Source: Quiz 7 Review, Principles of Statistics I (Texas A&M University)
Tags: hypothesis testing, null hypothesis, alternative hypothesis, t-test, z-test, one-sample test, test statistic, p-value, significance level, alpha, reject null, fail to reject, STAT 211, proportions test
Hypothesis testing is a structured method for deciding whether sample data provides enough evidence to reject a claim about a population parameter. You set up competing hypotheses (null vs. alternative), compute a test statistic, find the p-value, and compare it to your significance level to reach a conclusion. The choice between a Z-test and a T-test depends on whether you know the population standard deviation and whether you are testing a mean or a proportion.
Null hypothesis (H₀)
The default claim about the population parameter. It always contains an equality (=, ≤, or ≥). You assume it is true unless the data gives you strong enough evidence to reject it.
Alternative hypothesis (H₁ or Hₐ)
The competing claim you are trying to find evidence for. It can be one-sided (< or >) or two-sided (≠).
Test statistic
A standardised value calculated from your sample data, assuming H₀ is true. It measures how far your sample result is from what H₀ predicts. Larger absolute values suggest stronger evidence against H₀.
P-value
The probability of observing a test statistic as extreme as (or more extreme than) the one you calculated, assuming H₀ is true. A small p-value means the observed data would be unusual if H₀ were correct.
Significance level (α)
The threshold you set before testing. If your p-value falls at or below α, you reject H₀. Common values are 0.05 and 0.01.
One-tailed test (one-sided)
Used when H₁ specifies a direction (greater than or less than). The entire rejection region sits in one tail of the distribution.
Two-tailed test (two-sided)
Used when H₁ specifies "not equal to." The rejection region is split across both tails.
Degrees of freedom (df)
For a one-sample t-test: df = n − 1. This determines which t-distribution you use to find the p-value.
Sample proportion (p̂)
The proportion of successes in your sample: p̂ = X / n, where X is the number of successes and n is the sample size.
There are three test statistics covered in this material, each suited to different conditions:
Z-test for a mean (σ known): use when the population standard deviation σ is given and the sample size is large, or the population is normal
Z* = (X̄ − μ₀) / (σ / √n)
T-test for a mean (σ unknown): use when the population standard deviation is not known and you rely on the sample standard deviation s instead; requires normality or large n
T* = (X̄ − μ₀) / (s / √n)
Degrees of freedom: df = n − 1
Z-test for a proportion: use when testing a population proportion p against a claimed value p₀
Z* = (p̂ − p₀) / √(p₀(1 − p₀) / n)
Conditions: np₀ ≥ 5 and n(1 − p₀) ≥ 5
State the hypotheses: write H₀ (the claim with equality) and H₁ (what you want evidence for)
Set the significance level (α): typically given in the problem
Compute the test statistic: plug your sample values into the appropriate formula, assuming H₀ is true
Find the p-value: the tail probability corresponding to your test statistic, matched to the direction of H₁
Make your decision: if p-value ≤ α, reject H₀; if p-value > α, fail to reject H₀
The wording of the research question tells you H₁:
"Is the true mean equal to 50?" → two-tailed: H₁: μ ≠ 50
"Do students drink more than 3 cups?" → right-tailed: H₁: μ > 3
"Is the proportion less than 49.4%?" → left-tailed: H₁: p < 0.494
"Has the average decreased?" → left-tailed: H₁: μ < 15.2
H₀ always takes the complementary form with equality included.
Setup: n = 6 steel beams, X̄ = 58 psi, s = 6 psi, α = 0.05. Test whether the true mean compressive strength is 50 psi. Normality assumed.
H₀: μ = 50
H₁: μ ≠ 50 (two-tailed, because the question says "whether" the mean is 50, not "greater" or "less")
σ is unknown, so use the t-test
T* = (58 − 50) / (6 / √6) = 8 / 2.449 ≈ 3.266
df = 6 − 1 = 5
P-value: find the two-tailed probability for T* = 3.266 with 5 df. This falls between 0.02 and 0.05 (approximately 0.022)
Since p-value < 0.05, reject H₀
Interpretation: At the 0.05 significance level, there is sufficient evidence to conclude that the true mean compressive strength of the steel is not 50 psi.
Setup: n = 49 students, X̄ = 3.7 cups, s = 1.4 cups, α = 0.05. National average μ₀ = 3 cups. Test whether university students drink more coffee than the national average.
H₀: μ ≤ 3 (or μ = 3)
H₁: μ > 3 (right-tailed, because the question asks "more than")
σ is unknown, so use the t-test
T* = (3.7 − 3) / (1.4 / √49) = 0.7 / 0.2 = 3.5
df = 49 − 1 = 48
P-value: right-tail probability for T* = 3.5 with 48 df. This is very small (approximately 0.0005)
Since p-value < 0.05, reject H₀
Interpretation: At the 0.05 significance level, there is sufficient evidence to conclude that university students drink more coffee on average than the national average of 3 cups per day.
Setup: n = 40 gym members, X = 28 met guidelines, p₀ = 0.494, α = 0.01. Test whether the proportion of gym members meeting CDC physical activity guidelines is less than the national average.
p̂ = 28 / 40 = 0.70
H₀: p ≥ 0.494 (or p = 0.494)
H₁: p < 0.494 (left-tailed, because the question asks "less than")
Z* = (0.70 − 0.494) / √(0.494 × 0.506 / 40) = 0.206 / 0.0791 ≈ 2.604
The test statistic is positive (sample proportion is above, not below, the claimed value)
P-value for a left-tailed test with Z* = 2.604 is very large (approximately 0.9954)
Since p-value > 0.01, fail to reject H₀
Interpretation: At the 0.01 significance level, there is not sufficient evidence to conclude that the proportion of gym members meeting the CDC guidelines is less than the national average of 49.4%. In fact, the sample proportion (70%) is well above the national figure, so the data runs in the opposite direction of the alternative hypothesis.
Setup: n = 35 calls, X̄ = 14.3 minutes, σ = 5 minutes (known), μ₀ = 15.2 minutes, α = 0.05. Test whether the average evening long-distance call has decreased.
H₀: μ ≥ 15.2 (or μ = 15.2)
H₁: μ < 15.2 (left-tailed, because the question asks "decreased")
σ is known, so use the z-test
Z* = (14.3 − 15.2) / (5 / √35) = −0.9 / 0.8452 ≈ −1.065
P-value: left-tail probability for Z* = −1.065 ≈ 0.1435
Since p-value > 0.05, fail to reject H₀
Interpretation: At the 0.05 significance level, there is not sufficient evidence to conclude that the average evening long-distance call duration has decreased from 15.2 minutes. Answer: No.
Z-test for a mean (σ known):
Z* = (X̄ − μ₀) / (σ / √n)
T-test for a mean (σ unknown):
T* = (X̄ − μ₀) / (s / √n), with df = n − 1
Z-test for a proportion:
Z* = (p̂ − p₀) / √(p₀(1 − p₀) / n)
where p̂ = X / n
⚠️ The single most common decision point: use a T-test when the population standard deviation σ is unknown and you are substituting the sample standard deviation s. Use a Z-test when σ is given or when testing a proportion.
⚠️ A two-tailed test doubles the tail area. If your one-tailed p-value is 0.011, the two-tailed p-value is 0.022. Forgetting to double is a frequent error.
⚠️ "Fail to reject H₀" is not the same as "accept H₀." You never prove the null is true; you simply say the data did not give enough evidence against it.
⚠️ Always interpret your conclusion in plain language, in context. A bare "reject H₀" without a sentence about what that means for steel beams, coffee, or gym members will lose you marks.
⚠️ Check the direction of the test statistic against the direction of H₁. In Problem 3, the sample proportion was above the claimed value while H₁ was testing for below. The p-value was enormous, and the conclusion was to fail to reject. The numbers have to make directional sense.
⚠️ When testing proportions, verify the conditions np₀ ≥ 5 and n(1 − p₀) ≥ 5 before proceeding.
Q: You have a sample of 25 observations with X̄ = 72, s = 8, and you want to test H₀: μ = 70 at α = 0.05 (two-tailed). What test statistic do you use, and what is its value?
A: Use a t-test because σ is unknown. T* = (72 − 70) / (8 / √25) = 2 / 1.6 = 1.25, with df = 24.
Q: A researcher claims that more than 60% of households own a pet. In a sample of 200 households, 130 own a pet. State H₀ and H₁, and compute the test statistic.
A: H₀: p ≤ 0.60, H₁: p > 0.60. p̂ = 130/200 = 0.65. Z* = (0.65 − 0.60) / √(0.60 × 0.40 / 200) = 0.05 / 0.03464 ≈ 1.443.
Q: If your calculated p-value is 0.03 and α = 0.01, what is your decision?
A: Fail to reject H₀. The p-value (0.03) exceeds the significance level (0.01), so the evidence is not strong enough at the 1% level.
Q: Why do we use a t-distribution instead of the standard normal when σ is unknown?
A: Replacing σ with the sample standard deviation s introduces additional uncertainty. The t-distribution accounts for this by having heavier tails, which makes extreme values more likely and produces wider confidence intervals. As n increases, the t-distribution approaches the standard normal.
Q: In Problem 4, the population standard deviation σ is known to be 5 minutes. If it were not known and you had s = 5 instead, what would change?
A: You would use a t-test with df = 34 instead of a z-test. For large samples (n = 35), the t-distribution is very close to the normal, so the p-value would be nearly identical, but the procedure and notation change.
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