Hypothesis Tests for a Mean with the t-Distribution · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Testing a claim about a mean

Mathematics · Data & Statistics · ages 20-21
Name ______________________   Date ____________

  1. A t-test summary gives sample mean 31, sample standard deviation 7, sample size 12 and null mean 29. What value is s in SE = s / sqrt(n)?

    Answer: ______________

  2. A sample has n = 4, mean 12 and sample standard deviation s = 4. For null mean 10, calculate the t statistic.

    Answer: ______________

  3. Before collecting data, a researcher plans to test whether the population mean recovery time is below a fixed benchmark. Which alternative matches this question?

    • Population mean differs from the benchmark
    • Population mean exceeds the benchmark
    • Population mean is below the benchmark

  4. A one-sample t-test uses 10 observations. How many degrees of freedom does its reference distribution have?

    • 9
    • 10
    • 8

  5. Independent normal bottle fills give n = 9, mean 54 mL and s = 6 mL. Test μ = 50 against μ > 50 at α = 0.05. The df = 8 table gives P(T ≥ 2) = 0.0403. There is evidence that the population mean fill exceeds 50 mL.

    Circle one:   True   False

  6. A matching 95% t interval is [18, 22]. A two-sided t-test of μ = 20 at α = 0.05 rejects that null value.

    Circle one:   True   False

  7. For a planned two-sided test, t = -2. The correct df row gives P(T ≥ 2) = 0.04. What is the p-value?

    • 0.04, using one tail
    • 0.08, using both tails
    • 0.96, using central area
    • 0.02, halving one tail

  8. A sample has n = 12 and sample standard deviation s = 5.196. Calculate its estimated standard error to three decimal places. Use sqrt(12) ≈ 3.464.

    Answer: ______________

  9. A planned test of mean repair time gives p = 0.06 at α = 0.05. H0 states that the population mean is 40 minutes. Which conclusion follows?

    • The population mean is established as exactly 40 minutes
    • The evidence does not cross the planned rejection threshold
    • The observed p-value sets a new rejection threshold of 0.06

  10. A sample has n = 8, mean 10 and sample standard deviation s = 4. For null mean 12, calculate t to three decimal places. Use sqrt(8) ≈ 2.828.

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Testing a claim about a mean W1-mt_ig2o39hAwh-s1

  1. 7 · The symbol s denotes the sample standard deviation, which is 7 in this summary.
  2. 1 · The signed gap is 2 and the estimated standard error is 2, so t = 1. This calculation alone does not check the sampling model.
  3. Population mean is below the benchmark · The planned question concerns a lower population mean, so the alternative uses < and the lower tail.
  4. 9 · The one-sample rule is df = n - 1, giving 9.
  5. True · SE = 2 mL and t = 2 with df = 8. The planned upper-tail p-value is 0.0403, below 0.05, supporting the stated population-mean conclusion.
  6. False · The null value 20 lies inside the matching 95% interval, so the corresponding two-sided test does not reject it at 0.05.
  7. 0.08, using both tails · The two-sided alternative uses both symmetric tails beyond |t| = 2, giving 2 × 0.04 = 0.08.
  8. 1.5 · Using the supplied approximation, SE = 5.196 / 3.464 = 1.5, or 1.500 to three decimal places.
  9. The evidence does not cross the planned rejection threshold · The test fails to reject because 0.06 exceeds 0.05. This does not prove H0 or mean that the data supply no evidence against it.
  10. -1.414 · The signed gap is -2. Dividing by SE = 4 / 2.828 gives t ≈ -1.414.
Worksheet · LightMySky