▶️ Answer/Explanation
(a)
No, it is not reasonable to believe the distribution is approximately normal.
The minimum time of \(4.40\) seconds is \(z = \frac{4.40 – 4.60}{0.15} = -1.33\) standard deviations below the mean.
In a normal distribution, about \(9.2\%\) of values would be below \(z = -1.33\), but the data show no values below \(4.40\) seconds, which contradicts what we would expect from a normal distribution.
(b)
\(z = \frac{370 – 310}{25} = 2.4\)
Interpretation: This player’s weight lifting ability is \(2.4\) standard deviations above the mean for all players in this position.
Answer: \(\boxed{2.4}\)
(c)
Calculate z-scores for both players:
Speed (lower is better):
Player A: \(z = \frac{4.42 – 4.60}{0.15} = -1.2\)
Player B: \(z = \frac{4.57 – 4.60}{0.15} = -0.2\)
Strength (higher is better):
Player A: \(z = \frac{370 – 310}{25} = 2.4\)
Player B: \(z = \frac{375 – 310}{25} = 2.6\)
Player A is much faster (\(1.0\) standard deviation better) and only slightly weaker (\(0.2\) standard deviation worse). Since speed and strength are equally important, the team should select Player A.
▶️ Answer/Explanation
(a)
This is a conditional probability. We are given that the voter is male, so we only look at the “Male” row. There are \(\(200\)\) males in total, and \(\(48\)\) of them are registered for Party Y. \[ P(\text{Party Y} \mid \text{Male}) = \frac{\text{Number of Males in Party Y}}{\text{Total Number of Males}} = \frac{48}{200} = 0.24 \]
(b)
No, the events are not independent. Two events A and B are independent if \(P(A \mid B) = P(A)\).
From part (a), we know: \[ P(\text{Party Y} \mid \text{Male}) = 0.24 \] Now, we calculate the overall probability of being registered for Party Y: \[ P(\text{Party Y}) = \frac{\text{Total Registered for Party Y}}{\text{Total Registered Voters}} = \frac{168}{500} = 0.336 \] Since \(P(\text{Party Y} \mid \text{Male}) \neq P(\text{Party Y})\) ( \(0.24 \neq 0.336\) ), the events “is a male” and “is registered for Party Y” are not independent.
(c)
If party registration is independent of gender, the distribution of parties must be the same for males, females, and the overall population. We use the marginal (overall) proportions from the “Total” row of the Franklin Township table.
- Party W: \(\frac{88}{500} = 0.176\)
- Party X: \(\frac{244}{500} = 0.488\)
- Party Y: \(\frac{168}{500} = 0.336\)
Both the “Male” and “Female” bars in the graph must be segmented identically using these proportions.
- Party W segment: from \(0.0\) to \(0.176\)
- Party X segment: from \(0.176\) to \(0.176 + 0.488 = 0.664\)
- Party Y segment: from \(0.664\) to \(0.664 + 0.336 = 1.0\)
Completed Graph:
▶️ Answer/Explanation
(a)
To get a sample of \(8\) apartments using floors as clusters, the owner must select \(2\) floors (since each floor has \(4\) apartments).
Step 1: Label the nine floors with unique integers from \(1\) to \(9\).
Step 2: Use a random number generator to select two different integers from \(1\) to \(9\). (For example, generate a random integer from \(1\) to \(9\), then generate another from \(1\) to \(8\) to select from the remaining floors, or use a generator to select two unique numbers simultaneously).
Step 3: The sample will consist of all \(4\) apartments on each of the \(2\) randomly selected floors, for a total of \(8\) apartments.
(b)
The primary statistical advantage of the stratified sample is that it guarantees representation from both strata (apartments with children and apartments without children).
The amount of wear on the carpet is the variable being studied, and it is very likely that the presence of children will affect how the carpet wears. The cluster sample described in part (a) could, by pure chance, select two floors with no children (e.g., Floor \(3\) and Floor \(6\)). This sample would provide no information about carpet wear in apartments with children, making the results unrepresentative.
The stratified sample ensures that both types of apartments are included, providing a more representative sample and more precise information about carpet wear for both groups.
▶️ Answer/Explanation
State:
Let \( \mu_A \) = mean cholesterol reduction for all such patients receiving placebo
Let \( \mu_B \) = mean cholesterol reduction for all such patients receiving drug
\( H_0: \mu_A = \mu_B \)
\( H_a: \mu_A < \mu_B \)
Plan:
Two-sample \( t \)-test for difference in means
Conditions:
1. Random assignment: Subjects were randomly assigned to groups A and B
2. Independence: Reasonable to assume observations are independent
3. Normality: Sample sizes are small (n=10 each), but no strong skewness or extreme outliers apparent
The following dotplots reveal slight skewness and a possible outlier for group B, but it appears reasonable to proceed with the two-sample t-test.
Do:
\[t = \frac{\bar{x}_A – \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} = \frac{10.20 – 16.40}{\sqrt{\frac{7.66^2}{10} + \frac{9.40^2}{10}}} = \frac{-6.20}{\sqrt{5.8676 + 8.836}} = \frac{-6.20}{\sqrt{14.7036}} = \frac{-6.20}{3.834} \approx -1.62\]
Degrees of freedom ≈ \(17.3\)
\( p \)-value ≈ \(0.062\)
Conclude:
Since \( p \)-value (\(0.062\)) > \( \alpha \) (\(0.01\)), we fail to reject \( H_0 \). There is not convincing evidence that the cholesterol drug produces a greater mean reduction than exercise and diet alone.
▶️ Answer/Explanation
(a)
The equation is found using the “Coef” (coefficient) column from the output.
- The intercept (Constant) is \(0.137\).
- The slope (Wind velocity) is \(0.240\).
The equation is: \[ \hat{y} = 0.137 + 0.240x \] Where:
- \(\hat{y}\) is the predicted electricity production in amperes.
- \(x\) is the wind velocity in mph.
(Alternatively: Predicted Production = \(0.137 + 0.240 \times\) Wind Velocity)
(b)
The slope of the line, \(0.240\), represents the expected increase in electricity production (in amperes) for each additional \(1\) mph of wind velocity.
The question asks for the expected difference between \(25\) mph and \(15\) mph, which is a difference of \(10\) mph.
Calculation: \[ (10 \text{ mph}) \times (0.240 \frac{\text{amperes}}{\text{mph}}) = 2.4 \text{ amperes} \] The windmill is expected to produce \(2.4\) amperes more electricity on a day with \(25\) mph wind velocity than on a day with \(15\) mph wind velocity.
(c)
The proportion of variation in the response variable (electricity production) that is explained by the linear relationship with the explanatory variable (wind velocity) is given by the coefficient of determination, \(R-Sq\).
From the computer output, \(R-Sq = 0.873\).
Therefore, \(0.873\) (or \(87.3\%\)) of the variation in electricity production is explained by its linear relationship with wind velocity.
(d)
Yes, there is statistically convincing evidence that electricity production is related to wind velocity.
Explanation: We perform a hypothesis test for the slope of the regression line, \(\beta_1\).
- Hypotheses: \(H_0: \beta_1 = 0\) (There is no linear relationship) vs. \(H_a: \beta_1 \neq 0\) (There is a linear relationship).
- Test Statistic and p-value: From the “Wind velocity” row of the output, the t-statistic is \(t = 12.63\) and the p-value is \(P = 0.000\).
- Conclusion: Assuming a standard significance level (e.g., \(\alpha = 0.05\)), our p-value (\(0.000\)) is less than \(\alpha\). We reject the null hypothesis.
Because the p-value is so small (essentially zero), we have strong evidence to conclude that a significant linear relationship exists between wind velocity and electricity production.
▶️ Answer/Explanation
(a)
State:
We construct a one-sample z-interval for a population proportion \(p\).
Plan:
Conditions:
1. Random: Random sample stated
2. Large Counts: \(n\hat{p} = 9,600 \times 0.28 = 2,688 \geq 10\) and \(n(1-\hat{p}) = 9,600 \times 0.72 = 6,912 \geq 10\)
Do:
\(\hat{p} = 0.28\), \(n = 9,600\), \(z^* = 2.576\)
Standard error: \(\sqrt{\frac{0.28 \times 0.72}{9,600}} \approx 0.00459\)
Margin of error: \(2.576 \times 0.00459 \approx 0.0118\)
Interval: \(0.28 \pm 0.0118 = (0.2682, 0.2918)\)
Conclude:
We are \(99\%\) confident that the true proportion of all U.S. twelfth-grade students who would answer this question correctly is between \(0.268\) and \(0.292\).
(b)
Completed tree diagram:
(c)
\(P(\text{correct}) = P(\text{knows and correct}) + P(\text{doesn’t know and guesses correctly})\)
\(= k \times 1 + (1-k) \times 0.25 = k + 0.25 – 0.25k = 0.25 + 0.75k\)
(d)
From part (c): \(P(\text{correct}) = 0.25 + 0.75k\)
From part (a): \(0.268 \leq P(\text{correct}) \leq 0.292\)
Solve for \(k\):
\(0.25 + 0.75k = 0.268 \Rightarrow 0.75k = 0.018 \Rightarrow k = 0.024\)
\(0.25 + 0.75k = 0.292 \Rightarrow 0.75k = 0.042 \Rightarrow k = 0.056\)
We are \(99\%\) confident that the true proportion of all U.S. twelfth-grade students who actually know the correct answer is between \(0.024\) and \(0.056\).