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Hypothesis Tests for a Mean with the t-Distribution

Test a population mean using a one-sample t statistic when the population standard deviation is unknown. Check the sampling model, use the planned tails and degrees of freedom, and interpret the p-value assuming the null hypothesis.

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What a learner can do afterwards

  • Carry out a one-sample t-test and state the conclusion in context
  • Choose between one-tailed and two-tailed alternatives before seeing the data
  • State what a p-value is and what it is not

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Calculate t as the sample mean minus null mean μ0, divided by SE = s / sqrt(n). Here s is the sample standard deviation, n the sample size and sqrt(n) its square root. Use df = n - 1, so n = df + 1. Under H0, the t reference is exact for independent observations from one normal population with positive variance. Large independent samples can allow a non-normal approximation; skewness and outliers matter. Calculating t alone does not validate this reference.

n = 25, s = 15sqrt(n) = 5SE = s / sqrt(n) = 3
Calculate the estimated standard error before dividing the gap from the null mean.

State the population mean, null value, alternative and level α before examining results. An alternative μ < μ0 uses the lower tail; μ > μ0 uses the upper tail; μ ≠ μ0 uses both tails. Choose this from the planned question, not the observed sign. A prediction alone does not require a one-tailed test if the planned question concerns departures in either direction.

Under H0 and the assumed model, the p-value is the probability of a statistic at least as extreme as observed. The planned alternative defines extremeness. Let T have the null t distribution with n - 1 degrees of freedom. Use its lower tail for μ < μ0 and upper tail for μ > μ0. For μ ≠ μ0, use both tails: by symmetry, double the upper-tail probability beyond |t|, the size of t without its sign. This is not the probability that H0 is true.

Independent normal fillsH0: μ = 100 mLH1: μ > 100 mLn = 16; s = 8 mLsample mean = 105 mLSE = 8 / sqrt(16) = 2 mLt = (105 - 100) / 2 = 2.5df = 15; upper p = 0.0123p < 0.05: reject H0evidence μ > 100 mL
Software gives the rounded upper-tail p-value. At planned α = 0.05, reject H0. This supports a mean-fill difference, not a cause.

Calculate SE = s / sqrt(n), then t = (sample mean - μ0) / SE. Keep the gap's sign. For n = 36, sample mean 18, s = 12 and μ0 = 24, sqrt(36) = 6, SE = 12 / 6 = 2 and t = (18 - 24) / 2 = -3. Under a symmetric null t model, negative t has upper-tail p > 0.5. Recover the sample mean as μ0 + t × SE. Use supplied square-root approximations and round only the final result.

At the level α chosen in advance, reject when p < α; otherwise fail to reject. Failure to reject means the evidence did not cross that threshold, not that H0 is true or that no evidence opposes it. Report the population, variable and planned direction. For a matching two-sided t-test at α = 0.05, null values within the closed 95% t interval, including its endpoints, are not rejected; values outside are rejected.

Check the model and plan the alternative and level. Calculate t, df and the matching tail probability. State the population conclusion; failing to reject does not prove H0.

Original lesson, questions and flashcards about one-sample t-tests, authored for this learning dataset.

The readings, questions and flashcards are original text authored for this learning dataset. The linked videos are by Khan Academy.

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Hypothesis Tests for a Mean with the t-Distribution · Mathematics, ages 20 to 21 · LightMySky