Optimizing Experimental Designs for Accurate Results

1. Randomized Block Design

(a) A randomized block design will be better for detecting the difference in filter effectiveness when there are factors in the field that can introduce variability, such as the proximity to a granite ledge & a burned forest. This design allows for controlling & minimizing the impact of external factors on the results.

(b) Wells in Each Block

Block 1: Wells 1, 2, 5, & 6.

Block 2: Wells 3, 4, 7, & 8.

(c) Creating a Randomized Block Design

Assign unique identifiers to each filter.

Randomly assign one filter from each type to each well within Block 1.

Randomly assign one filter from each type to each well within Block 2.

This ensures random allocation of filters within each block, accounting for potential variations caused by the proximity to the granite ledge or the burned forest.

2. Blocking by Classification

(a) Blocking by the classification of runners in the experiment provides a statistical advantage by controlling for potential confounding variables related to skill level or experience. It ensures that both types of shoes are represented within each classification, allowing for a more accurate comparison of the shoe types.

(b) Randomizing the type of shoe the runner will wear is important to eliminate bias introduced by self-selection. Allowing runners to choose their shoes could lead to preferences or biases that are unrelated to the effectiveness of the shoes. Randomization ensures unbiased assignment & reduces confounding effects, making the comparison between shoe types more valid.

(c) The experiment incorporates replication by randomly selecting 50 professional runners & 50 recreational runners. Within each classification, runners are randomly assigned to wear either type A or type B shoes. Replication with multiple runners wearing each shoe type increases the reliability of the results, accounting for individual differences & sources of variability. It allows for more precise estimates & confident conclusions about the performance of the two shoe types.

3. Exit Times

(a) There were 3 out of the 60 sites that had an exit time before 8:30 A.M.

There were 18 out of the 60 sites that had an exit time of 11:00 A.M. or later.

(b) The 'Without Young Children' histogram has a wider range & higher frequencies, suggesting that those without young children tend to have earlier exit times.

(c) Based on the histograms, a reasonable estimate of the median exit time for the random sample of 60 sites is around 90 minutes relative to 9 A.M. This estimate is derived from the interval with the highest concentration of bars, which falls within the range of 75 to 105 minutes relative to 9 A.M.

4. Identifying Potential Outliers

(a) Procedure for identifying potential outliers:

Calculate the 5-number summary: minimum, Q1, median, Q3, & maximum.

Calculate the interquartile range (IQR) by subtracting Q1 from Q3.

Determine the lower fence (Q1 - 1.5 * IQR) & the upper fence (Q3 + 1.5 * IQR).

Values below the lower fence or above the upper fence are considered potential outliers.

Applying the procedure to the stemplot:

The 5-number summary is minimum = 123, Q1 = 146, median = 164, Q3 = 181, & maximum = 245.

The IQR is 181 - 146 = 35.

The lower fence is 146 - 1.5 * 35 = 93.5, and the upper fence is 181 + 1.5 * 35 = 233.5.

There are no values below the lower fence or above the upper fence, indicating no potential outliers.

(b) Based on the stemplot:

The distribution of money spent on textbooks is unimodal & slightly right-skewed.

The majority of the students fall within the range of 130 to 190.

There is a gradual decrease in the number of responses as the amount spent on textbooks increases.

The stemplot suggests that most students spent between 130 and 190, with a few spending less & a few spending more.

5. Population Analysis

(a) No, because in Country A, the bars for the age groups 60 & above are shorter.

(b) Country A experienced an increase in the birth rate about 20 years prior to 2015. This can be seen in the population pyramid of Country A, where the bar in the age group of 40-44 is wider & taller compared to the bars in the age groups of 20-24 & 25-29. This indicates a higher birth rate in that period.

(c) The median age for the males in Country A in 2015 is in the 30 to 39 age group. The median age group corresponds to the bar such that at least 50% of the population is in that age group or higher, & at least 50% of the population is in that age group or lower. Adding the lengths of the bars either above or below the 30 to 39 age group shows that this group satisfies the condition.

6. Histogram Analysis

(a) Histogram II is more inclined to depict the completion times of Group R. As mentioned, students in Group S generally required less time to finish the assignment. Although both histograms have the same range, the values in Histogram I tend to be smaller compared to those in Histogram II. Consequently, it is probable that Histogram I represents students in Group S, while Histogram II represents students in Group R.

(b) If we were to merge the two histograms, the distribution of completion times would exhibit a bimodal pattern. All values would fall within the range of 35 to 115. The intervals from 35 to 55 & 95 to 115 would contain a higher frequency of completion times compared to the middle interval of 65 to 85.

(c) The sampling distribution of the sample mean will exhibit an approximately normal distribution with a mean of μx¯ = 70 minutes & a standard deviation of σx¯ = 3.75 minutes, calculated using the formula σx¯ = σ/√n where σ represents the standard deviation of the original distribution & n is the sample size. Despite the original completion time distribution being bimodal, the Central Limit Theorem can be applied in this scenario since the sample size of 50 is relatively large, particularly when there are no significant outliers or skewness present. Thus, the sampling distribution can be approximated as normal.

7. Race Winner Analysis

(a) The race winner would be determined by the runner with the smallest sum of the two variables, specifically the runner whose reaction time was approximately 0.152 seconds & running time was around 9.61 seconds, resulting in a total of approximately 9.762 seconds.

(b) It is not valid to assume that reaction time & running time are independent variables. There exists a strong linear relationship between them, as evident from the scatter plot.

(c) Predicting the running time for a runner with a reaction time of 0.30 seconds might not be suitable. This is because the highest recorded reaction time in the graph is approximately 0.202 seconds, & 0.30 seconds is considerably slower than 0.202 seconds. Additionally, extrapolating beyond the value of 0.202 seconds may not be appropriate because the relationship between the reaction time (x-variable) & running time (y-variable) could potentially differ for higher reaction time values.

8. Sampling Plan & Estimation

(a) Sampling Plan: To obtain a simple random sample of 1,000 customers from the list, the manufacturer can assign unique identifiers to each customer, use a random number generator to select 1,000 random numbers, match those numbers to the corresponding customers on the list, & include those 1,000 customers in the sample.

(b) Explanation: The proportion 0.325 should not be used to estimate the population proportion of 30,000 new cars with power door lock problems because it represents the proportion of customers who reported power door lock problems out of those who reported any mechanical problem. It does not account for customers who did not report any mechanical problems or those who reported problems other than power door locks. Thus, it is not a representative estimate for the entire population of new cars sold.

(c) Point Estimate: Based on the sample, the point estimate for the number of new cars sold with power door lock problems within the first 5,000 miles is 9,750. This estimate is obtained by multiplying the sample proportion (0.325) by the total number of new cars sold (30,000). However, it's important to note that this is a point estimate & not an exact value, & there is some level of uncertainty associated with it.

10. Outcome Analysis

(a) The potential outcomes are presented below, grouped based on the winner of the match. Under each category, the winner of each set is indicated.

(b) The ways in which Player V can win a match against Player M & the corresponding probabilities are shown below. Adding the probabilities for the various ways Player V wins the match yields the overall probability of 0.4575.

(d) The number of sets played must be either 2 or 3. The probability of exactly 2 sets is

P(VV) + P(MM) = (0.5)(0.6) + (0.5)(0.7) = 0.3 + 0.35 = 0.65

Therefore, the probability of 3 sets is 1 - 0.65 = 0.035

The expected value is (2)(0.65) + (3)(0.35) = 1.3 + 1.05 = 2.35 sets.

11. Age Requirement Analysis

(a) The selected woman will not meet the age requirement if she is 17, 18, or 19 years old. Therefore, the probability that the selected woman will not meet the age requirement is 0.005 + 0.107 + 0.111 = 0.223

(b) Let X represent the number of women in the sample who do not meet the age requirement. X is a binomial random variable with n = 100 & p = 0.223 as found in part (a). At least 30% of the sample will not meet the age requirement if X ≥ 30. Using an exact binomial probability gives P(X ≥ 30) = 1 - P(X ≤ 29) = 1 - 0.9547 = 0.0453

(c) As shown in part (a), the proportion of women in the population who do not meet the age requirement is 0.223

With a simple random sample of 100, the expected percent who do not meet the age requirement is 22.3%

But with the stratified sample, the actual percent who do not meet the age requirement is set at 30%

Therefore, a woman who does not meet the age requirement is more likely to make it into the stratified sample than the simple random sample.