Final exam, fully explained.
Four worked problems with live graphs and short checks. Follow each calculation, then change a variable to see what the result means.
1. Sampling distribution & CLT
How much does the average fuel consumption change from sample to sample?
Choose the right quantity
The population standard deviation describes the spread of single flights. The standard error describes the spread of averages from repeated samples of 16 flights. The question asks for the second quantity.
Worked solution
- (a) Calculate standard error.
Averaging 16 independent flights reduces the sampling spread by √16 = 4.
SE = 380 / √16 = 380 / 4 = 95 kg - (b) Standardize the interval.
Using one sampling SD, the interval is 2,450 ± 95, or 2,355 to 2,545 kg. Convert each bound to a Z-score.
Z = (x̄ − μ) / SE
Z lower = −1; Z upper = +1From the standard normal distribution:
P(−1 ≤ Z ≤ 1) = Φ(1) − Φ(−1)
= 0.84134 − 0.15866 = 0.68269 - (c) Judge the approximation.
The population is right-skewed. A sample of 16 may be too small for the sample mean to follow a normal distribution closely. We cannot confirm the accuracy without more detail about the skewness.
Explore the sampling distribution
Change the sample size. The population values remain fixed at the figures in the paper.
From 2,355 to 2,545 kg. The shaded area is the probability that a sample mean falls in this range, if the normal approximation applies.
Why can skewness matter?
The CLT concerns the distribution of sample means, not the shape of individual measurements. With larger independent samples, the sample-mean distribution tends toward a normal shape, provided its conditions hold.
A sample size of 16 does not guarantee a good approximation for a skewed population. The next experiment is only an illustration: the exam gives no actual flight measurements or detailed distribution.
Matches μ = 2,450 kg and σ = 380 kg, but is not flight data from the paper.
Run the simulation to compare observed sample means with the normal approximation.
2. Confidence interval for shaft stress
How precisely do 36 measurements estimate the true mean maximum shear stress?
Why use a Z interval?
The paper gives a known standard deviation of 6 MPa from instrument calibration. For a normally distributed measurement model, we use a Z-based confidence interval for the population mean.
Worked solution
- (a) Find the standard error.SE = 6 / √36 = 6 / 6 = 1 MPa
The sample mean has less sampling variability than an individual measurement.
- Find the 95% margin of error.
For 95% confidence, z* = 1.96.
Margin = 1.96 × 1 = 1.96 MPa - Build the interval.142 − 1.96 = 140.04 MPa
142 + 1.96 = 143.96 MPaAbout 95% of intervals constructed this way would contain the true population mean across repeated sampling.
- (b) Compare with the allowable stress.Upper endpoint = 143.96 MPa < 155 MPa
The full interval for the mean lies below the allowable limit. That supports the stated mean-stress check, under the model assumptions.
Adjust the confidence interval
Move the sample mean or sample size, or choose a confidence level. The known σ = 6 MPa and limit = 155 MPa stay fixed.
The bar estimates the population mean, not the maximum of every individual stress measurement.
Why does confidence change the width?
A higher confidence level uses a larger critical value z*. That makes the interval wider when the mean, σ, and n stay fixed.
Increasing n reduces the standard error because the denominator √n grows. Larger samples usually produce a narrower interval with the same confidence level.
3. One-sample t-test
Does the aluminum alloy's mean Young's modulus fall below the required 70 GPa?
Start with the hypotheses
We test for failure of the minimum specification. The alternative hypothesis therefore points left: the true mean is less than 70 GPa.
H₁: μ < 70 GPa Evaluate the test statistic at the boundary μ₀ = 70. Significance level: α = 0.05.
Worked solution
- (a) Use the first six measurements.
The sample mean is 69.85 GPa, and the sample SD is about 0.5010 GPa. We estimate the unknown population SD from the data, so we use Student's t distribution.
t = (x̄ − 70) / (s / √n)
= (69.85 − 70) / (0.5010 / √6)
= −0.7334, df = 5The left-tail p-value is 0.2481. Since 0.2481 > 0.05, fail to reject H₀. There is not enough evidence to say the true mean is below 70 GPa.
- (b) Include three new measurements.
The updated mean falls to 69.6111 GPa. The sample SD becomes 0.5904 GPa.
t = (69.6111 − 70) / (0.5904 / √9)
= −1.9760, df = 8The left-tail p-value is now 0.0418. Since 0.0418 < 0.05, reject H₀. There is statistically significant evidence that the true mean is below the requirement.
- Explain why the decision changes.
The three added values (68.6, 69.6, 69.2) lower the sample mean. The resulting t-statistic becomes more negative, and its left-tail p-value falls below 0.05.
Show the sample variance calculation
Use the formula printed on the exam sheet:
Initial six measurements: the sum of squared deviations is 1.255 GPa². Divide by 5 to obtain s² = 0.2510 GPa². Therefore s = √0.2510 = 0.5010 GPa.
All nine measurements: s² = 0.348611 GPa² and s = 0.590433 GPa.
Compare the preliminary and revised tests
Switch between the original samples. Changing α changes the decision rule, not the data or p-value.
Young's modulus (GPa, original measurements)
4. Pooled two-sample t-test
Is the new teaching method associated with a higher mean score?
State the comparison first
The claim says the new method produces higher mean scores. Define the difference as new minus old, so a positive value supports the alternative hypothesis.
H₁: μnew > μold Right-tailed test with α = 0.05, using pooled variance as requested by the paper.
Worked solution
- (a) Set the hypotheses.
We test whether μnew − μold is greater than zero. The reference difference under the null boundary is zero.
- (b) Compute means and sample variances.x̄ old = 84.24; s² old = 8.4204
x̄ new = 85.54; s² new = 13.3249The observed mean difference is 1.30 points.
- Pool the variances.
This method assumes equal population variances. Weight each sample variance by its degrees of freedom, then divide by the total degrees of freedom.
s² pooled = [9(8.4204) + 9(13.3249)] / 18
= 10.8727; s pooled = 3.2974 - Calculate the t-statistic.t = (85.54 − 84.24) / [3.2974√(1/10 + 1/10)]
= 1.30 / 1.4746 = 0.8816The degrees of freedom are 10 + 10 − 2 = 18.
- (c) Use the right-tail p-value.p = P(T₁₈ ≥ 0.8816) = 0.1948
Because p = 0.1948 > 0.05, fail to reject H₀. The observed improvement does not give enough evidence for a higher population mean.
Check the original scores and variances
Old method: 89.7, 81.4, 84.5, 84.8, 87.3, 79.7, 85.1, 81.7, 83.7, 84.5.
New method: 84.7, 86.1, 83.2, 91.9, 86.3, 79.3, 82.6, 89.1, 83.7, 88.5.
Sum each set and divide by ten to obtain 84.24 and 85.54. For the sample variances, the sums of squared deviations are 75.784 (old) and 119.924 (new). Divide each by 9 to obtain 8.4204 and 13.3249.
Why pooled instead of Welch's test?
The exam explicitly requests a pooled two-sample t-test. That method assumes the two population variances are equal, even though their sample variances can differ. The pooled estimate combines both sample variances using their degrees of freedom.
Welch's test does not require equal population variances and uses a different standard error and degrees-of-freedom formula. We do not replace the method named in this paper.
See the effect of an improved score
Add the same number of points to every new-method score. This is a hypothetical exercise, not a change to the exam data.
Practice & review
Check whether you can choose the method, interpret the result, and avoid common mistakes.
Answer without referring to the steps above. After each choice, the page explains the decision. Incorrect answers can be retried when you restart.
Formula reference
These are the core formulas used in the solutions. Check the conditions before using any formula.
Known population SD. Used in Problem 1 and the Z interval in Problem 2.
Population SD known; normal model or an appropriate large-sample approximation.
Unknown population SD. Degrees of freedom: n − 1.
Estimated from data, as in Problem 3.
Requires the equal-population-variance assumption.
At null boundary Δ₀ = 0. Degrees of freedom: n₁ + n₂ − 2.