Let count the defects on bolt . Each count has Poisson mean 5. Assuming independent bolts, their total also follows a Poisson distribution. Across 125 bolts we expect defects.
The sample mean is . Multiply the cutoff by 125 to express it as a total. This preserves the strict inequality.
Check the two counts nearest the boundary. With 687 defects, the mean is . One more defect raises it to . The first mean falls below 5.5, so we add the probabilities from zero through 687.
With an expected total of 625, a normal curve closely approximates the distribution of the total. A Poisson count has variance equal to its mean. For one bolt the standard deviation is . Averaging 125 independent counts reduces the standard deviation to defects per bolt.
The cutoff lies 2.5 standard errors above the expected mean. Read the area to the left of 2.50 in the normal table. It gives 99.38%, close to the exact Poisson result of 99.32%.
Each value is the total luggage weight on one Monday. The 39,500 kgf entry stands apart from the other seven, which range from 23,500 to 24,900 kgf. The exam asks us to omit the suspected recording error. Use the seven remaining trips throughout the calculation.
24,100 · 24,900 · 23,500 · 24,400 · 39,500 · 23,800 · 24,700 · 24,200 kgf
The seven retained loads sum to 169,600 kgf. To find their sample variance, sum their squared distances from the mean and divide by six.
The sample mean is kgf. The sample standard deviation, kgf, describes the spread of individual loads. For uncertainty in the mean of seven trips, use its standard error, kgf.
Use Student’s because we estimate the population spread from a small sample. The reference distribution has degrees of freedom. Assume independent trips and approximately normal Monday loads after removing the suspect entry.
We need both a lower and an upper endpoint. Split equally between the two tails, leaving 0.005 in each. In the exam’s upper-tail table, row 6 and column 0.005 give the critical value .
Under the model, about 99% of intervals built this way from repeated samples would contain the true mean Monday load.
The company wants an upper limit on the mean, so we add one margin to the sample mean. Read row 6 of the upper-tail table again, this time in column 0.05. This gives and a margin of 359.09 kgf.
This bound estimates the long-run mean total per Monday. A single sailing can carry more. To assess a future sailing, use a prediction bound and compare it with the ferry’s capacity.
Let be the population mean yield for farms growing this variety. The claim asks whether this variety beats the standard yield of 20. A higher mean supports the claim, so the test uses the right tail.
Assume independent farms and approximately normal yields. We estimate the unknown population standard deviation from the 12 farms, so we use a one-sample test with degrees of freedom.
21.5, 24.5, 18.5, 17.2, 14.5, 23.2, 22.1, 20.5, 19.4, 18.1, 24.1, 18.5
The sample mean exceeds 20 by . Individual yields have a sample standard deviation of . To judge the gain in an average of 12 farms, divide that spread by . This gives the standard error of the mean, .
In the table, 11 degrees of freedom and an upper-tail area of 0.05 give a cutoff of . At the observed spread, reaching that cutoff would require a sample mean of about . The actual gain amounts to just 0.20 standard errors. If the true mean were 20, the chance of a test statistic at least this large would be
Fail to reject . With , these 12 farms give insufficient evidence that the mean yield exceeds 20.
Let be the population variance of the fill volumes. Variance uses squared volume units, so the stated limit is .
Two batches can have the same average fill and very different variation from drink to drink. The machine’s rule concerns that variation. The exam assumes an independent random sample of approximately normal fills. Under this model, test the variance with a chi-square statistic at the boundary . We used the sample mean to compute the 25 deviations, leaving 24 degrees of freedom.
Use the sample variance and the boundary value directly in the ratio.
The observed variance exceeds the limit by about 77%. To judge how unusual that excess is for a sample of 25, look under 24 degrees of freedom and an upper-tail area of 0.05. The table gives a cutoff of , equivalent to a sample variance of . The observed exceeds that cutoff.
Reject . At the 5% level, the sample gives evidence that the fill variance exceeds . This meets the machine's out-of-control rule.
Let be the difference in mean heat measured, in cal/g. Either method could read higher, so both parts use the same two-sided test.
| Method | Count | Mean | Sample SD |
|---|---|---|---|
| A | 13 | 80.02077 | 0.02397 |
| B | 8 | 79.95375 | 0.09054 |
A
79.98, 80.04, 80.02, 80.04, 80.03, 80.03, 80.04, 79.97, 80.05, 80.03, 80.02, 80.00, 80.02
B
80.02, 79.74, 79.98, 79.97, 79.97, 80.03, 79.95, 79.97
Seven method B readings cluster between 79.95 and 80.03 cal/g. The eighth, 79.74, lowers the mean and accounts for about 80% of B’s sum of squared deviations. The exam gives no reason to discard it.
The sample difference is . Both tests assume independent readings from approximately normal populations.
The exam assumes the two populations share a common variance. Estimate it using both groups. Multiply each sample variance by its own degrees of freedom to recover its sum of squared deviations. Add those sums and divide by .
Independence lets us add the two means’ sampling variances. Their difference has a standard error of 0.02614 cal/g. Allow 2.5% in each tail. With 19 degrees of freedom the cutoffs are .
Reject . The data support a difference in the mean readings. Method A has the higher sample mean.
Past measurements give and cal/g. Now each mean uses its own known population spread. Divide each population variance by its sample size, add the two contributions, and take the square root.
The two-sided 5% cutoffs are . Add the probabilities in both tails to get the p-value.
Reject at 5%. Both tests use the same mean difference. The known standard deviations give a smaller standard error, so the difference lies farther from zero in standard-error units.
| Problem | Exam result |
|---|---|
| 01 Cloth defects | 99.32% exact probability that the sample mean is below 5.5. The normal approximation gives 99.38%. |
| 02 Ferry luggage | After excluding 39,500 kgf, the 99% interval for the mean is 23,543.45 to 24,913.69 kgf. The 95% upper bound is 24,587.66 kgf. |
| 03 Potato yield | . The sample gives insufficient evidence that the mean yield exceeds 20. |
| 04 Drink dispenser | . The data support variance above the limit. |
| 05 Heat of fusion | Equal unknown spreads give . Known spreads give . Both tests detect a difference. Method A reads higher on average. |
The exam gives the questions on page 1 and the test formulas on page 2. Select a problem to see how its data lead to the result.