WF3102 final exam | Interactive solutions

The exam answers

All four questions and all ten parts are covered. The worked answers stay fixed when you explore other values.

Fuel use

Standard error: 95 kg. Normal approximation: 68.27%. With 16 flights and a right-skewed population, the probability estimate needs caution.

Shear stress

95% confidence interval: 140.04 to 143.96 MPa. The entire interval lies below 155 MPa. This supports the mean being below the limit under the test model.

Young’s modulus

Six rods: do not reject. Nine rods: reject at 5%. The left-tail p-value changes from 0.2481 to 0.0418. The added data support a mean below 70 GPa.

Teaching methods

Pooled $t=0.8816$ , 18 degrees of freedom, $p=0.1948$ . The observed 1.30-point increase does not give enough evidence of a higher population mean at 5%.

A p-value measures how unusual a result at least this extreme would be under the null model. It is not the probability that the null hypothesis is true. A small p-value and a large practical effect are separate questions.

1. Fuel use and the sample mean

Known. Individual flights have mean fuel use $\mu=2450\,\mathrm{kg}$ and standard deviation $\sigma=380\,\mathrm{kg}$ . Their distribution is right-skewed. The random sample contains $n=16$ flights.

Find. The spread of the sample mean, a probability under the normal approximation, and the limits of that approximation.

Model. Treat the flights as independent draws from the same population. For sampling without replacement, assume the sample is a small fraction of the population.

(a) Find the standard error

The sample mean changes from one random sample to another. Its standard deviation is the standard error. For independent flights, averaging divides the variance by the sample size:

\operatorname{Var}(\bar X)=\frac{\sigma^2}{n}, \qquad \sigma_{\bar X}=\frac{\sigma}{\sqrt n}.

Substitute the given values:

\sigma_{\bar X}=\frac{380}{\sqrt{16}} =\boxed{95\,\mathrm{kg}}.

The 380 kg value describes individual flights. The 95 kg value describes means of 16 flights. This standard-error formula does not require a normal population.

(b) Find the probability within one standard error

Following part (a), “one standard deviation” means one standard deviation of the sample mean. We seek the probability that its distance from 2450 kg is at most 95 kg:

2450-95\leq\bar X\leq2450+95.

The question asks us to assume that the Central Limit Theorem gives a suitable normal approximation. Use the Z formula from the paper:

Z=\frac{\bar X-\mu}{\sigma/\sqrt n}.

The lower limit gives $(2355-2450)/95=-1$ ; the upper limit gives $(2545-2450)/95=1$ . Therefore,

\begin{aligned} P(2355\leq\bar X\leq2545) &\approx P(-1\leq Z\leq1)\\ &=\Phi(1)-\Phi(-1)\\ &=0.8413447-0.1586553. \end{aligned}

Here, $\Phi(z)$ is the area under the standard normal curve to the left of $z$ .

\boxed{P\approx0.6827=68.27\%}

The empirical rule gives the rounded answer of about 68%.

(c) Assess the normal approximation

With only 16 flights, the sample mean may still show right skew. Strong skew or rare, very high fuel-use values can slow its approach to a normal distribution.

The exam gives the direction of skew, but not its size or the full distribution. We therefore cannot guarantee that 68.27% is accurate. Mild skew may allow a useful approximation at this sample size. Stronger skew can require more flights. No single sample-size threshold works for every population.

The standard error remains 95 kg under the stated assumptions. Only the probability calculation depends on the normal approximation.

Why use 95 kg instead of 380 kg in part (b)?

Part (a) asks for the standard deviation of the sample mean, so that is the natural reference in part (b). If the wording instead meant one population standard deviation, the interval would be 2070 to 2830 kg. Its Z limits would be $-4$ and $4$ , giving about 99.9937% under the normal approximation. The sequence of parts and the empirical-rule hint support 68.27% as the intended answer.

About the illustrative skewed distributions

The comparison curves are synthetic examples, not observed flight data. Each uses the given mean and standard deviation while changing the amount of skew. Let $G$ follow a gamma distribution with shape $k$ and scale $380/\sqrt{k}$ :

X=2450-380\sqrt{k}+G, \qquad G\sim\operatorname{Gamma}\!\left(k,\frac{380}{\sqrt{k}}\right).

For independent flights, the sample mean has this exact illustrative distribution:

\bar X=2450-380\sqrt{k}+\bar G, \qquad \bar G\sim\operatorname{Gamma}\!\left(nk,\frac{380}{n\sqrt{k}}\right).

Its skewness is $2/\sqrt{nk}$ , so increasing the sample size reduces skew. This comparison explains why the population shape matters; it cannot recover the actual probability from the exam's limited information.

Source: final exam, page 1, Question 1; Z formula on page 2.

2. A confidence interval for shear stress

Known. There are $n=36$ independent measurements, with sample mean $\bar\tau=142\,\mathrm{MPa}$ and known error standard deviation $\sigma=6\,\mathrm{MPa}$ . The allowable maximum shear stress is 155 MPa.

Model. Use the exam's normal measurement-error model. Assume calibrated readings have no systematic bias and the 6 MPa error SD describes the variation relevant to these repeated measurements at the tested condition.

(a) Construct and interpret the 95% interval

The population standard deviation is known, so use a Z interval. We would use a t interval if we estimated that standard deviation from the sample.

First, calculate the standard error:

\mathrm{SE}=\frac{\sigma}{\sqrt n} =\frac{6}{\sqrt{36}}=1\,\mathrm{MPa}.

A central 95% normal interval leaves 2.5% in each tail. Its critical value is $z_{0.975}\approx1.96$ . Rearranging the paper's Z formula gives:

\begin{aligned} \mu_\tau&\in\bar\tau\pm z_{0.975}\frac{\sigma}{\sqrt n}\\ &=142\pm1.96(1)\,\mathrm{MPa}. \end{aligned}

\boxed{\mu_\tau\in[140.04,143.96]\,\mathrm{MPa}}

We are 95% confident that this interval contains the true mean maximum shear stress, under the model assumptions. The 95% describes the method: over many repeated samples, about 95% of intervals constructed this way would contain the fixed true mean. It does not describe the fraction of individual readings inside these limits.

(b) Compare the interval with the allowable stress

Compare the upper confidence limit with 155 MPa:

143.96<155\,\mathrm{MPa}.

Even the upper limit is 11.04 MPa below the allowable value. Under the exam’s mean-stress criterion, this supports assessing the shaft as safe at the 95% confidence level.

\boxed{\mu_\tau<155\,\mathrm{MPa}\text{ is supported.}}

This is evidence about the mean under the tested conditions. It does not guarantee that every future stress maximum stays below 155 MPa. The given 6 MPa describes measurement error; it does not, by itself, describe real changes in operating stress. Extra variation from changing loads would require a separate model.

Source: final exam, page 1, Question 2; Z formula on page 2.

3. Does the alloy meet the mean modulus requirement?

Let $\mu$ be the population mean Young’s modulus of the production batch. The specification requires a mean of at least $70\,\mathrm{GPa}$ . We test for a shortfall:

H_0:\mu\geq70\,\mathrm{GPa},\qquad H_1:\mu\lt70\,\mathrm{GPa}.

This is a left-tailed test: a sufficiently low sample mean counts against the design claim. We calculate the reference distribution at the null boundary, $\mu_0=70\,\mathrm{GPa}$ , and use $\alpha=0.05$ .

(a) Preliminary conclusion from six rods

Original measurements from the exam
Rod	1	2	3	4	5	6
Modulus (GPa)	69.7	70.2	68.9	70.2	70.1	70.0

Model. The population standard deviation is unknown, so we estimate it from the sample. We use the paper’s one-sample Student t formula. An exact t test assumes independent observations from a normal population. The exam gives a random sample but does not establish normality; with only six rods, this remains a needed assumption.

Find the mean and spread. The sample mean uses all six values:

\bar{x}=\frac{\sum x_i}{n}=\frac{419.1}{6}=69.85\,\mathrm{GPa}.

The squared deviations sum to $1.255\,\mathrm{GPa}^2$ . We divide by $n-1$ because we have already estimated the mean from these data:

\begin{aligned} s^2&=\frac{\sum(x_i-\bar{x})^2}{n-1}=\frac{1.255}{5}=0.251\,\mathrm{GPa}^2,\\ s&=\sqrt{0.251}=0.500999\,\mathrm{GPa},\\ \mathrm{SE}&=\frac{s}{\sqrt n}=\frac{0.500999}{\sqrt6}=0.204532\,\mathrm{GPa}. \end{aligned}

The standard error, or SE, estimates how much the sample mean varies across repeated samples of the same size.

Check the squared deviations

\begin{aligned} \sum(x_i-69.85)^2={}&(-0.15)^2+(0.35)^2+(-0.95)^2\\ &+(0.35)^2+(0.25)^2+(0.15)^2\\ ={}&1.255\,\mathrm{GPa}^2. \end{aligned}

Measure the gap in standard errors. Using the supplied test formula:

t=\frac{\bar{x}-\mu_0}{s/\sqrt n} =\frac{69.85-70}{0.204532}=-0.733382.

The test has $\nu=n-1=5$ degrees of freedom, the number of independent pieces of information left after estimating the mean. Its left-tail probability is

p=P(T_5\leq-0.733382)=0.248125.

Since $p\gt0.05$ , we fail to reject $H_0$ . The same decision follows from the critical-value rule: reject only below $-2.015048$ , while the observed statistic is $-0.733382$ .

The p-value measures compatibility with the null boundary. If the true mean were 70 GPa and the model assumptions held, about 24.8% of repeated samples would give a t statistic this low or lower. It does not give the probability that the batch meets the specification.

\boxed{\text{Fail to reject }H_0.}

The first six rods give insufficient evidence that the batch mean falls below 70 GPa. This does not prove that the batch meets the requirement.

(b) Revised conclusion after three more rods

Add $68.6,\ 69.6,\ 69.2\,\mathrm{GPa}$ to the original sample. We now have nine rods. The hypotheses, test direction, and significance level stay the same.

\begin{aligned} \bar{x}&=\frac{419.1+68.6+69.6+69.2}{9}\\ &=\frac{626.5}{9}=69.611111\,\mathrm{GPa},\\ s^2&=\frac{2.788889}{9-1}=0.348611\,\mathrm{GPa}^2,\\ s&=0.590433\,\mathrm{GPa},\\ \mathrm{SE}&=\frac{0.590433}{\sqrt9}=0.196811\,\mathrm{GPa}. \end{aligned}

Here $2.788889\,\mathrm{GPa}^2$ is the sum of squared deviations of all nine observations from their new mean. Recalculate the variance; the first sample’s variance no longer describes the full set.

\begin{aligned} t&=\frac{69.611111-70}{0.196811}=-1.975951,\\ \nu&=9-1=8,\\ p&=P(T_8\leq-1.975951)=0.041785. \end{aligned}

Now $p\lt0.05$ . The statistic also falls below the revised critical value, $-1.859548$ . Both rules lead to rejection.

\boxed{\text{Reject }H_0\text{ at }\alpha=0.05.}

The nine-rod sample gives evidence that the population mean modulus falls below the design requirement, under the stated model.

The requirement concerns the population mean. A rod below 70 GPa does not, by itself, show that the population mean falls below 70 GPa. The test combines the size of the mean shortfall, the observed spread, and the sample size.

Why the conclusion changes

Original and revised tests
Quantity	First 6 rods	All 9 rods
Mean (GPa)	69.8500	69.6111
Sample SD (GPa)	0.5010	0.5904
Standard error (GPa)	0.2045	0.1968
t statistic	−0.7334	−1.9760
Degrees of freedom	5	8
Left-tail p-value	0.2481	0.0418
Decision at 5%	Fail to reject	Reject

All three new values lie below 70 GPa, lowering the mean. The sample SD rises, while the larger sample reduces the SE slightly. Most of the stronger evidence comes from the larger mean shortfall, which grows from $0.1500$ to $0.3889\,\mathrm{GPa}$ . The revised p-value lies near the chosen cutoff, so the decision does not imply certainty.

A note for an actual study

The exam asks us to calculate each stated sample. In a real study, adding data and retesting until a result becomes significant can raise the overall false-positive rate. Plan the sample size or account for repeated tests. The paper does not say whether the first result drove the decision to collect more data.

Source: Final exam, Question 3 on page 1 and useful equations on page 2. Calculations use the printed measurements.

4. Does the new teaching method raise the mean score?

(a) Null and alternative hypotheses

Let $\mu_N$ and $\mu_O$ be the population mean scores under the new and old methods. The claim says that the new method gives a higher mean:

H_0:\mu_N-\mu_O\leq0,\qquad H_1:\mu_N-\mu_O\gt0.

We use a right-tailed test at $\alpha=0.05$ , with a hypothesized difference of zero at the null boundary.

The scores printed in the exam
Group	Scores
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

Model. The requested pooled two-sample t test assumes independent observations within and between groups, normal score populations, and equal population variances. “Pooled” means that we combine both sample variances to estimate that common variance. Equal sample sizes do not establish matched pairs; the exam requests an independent-groups test.

Equal population variances do not require identical sample variances. Each sample estimate varies. Here, we follow the equal-variance model requested by the paper; the scores alone do not establish that its assumptions hold.

(b) Test statistic and degrees of freedom

Use the paper’s formulas $\bar{x}=\sum x_i/n$ and $s^2=\sum(x_i-\bar{x})^2/(n-1)$ for each group. Assign group 1 to the new method, so a positive difference agrees with the claim.

Group summaries
Quantity	New method	Old method
Sample size	10	10
Sum of scores	855.4	842.4
Mean	85.54	84.24
Sum of squared deviations	119.924	75.784
Sample variance	13.324889	8.420444

The observed mean difference is $85.54-84.24=1.30$ points. Pool the two variance estimates using their degrees of freedom as weights:

\begin{aligned} s_p^2&=\frac{(n_N-1)s_N^2+(n_O-1)s_O^2}{n_N+n_O-2}\\ &=\frac{9(13.324889)+9(8.420444)}{18}\\ &=10.872667\text{ points}^2,\\ s_p&=\sqrt{10.872667}=3.297373\text{ points}. \end{aligned}

The standard error of the difference accounts for sampling variation in both independent group means:

\begin{aligned} \mathrm{SE}_{N-O}&=s_p\sqrt{\frac1{n_N}+\frac1{n_O}}\\ &=3.297373\sqrt{\frac1{10}+\frac1{10}}\\ &=1.474630\text{ points}. \end{aligned}

Insert the observed difference and the null difference of zero into the supplied pooled t formula:

\begin{aligned} t&=\frac{(\bar{x}_N-\bar{x}_O)-0}{s_p\sqrt{1/n_N+1/n_O}}\\ &=\frac{1.30}{1.474630}=0.881577,\\ \nu&=n_N+n_O-2=18. \end{aligned}

\boxed{t=0.8816,\qquad\nu=18.}

(c) p-value and conclusion

The alternative points upward, so use the area to the right of the observed statistic:

p=P(T_{18}\geq0.881577)=0.194813.

Under equal population means and the test assumptions, this is the probability of a t statistic at least as large as the one observed. It is not the probability that the null hypothesis is true.

Since $p\gt0.05$ , we fail to reject $H_0$ . The critical-value check agrees: $0.881577$ is below the right 5% cutoff, $1.734064$ .

\boxed{\begin{aligned}p&\approx0.1948,\\&\text{fail to reject }H_0.\end{aligned}}

The new-method sample mean is 1.30 points higher. However, these data give insufficient evidence that the new method raises the population mean score at the 5% level. This result does not prove that both methods have equal effects.

The observed increase is only about 0.88 standard errors above zero. A positive sample difference can therefore coexist with insufficient evidence of a positive population difference. Keep the measured effect and the test decision distinct.

Source: Final exam, Question 4 and useful equations on page 2. Results follow the requested pooled test.

Choose the formula by what is known

The paper supplies the formulas used below. The choice depends on whether the population spread is known and whether one or two means are involved.

Question	Use	Why it fits
1 · Sampling spread	$\sigma_{\bar X}=\sigma/\sqrt n$	The population SD is known. Independence gives the variance of the mean.
2 · Mean interval	$\bar x\pm z_{1-\alpha/2}\sigma/\sqrt n$	The normal error SD is known. For 95%, use 1.96.
3 · One mean	$t=(\bar x-\mu_0)/(s/\sqrt n)$	Estimate the unknown population SD from the sample. Use $n-1$ degrees of freedom.
4 · Two means	$t=\dfrac{\bar x_N-\bar x_O}{s_p\sqrt{1/n_N+1/n_O}}$	The exam requests pooled variances. Use $n_N+n_O-2$ degrees of freedom.

The alternative sets the tail

In Question 3, a mean below the requirement matters, so use the left tail. In Question 4, a mean above the old method matters, so use the right tail. Both tests use one tail. Choose the direction from the claim before looking at the result.

Make the decision, then state its limit

At the chosen significance level, reject $H_0$ when $p\leq\alpha$ . Otherwise, fail to reject it. Failure to reject means the data do not give enough evidence against the null; it does not prove equality or compliance with a requirement.

Sources and model notes

UAS WF3102 Sem 1 20252026.pdfPrimary source. Page 1: Questions 1–3. Page 2: Question 4 and useful equations. Two scanned pages, checked visually. This guide paraphrases the prompts and preserves the numerical data.
NIST: What are confidence intervals?Known-SD normal interval and the repeated-sampling meaning of a confidence level.
NIST: Confidence limits and one-sample t testsOne-sample statistic, degrees of freedom, and one-sided rejection regions.
NIST: Two-sample t testsPooled variance, independent versus paired data, and right-tail rejection.
Penn State STAT 414: The Central Limit TheoremHow the population shape and sample size affect the normal approximation.

The exam gives no full fuel-use distribution. The gamma distributions in Question 1 are clearly marked teaching models. Question 2 treats the given error SD as the variation relevant to repeated readings at the tested condition. Questions 3 and 4 rely on the stated normal and independence assumptions; Question 4 also assumes equal population variances.

Explorations recompute their plots and results from one set of inputs. Moving the new scores in Question 4 is a hypothetical change, not an edit to the exam data. Changing the significance level changes the decision threshold; it does not change the p-value for fixed data.

Math, fonts, plots, and calculations are included in this file. Source links need an internet connection; the study guide itself does not. Values shown in the written solutions are rounded only after calculation.