- 2074 Asoj · 5 marks
Estimation vs testing; CI for hardness
Short answer
Estimation gives a parameter value or interval from a sample; hypothesis testing decides if a claim is supported. For hardness: mean = 64.68, t(0.005,9)=3.250, 99% confidence interval is (62.86, 66.50).
Discuss difference between estimation and hypothesis test of significance of population. The following hardness readings were obtained: 66.3, 63.5, 64.9, 61.9, 64.3, 64.7, 65.1, 64.5, 68.4, 63.2. Find 99% confidence interval for true hardness of magnesium alloy.
Estimation versus testing of hypothesis
Estimation finds a value, or an interval of values, for an unknown population parameter from a sample. Testing of hypothesis decides, using sample evidence, whether a stated claim about the parameter should be accepted or rejected.
| Basis | Estimation | Hypothesis testing |
|---|---|---|
| Purpose | To find the value of the parameter | To check a claim about the parameter |
| Starting point | No prior claim about the parameter | A definite claim (H0) is made first |
| Result | Point value or confidence interval | Decision: reject or fail to reject H0 |
| Question answered | "What is the value of θ?" | "Is θ equal to (or greater/less than) θ₀?" |
| Error control | Confidence level (1 − α) | Level of significance α and power |
| Example | 99% interval for mean hardness | Test whether mean hardness = 65 |
Both use the sampling distribution of the statistic and are closely linked: a two-tailed test at level α rejects H0: μ = μ₀ exactly when μ₀ lies outside the (1 − α) confidence interval.
99% confidence interval for mean hardness
Population SD is unknown and n is small (10), so the t-distribution is used.
Σx = 646.8, n = 10
x̄ = 64.680
s² = Σ(x−x̄)²/(n−1) = 28.1760/9 = 3.1307
s = 1.7694
SE = s/√n = 1.7694/√10 = 0.5595
t(0.005, 9) = 3.250
margin = 3.250 × 0.5595 = 1.818
limits = 64.680 ± 1.818
Answer: the 99% confidence interval for the true hardness is (62.86, 66.50), i.e. about 62.9 to 66.5.
✓ Answer checked by Claude Sonnet 5.5
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