▶️ Answer/Explanation
(a)
From the stemplot, crows with lead levels > \(6.0\) ppm are: \(6.3\), \(6.4\), \(6.6\), \(6.8\).
Number of unhealthy crows: \(4\)
Total sample size: \(23\)
Proportion: \(\frac{4}{23} \approx 0.174\)
Answer: \(\boxed{0.174}\)
(b)
Step 1: Identify procedure and check conditions
We use a one-sample t-interval for a population mean.
• Random: Random sample stated
• Normality: Sample size \(n = 23\) is small, but stemplot shows no strong skewness or outliers
Step 2: Calculate the interval
Formula: \(\bar{x} \pm t^* \frac{s}{\sqrt{n}}\)
\(\bar{x} = 4.90\), \(s = 1.12\), \(n = 23\)
Degrees of freedom: \(df = 22\)
Critical value: \(t^* = 2.074\)
Margin of error: \(2.074 \times \frac{1.12}{\sqrt{23}} \approx 0.484\)
Interval: \(4.90 \pm 0.484 = (4.416, 5.384)\)
Step 3: Interpretation
We are \(95\%\) confident that the true population mean lead level of all crows in this region is between \(4.416\) ppm and \(5.384\) ppm.
Answer: \(\boxed{(4.416, 5.384)}\)
▶️ Answer/Explanation
(a)
This is a convenience sample. The first \(500\) students arriving at a football game are not likely to be representative of the entire student population. These students might have more school pride than the average student. This pride could be associated with having a more positive opinion about the university’s appearance. If so, the sample proportion of satisfied students would likely be higher than the true population proportion, leading to a biased estimate.
(b)
A simple random sample (SRS) can be implemented as follows:
- Obtain a complete list of all \(70,000\) students at the university.
- Assign a unique identification number to each student, from \(1\) to \(70,000\).
- Use a computer’s random number generator to select \(500\) unique integers between \(1\) and \(70,000\) (sampling without replacement).
- The \(500\) students corresponding to the selected numbers will form the sample.
(c)
Stratification provides a more precise estimate when the variability of responses within each stratum is low, and the variability between strata is high.
Therefore, stratification by campus would be more precise if the opinions on appearance differ significantly between the two campuses, but are similar within each campus. This strategy would be advantageous if the difference in opinions between the two campuses is greater than the difference in opinions between genders.
▶️ Answer/Explanation
(a)
Let \(W\) be the weight of a randomly selected full carton. We are given that \(W\) follows a Normal distribution with a mean \(\mu_W = 840\) grams and a standard deviation \(\sigma_W = 7.9\) grams. We want to find \(P(W > 850)\).
1. Find the z-score: \[ z = \frac{W – \mu_W}{\sigma_W} = \frac{850 – 840}{7.9} \approx 1.27 \]
2. Find the probability: Using a standard normal table or calculator: \[ P(W > 850) = P(Z > 1.27) \approx 1 – 0.8980 = 0.1020 \] The probability that a randomly selected carton weighs more than \(850\) grams is approximately \(0.1020\).
(b)
Let \(P\) be the weight of the packaging and \(X_i\) be the weight of the \(i\)-th egg. The total weight of the carton is \(W = P + X_1 + X_2 + \dots + X_{12}\).
i) Mean of \(X\): The rule for means is \(E(A+B) = E(A) + E(B)\). \[ E(W) = E(P + X_1 + \dots + X_{12}) \] \[ E(W) = E(P) + E(X_1) + \dots + E(X_{12}) \] Since all eggs have the same mean, \(E(X_i) = E(X)\): \[ E(W) = E(P) + 12 \times E(X) \] We are given \(E(W) = 840\) and \(E(P) = 20\). \[ 840 = 20 + 12 \times E(X) \] \[ 820 = 12 \times E(X) \] \[ E(X) = \frac{820}{12} \approx 68.33 \text{ grams} \]
ii) Standard deviation of \(X\): The rule for variances of independent variables is \(Var(A+B) = Var(A) + Var(B)\). \[ Var(W) = Var(P + X_1 + \dots + X_{12}) \] \[ Var(W) = Var(P) + Var(X_1) + \dots + Var(X_{12}) \] Since all eggs have the same variance, \(Var(X_i) = Var(X)\): \[ Var(W) = Var(P) + 12 \times Var(X) \] We must use variances (standard deviation squared). We are given \(\sigma_W = 7.9\) and \(\sigma_P = 1.7\). \[ Var(W) = (7.9)^2 = 62.41 \] \[ Var(P) = (1.7)^2 = 2.89 \] Substitute these values into the equation: \[ 62.41 = 2.89 + 12 \times Var(X) \] \[ 59.52 = 12 \times Var(X) \] \[ Var(X) = \frac{59.52}{12} = 4.96 \] The standard deviation is the square root of the variance: \[ SD(X) = \sqrt{Var(X)} = \sqrt{4.96} \approx 2.23 \text{ grams} \]
▶️ Answer/Explanation
State:
We will perform a chi-square test for independence at significance level \(\alpha = 0.05\).
\(H_0\): There is no association between age-group and fruit/vegetable consumption.
\(H_a\): There is an association between age-group and fruit/vegetable consumption.
Plan:
We check the conditions for inference:
1. Random: Random sample stated.
2. Large Counts: All expected counts ≥ 5.
Expected counts table:
| Age Group | Yes (Expected) | No (Expected) |
|---|---|---|
| 18–34 | 240.2 | 731.8 |
| 35–54 | 719.4 | 2,191.6 |
| 55+ | 1,231.4 | 3,751.6 |
All expected counts > 5, so condition is met.
Do:
Calculate test statistic:
\(\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(231-240.2)^2}{240.2} + \frac{(741-731.8)^2}{731.8} + \frac{(669-719.4)^2}{719.4} + \frac{(2242-2191.6)^2}{2191.6} + \frac{(1291-1231.4)^2}{1231.4} + \frac{(3692-3751.6)^2}{3751.6} \approx 8.983\)
Degrees of freedom: \((3-1)(2-1) = 2\)
p-value: \(P(\chi^2 \geq 8.983) \approx 0.011\)
Conclude:
Since p-value (\(0.011\)) < \(\alpha\) (\(0.05\)), we reject \(H_0\).
There is convincing statistical evidence that there is an association between age-group and fruit/vegetable consumption for U.S. adults.
▶️ Answer/Explanation
(a)
No, it would not be reasonable to conclude causation. Because this is an observational study and not a randomized experiment, a cause-and-effect relationship cannot be established. The men self-selected whether to meditate. It is possible that men who choose to meditate are different in other lifestyle ways (e.g., diet, exercise) that also reduce blood pressure. These other factors are potential confounding variables.
(b)
It is not reasonable because the Large Counts Condition for normality is not met. For a hypothesis test, we check the expected counts using the pooled proportion of successes, \(\hat{p}_{pooled}\).
1. Find pooled proportion: \[\hat{p}_{pooled} = \frac{\text{total successes}}{\text{total } n} = \frac{0 + 8}{11 + 17} = \frac{8}{28} \approx 0.286\] 2. Check expected counts: We must check that all four expected counts (successes and failures for both groups) are at least \(10\).
- Expected successes (meditators): \(n_m \hat{p}_{pooled} = 11\left(\frac{8}{28}\right) \approx 3.14\)
- Expected successes (non-meditators): \(n_c \hat{p}_{pooled} = 17\left(\frac{8}{28}\right) \approx 4.86\)
Since \(3.14 < 10\) and \(4.86 < 10\), the condition is not met. (Note: The number of observed successes in the meditation group, \(0\), is also less than \(10\), which is another valid check).
(c)
1. Calculate observed statistic: The observed difference in proportions is: \[\hat{p}_m – \hat{p}_c = \frac{0}{11} – \frac{8}{17} \approx -0.4706\]
2. Find p-value from simulation: The simulation was run \(10,000\) times assuming the null hypothesis (\(H_0: p_m – p_c = 0\)) is true. We need to find the probability of getting a result as or more extreme than our observed statistic (\(-0.4706\)). According to the histogram, the value \(-0.47\) (which represents the bin for \(\le -0.47\)) occurred \(76\) times.
The approximate p-value is \(P(\hat{p}_m – \hat{p}_c \le -0.47) = \frac{76}{10000} = 0.0076\).
3. Conclude: Because the p-value of \(0.0076\) is very small (e.g., smaller than \(\alpha = 0.05\)), we reject \(H_0\). There is convincing statistical evidence that men in this retirement community who meditate are less likely to have high blood pressure than men who do not meditate. However, as stated in part (a), this is an observational study, so we can only conclude an association, not a causal relationship.
▶️ Answer/Explanation
(a)
1. Center: The Western Pacific had more typhoons than the Eastern Pacific in almost all years. The average was about \(31\) typhoons/year for Western Pacific vs. about \(19\) typhoons/year for Eastern Pacific.
2. Variability: The Western Pacific showed more variability (range ≈ \(21\) typhoons) than the Eastern Pacific (range ≈ \(10\) typhoons).
3. Context: The Western Pacific region consistently experiences more typhoons annually with greater year-to-year fluctuation.
(b)
• Western Pacific: Showed a clear decreasing trend over the time period, especially from around \(2001\) to \(2010\).
• Eastern Pacific: Remained relatively consistent over the time period, with a slight increasing trend in the later years (\(2005\)-\(2010\)).
(c)
The 4-year moving average for Western Pacific in \(2010\) uses data from \(2007\), \(2008\), \(2009\), and \(2010\):
\[ \frac{28 + 27 + 28 + 18}{4} = \frac{101}{4} = 25.25 \]
The value \(25.25\) should be written in the table for Western Pacific 4-year moving average for \(2010\).
(d)
On Graph B, plot the point for Western Pacific 4-year moving average at year \(2010\) and frequency \(25.25\), then connect this point to the previous moving average point at \(2009\).
(e)
i) More apparent: The overall long-term trends are more visible. The moving averages clearly show the decreasing trend in Western Pacific and the slight increasing trend in Eastern Pacific.
ii) Less apparent: The year-to-year variability is smoothed out, making individual yearly fluctuations less noticeable.