-value with 999 degrees of freedom is 1.962. The formula in Excel is =TINV(0.05,999). You cannot find this value on the Student-t table.
- What is the point estimate for the population mean, μ?
Point Estimate for the population mean, μ = |
|
What is the appropriate z-value to use when constructing a confidence interval at a 95% and 99% confidence levels? Use the z-value table and Microsoft Excel.
Confidence Level |
z-Value (Table) |
z-Value (Excel) |
Excel Formula |
95% (0.90) |
|
|
=ABS(NORM.S.INV(0.025)) |
99% (0.99) |
|
|
=ABS(NORM.S.INV(0.005)) |
- What is the appropriate t-value to use when constructing a confidence interval at a 95% and 99% confidence levels using Excel TINV
function . Use Microsoft Excel to find the t-values as they will not be on the Student-t table.
Confidence Level |
t-Value (Excel) |
Excel Formula |
95% (0.95) |
|
=TINV(1-.95,df) |
99% (0.99) |
|
=TINV(1-.99,df) |
Using a z-values calculate the confidence interval using at a 95% and 99% confidence levels. Do this by hand and use Excel’s CONFIDENCE.NORM function.
Using a t-values calculate the confidence interval using at a 95% and 99% confidence levels. Do this by hand and use Excel’s CONFIDENCE.T function. Assume the sample standard deviation is $9,250.
What does these confidence intervals you constructed tell you?
Compare the confidence intervals created using z-values and student-t. What are the big differences?
Confidence Intervals are used to estimate unknown population parameters. Which standardized score is more appropriate: z-values or student-t? Explain your answer.
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