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TINV in WPS Spreadsheet: Two-Tailed Student t Critical Values

Statistical Intermediate WPS Spreadsheet

Calculate a two-tailed t cutoff with explicit degrees of freedom and avoid one-tail probability mistakes.

Verification status: See the evidence and test scope described in this guide. How formula examples are checked

Quick answer

=TINV(0.05,10) has an independent expected result of 2.228138851986. The probability of a Student t variable with 10 degrees of freedom falling below −2.22813885 or above +2.22813885 is 0.05 in total.

Syntax and arguments

TINV(probability,degrees_freedom)

Both arguments are required numeric values. probability is the combined two-tail area, P(|T| > t), not a lower-tail CDF. The lesson’s walkthrough specifies 0 < probability ≤ 1; the worked examples use strict interior values. degrees_freedom must be at least 1; fractional values are truncated according to the source. Use integer df derived from your analysis.

Worked worksheet: a two-sided 5% cutoff

Enter 0.05 in B2 and 10 in B3, labeled Combined tail probability and Degrees of freedom. Enter =TINV(B2,B3) in D2. Expect approximately 2.228138851986. The symmetric cutoffs are negative and positive versions of that value; the function returns the nonnegative magnitude.

Independent verification integrates the Student t density outside ±t, or equivalently solves the upper survival probability t.sf(t,10)=0.025. This is independent mathematics, not a WPS runtime test.

The first-party lesson prints 2.28139 for this example. That conflicts with the defining two-tail equation and independent integration. This guide uses 2.228138851986 and records the source typo rather than repeating it.

Advanced example: a one-sided cutoff

For an upper-tail probability of 0.05 with df=10, double that probability when calling this two-tail inverse:

=TINV(2*0.05,10)

Expected result: 1.812461122812. The positive cutoff has 5% above it; the corresponding negative cutoff has 5% below it. This doubling recipe is for one-tail areas at most 0.5; it is not a general inverse-CDF replacement. For common small significance levels, it preserves the required tail direction.

Applying a critical value carefully

For a conventional two-sided confidence interval for a normal-population mean with unknown variance, n=11 gives df=10. With sample mean 50 and sample standard deviation 3, the illustrative margin is =TINV(0.05,10)*3/SQRT(11), approximately 2.015427423505. The interval is approximately 47.984572576495 to 52.015427423505. This requires an appropriate independent sample and model assumptions; TINV does not check them.

Troubleshooting

WPS documents #VALUE! for nonnumeric inputs, #NUM! for df below 1, and an iterative-search failure #N/A. The formal probability-error line is incomplete, so do not infer its exact boundary errors from that line. Use the walkthrough’s bounded domain and interior controls.

A lower-CDF Student t inverse at probability 0.05 and df=10 instead gives approximately −1.812461122812. If that negative value appears where you expected the positive two-tail cutoff 2.228138851986, check whether a CDF inverse has been substituted. If the cutoff is 1.81 instead of 2.23, check whether 0.10 was supplied as the combined tail area. Formatting probability as 5% is equivalent to 0.05, not 5. Historical legacy spelling does not itself establish retirement or universal equivalence with modern t-inverse names.

Availability and verification

The official WPS TINV lesson was checked on October 19, 2026. The captured source does not establish a specific edition or numeric build. This Chinese first-party documentation is not proof of universal English-global, Mac, Linux, web or mobile support. Introduction and retirement versions remain unknown. All numerical results in this guide are independently calculated expectations, not native WPS test results or borrowed Excel outputs. Current GUI access was unavailable; no current-application screenshots or native execution are claimed.