Chapter 9 Questions




Question 1

You run an IT support firm and each technician requires varying numbers of rolls of specialized cabling tape each week to complete network installation jobs. Previous long-term experience tells you that the population distribution of rolls of tape used per technician per week matches the following table:

X = 0 1 2 3
P[X = x] 0.4 0.3 0.2 0.1
  1. What is the mean number of rolls used on each technician's weekly schedule?
  2. What is the variance of the number of rolls used on each schedule?
  3. Suppose you track a random sample of 100 technician schedules over a quarter and find a sample average of 1.15 rolls used per week. Construct a 95% confidence interval for the true mean number of rolls used.
  4. Based on your confidence interval from part (c), is it plausible that the true population mean has remained unchanged from the historical mean calculated in part (a)?

Answer

  1. $E[X] = \sum (x * P[X = x]) = 0*0.4 + 1*0.3 + 2*0.2 + 3*0.1 = 0 + 0.3 + 0.4 + 0.3 = 1.$
  2. $Var(X) = \sum ((x - \mu)^2 * P[X = x]) = (0-1)^2*0.4 + (1-1)^2*0.3 + (2-1)^2*0.2 + (3-1)^2*0.1 = 1*0.4 + 0*0.3 + 1*0.2 + 4*0.1 = 1.$
  3. Our confidence interval is of the form {$\bar{x} \pm cv * se(\bar{x})$}. Here $\bar{x} = 1.15$. The standard error is $\sigma/\sqrt{n} = 1/\sqrt{100} = 1/10 = 0.1$. For a 95% confidence interval, the critical value is 1.96 (from standard normal tables, P[Z > 1.96] = 0.025). So our confidence interval is:
    {1.15 ± 1.96 * 0.1} = {1.15 ± 0.196} = {0.954, 1.346}
  4. Yes, it is highly plausible. The historical mean calculated in part (a) is 1, which lies comfortably inside our 95% confidence interval of {0.954, 1.346}. Therefore, we fail to reject the possibility that the true underlying mean is still equal to 1.

Question 2

A regional logistics warehouse tracks the number of heavy-duty wooden pallets damaged during cargo loading operations per shift. Historical tracking logs show that the probability distribution of damaged pallets per shift follows this table:

X = 0 2 4 6
P[X = x] 0.4 0.3 0.2 0.1
  1. What is the mean number of damaged pallets per shift?
  2. What is the variance of the number of damaged pallets per shift?
  3. Management reviews a random sample of 64 shifts during a busy seasonal rush and finds a sample average of 2.55 damaged pallets per shift. Construct a 90% confidence interval for the true mean number of damaged pallets during this period.
  4. Based on your interval from part (c), does it look like pallet damage rates significantly increased during this seasonal rush compared to the historical baseline?

Answer

  1. $E[X] = \sum (x * P[X = x]) = 0*0.4 + 2*0.3 + 4*0.2 + 6*0.1 = 0 + 0.6 + 0.8 + 0.6 = 2.$
  2. $E[X^2] = \sum (x^2 * P[X = x]) = 0^2*0.4 + 2^2*0.3 + 4^2*0.2 + 6^2*0.1 = 0 + 4*0.3 + 16*0.2 + 36*0.1 = 1.2 + 3.2 + 3.6 = 8.$
    $Var(X) = E[X^2] - (E[X])^2² = 8 - (2)^2 = 8 - 4 = 4$.
  3. Our confidence interval is of the form {$\bar{x} \pm cv * se(\bar{x})$}. Here $\bar{x} = 2.55$. Since $Var(X) = 4$, the standard deviation $\sigma = 2$. The standard error is $\sigma/\sqrt{n} = 2/\sqrt{64} = 2/8 = 0.25$. For a $90\%$ confidence interval, the critical value is $1.645$. So our confidence interval is:
    {$2.55 \pm 1.645 * 0.25$} = {$2.55 \pm 0.411$} = {$2.139, 2.961$}
  4. Yes, it does. The historical baseline mean is $2$. The entire 90% confidence interval {2.139, 2.961} sits strictly above 2. This suggests that the true mean damage rate during the seasonal rush is significantly higher than the standard historical baseline.

Question 3

A university installs a new automated coffee kiosk in the library, intending to fund its maintenance through sales. After a few months, they analyze sales from a month they believe represents typical student traffic. Across a sample of 36 days, the daily average revenue is \$840 with an estimated standard deviation of $90.

  1. Construct a 90% confidence interval for the daily average revenue.
  2. A student government report had predicted that the kiosk would bring in $865 on average daily. Could the report be right?
  3. What is the representativeness problem mentioned above likely to do to the answer (suppose we just took the first month right after installation, why might that not be representative?)

Answer

  1. Our confidence interval is of the form $\{\bar{x} \pm cv * se(\bar{x})\}$. Here x̄ = 840 (given in question). The standard error is σ/√n = 90/√36 = 90/6 = 15. For a 90% confidence interval we want 5% probability in each tail, so the critical value is 1.645 (from the tables, P[Z > z] = 0.05 gives z = 1.645). So our confidence interval is:
    {840 ± 1.645 * 15} = {815.325, 864.675}
  2. Since 865 is just outside the upper end of the interval (864.675), we would reject this value with a two-sided test with size 10%. So in this sense, the student government's prediction is statistically unlikely to be completely accurate based on this representative period. Of course the number is very close to the critical value, meaning that the t statistic is very close to the cutoff.
  3. It might be that students were initially excited by the novelty of the new kiosk, leading to an initial surge in usage, or conversely, students didn't know it existed yet. If it was due to novelty, the first month would overstate long-run revenue; if due to lack of awareness, it would understate it. Hence, the initial period is unlikely to reflect steady-state behavior.

Question 4

A regional transit authority introduces an express shuttle service between the airport and downtown, funded by rider fares. To evaluate performance, they review a sample of 64 days from a mid-season month considered representative of standard travel volume. The daily average ridership revenue is \$2,450 with an estimated standard deviation of $320.

  1. Construct a 90% confidence interval for the daily average revenue.
  2. An urban planning firm hired by the city predicted that the daily average revenue would be $2,510. Could the firm be right?
  3. What is the representativeness problem mentioned above likely to do to the answer (suppose we collected data exclusively during the first two weeks of operation, why might that not be representative?)

Answer

  1. Our confidence interval is of the form {x̄ ± cv * se(x̄)}. Here x̄ = 2450. The standard error is σ/√n = 320/√64 = 320/8 = 40. For a 90% confidence interval we want 5% probability in each tail, so the critical value is 1.645 (P[Z > z] = 0.05 gives z = 1.645). So our confidence interval is:
    {2450 ± 1.645 * 40} = {2384.2, 2515.8}
  2. Since 2510 is inside the interval, we would not reject this value with a two-sided test with size 10%. So in this sense it is entirely possible that the firm's estimate was accurate. But it is close to the boundary of the confidence interval, perhaps we would want some more observations to be sure.
  3. In the first two weeks, travelers might not be aware of the new shuttle routes, schedules, or pickup points. Therefore, we would expect ridership to be depressed initially compared to when the service becomes a well-known, established option, making the early weeks unrepresentative of long-term revenue.

Question 5

A national polling organization conducts surveys to estimate the proportion of adult citizens who support a newly proposed healthcare reform bill. They use a random sample of 40,000 households, translating into about 90,000 individuals, to compute the support rate based upon that sample.

  1. Suppose they report an estimated approval rate of 45%. What is the 95% confidence interval on this number?
  2. If the approval rate is estimated to have decreased by 0.05% compared to last week, is this likely evidence of a shifting public opinion against the bill?
  3. If this survey were conducted exclusively via digital email invitations sent during typical working hours, how might this sample not be representative?

Answer

  1. The standard error for a proportions question is $\sqrt{\bar{x}(1-\bar{x})/n}$. Here $\bar{x} = 0.45$ and n = 90000, so this is √[0.45(0.55)/90000] = 0.00166. The confidence interval is $\bar{x} \pm 1.96 * s.e. = 0.45 \pm 1.96 * 0.00166 = (44.67\%, 45.33\%)$.
  2. No, since this is within the sampling error. A decrease of this small size is statistically negligible and may indicate no real change in the public's stance, as our estimates are not precise enough to detect a 0.05% shift.
  3. The issue with representativeness here might be that elderly or lower-income individuals who lack consistent internet access or digital literacy would be underrepresented. Since older demographics often hold distinct political views on healthcare, excluding them could introduce a systematic bias into the final support estimate. Some of these observations were within households, so we might also think that the observations were not independent.

Question 6

The Department of Transportation analyzes the proportion of daily commuters who utilize public transit over personal vehicles during peak morning hours. They conduct a large-scale travel survey across a sample of 40,000 individual commuters, to estimate this usage rate.

  1. Suppose they report that 12% of commuters use public transit. What is the 95% confidence interval on this number?
  2. If a subsequent monthly flash poll suggests transit usage rose by 0.02%, is this definitive evidence that recent city transit subsidies are successfully increasing ridership?
  3. If this survey was administered by interviewing commuters in-person exclusively at downtown urban transit hubs, how might this data fail to be representative?

Answer

  1. The standard error for a proportions question is $\sqrt{\bar{x}(1-\bar{x})/n}$. Here $\bar{x} = 0.12$ and n = 40000, so this is √[0.12(0.88)/40000] = 0.001625. The confidence interval is $\bar{x} \pm 1.96 * s.e. = 0.12 \pm 1.96 * 0.001625 = (11.68\%, 12.32\%)$.
  2. No, an increase of 0.02% falls completely inside the expected margin of sampling error. We cannot confidently attribute this tiny fluctuation to the new subsidies, as the change is too small for our measurement precision to confirm an actual upward trend.
  3. Sampling only at major downtown urban transit hubs creates a severe selection bias, over-representing individuals who already live or work in high-density areas with excellent transit access. It completely misses suburban or rural commuters who rely entirely on driving, causing the calculated proportion to be heavily biased upwards.

Copyright © Graham Elliott

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