▶️ Answer/Explanation
(a)
The distribution of room sizes is bimodal, with two distinct clusters of data. One cluster is between \(150\) and \(200\) square feet, and the other is between \(250\) and \(300\) square feet. The distribution is roughly symmetric with a center between \(200\) and \(250\) square feet. The range of the data is approximately \(350 – 100 = 250\) square feet.
(b)
First, determine potential outliers using the \(1.5 \times IQR\) rule.
– \(IQR = Q_3 – Q_1 = 292 – 174 = 118\).
– Lower Fence: \(Q_1 – 1.5(IQR) = 174 – 1.5(118) = 174 – 177 = -3\).
– Upper Fence: \(Q_3 + 1.5(IQR) = 292 + 1.5(118) = 292 + 177 = 469\).
Since the minimum value (\(134\)) is greater than the lower fence and the maximum value (\(315\)) is less than the upper fence, there are no potential outliers.
The boxplot is sketched below:
(c)
The most apparent characteristic visible in the histogram but not the boxplot is the **bimodal shape** of the distribution. The histogram clearly shows two distinct peaks, while the boxplot only shows a symmetric distribution with a wide interquartile range, hiding the two separate clusters of data.
▶️ Answer/Explanation
(a)
- Treatments: The four different concentrations of the fungus mixture: \(0\) ml/L, \(1.25\) ml/L, \(2.5\) ml/L, and \(3.75\) ml/L.
- Experimental units: The \(20\) individual containers, each containing an equal number of insects.
- Response variable: The number of insects that are still alive in each container one week after spraying.
(b)
Yes, the experiment has a control group. The group of containers that will be sprayed with the mixture containing \(0\) ml/L of fungus serves as the control group. This is because this group is treated in the same way as all other groups but does not receive the active ingredient (the fungus). This provides a baseline to which the effects of the other fungus concentrations can be compared.
(c)
To randomly assign the treatments, we can use the following process:
- Label each of the \(20\) containers with a unique number from \(1\) to \(20\).
- Create \(20\) identical slips of paper. On five slips, write “\(0\) ml/L”. On another five slips, write “\(1.25\) ml/L”. On another five, write “\(2.5\) ml/L”, and on the final five, write “\(3.75\) ml/L”.
- Place all \(20\) slips of paper into a hat or a box and mix them thoroughly.
- For each container (from \(1\) to \(20\)), draw one slip of paper from the hat without replacement. Assign the treatment written on the slip to that container.
This process ensures that each treatment is assigned to exactly five containers, and the assignment is completely random.
▶️ Answer/Explanation
(a)
(i) The probability that the person selected is a woman who responds “never” is the joint relative frequency found in the intersection of the “Women” row and the “Never” column.
\(P(\text{Never} \cap \text{Woman}) = 0.0636\).
(ii) The probability that the person selected is a woman or responds “never” is found using the general addition rule: \(P(A \cup B) = P(A) + P(B) – P(A \cap B)\).
From the table, \(P(\text{Never}) = 0.1200\) and \(P(\text{Woman}) = 0.5300\). We found \(P(\text{Never} \cap \text{Woman})\) in part (i). \[ P(\text{Never} \cup \text{Woman}) = 0.1200 + 0.5300 – 0.0636 = 0.5864 \]
(iii) The probability that the person responds “never” given that the person is a woman is a conditional probability. \[ P(\text{Never} | \text{Woman}) = \frac{P(\text{Never} \cap \text{Woman})}{P(\text{Woman})} \] \[ P(\text{Never} | \text{Woman}) = \frac{0.0636}{0.5300} = 0.12 \]
(b)
Yes, the events are independent. Two events A and B are independent if \(P(A|B) = P(A)\).
From part (a-iii), we found \(P(\text{Never} | \text{Woman}) = 0.12\).
From the “Total” row of the table, the marginal probability \(P(\text{Never}) = 0.1200\).
Since \(P(\text{Never} | \text{Woman}) = P(\text{Never})\) (\(0.12 = 0.12\)), the event of being a person whose response is “never” is independent of the event of being a woman.
(c)
Let \(X\) be the random variable for the number of people who will always take their medicine as prescribed. This scenario follows a binomial distribution because there is a fixed number of trials (\(n=5\)), each trial is independent, there are two outcomes (always takes medicine or not), and the probability of success is constant (\(p=0.54\)).
So, \(X \sim \text{Binomial}(n=5, p=0.54)\). We want to find the probability that at least \(4\) people always take their medicine, which is \(P(X \ge 4) = P(X=4) + P(X=5)\).
We use the binomial probability formula: \(P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}\).
For \(X=4\): \[ P(X=4) = \binom{5}{4}(0.54)^4(1-0.54)^{5-4} = 5(0.54)^4(0.46)^1 \approx 0.19557 \] For \(X=5\): \[ P(X=5) = \binom{5}{5}(0.54)^5(1-0.54)^{5-5} = 1(0.54)^5(0.46)^0 \approx 0.04592 \] Now we add the probabilities: \[ P(X \ge 4) = 0.19557 + 0.04592 = 0.24149 \]
▶️ Answer/Explanation
State:
We will perform a two-sample z-test for a difference in proportions at a significance level of \(\alpha=0.05\). Let \(p_{14}\) be the true proportion of all kochia plants resistant to glyphosate in \(2014\), and let \(p_{17}\) be the true proportion in \(2017\).
The hypotheses are:
\(H_0: p_{17} – p_{14} = 0\)
\(H_a: p_{17} – p_{14} > 0\)
Plan:
We must check the conditions for inference.
1. Random: The data come from two independent random samples of kochia plants.
2. Independence (10% condition): It is reasonable to assume the population of all kochia plants is much larger than \(10 \times 61 = 610\) in \(2014\) and \(10 \times 52 = 520\) in \(2017\).
3. Normality (Large Counts):
– \(\hat{p}_{14} = 0.197\), \(\hat{p}_{17} = 0.385\).
– Pooled proportion: \(\hat{p}_c = \frac{(61)(0.197) + (52)(0.385)}{61+52} = \frac{12.017 + 20.02}{113} \approx 0.2835\).
– Expected counts: \(n_{14}\hat{p}_c = 61(0.2835) \approx 17.29\), \(n_{14}(1-\hat{p}_c) \approx 43.71\), \(n_{17}\hat{p}_c = 52(0.2835) \approx 14.74\), \(n_{17}(1-\hat{p}_c) \approx 37.26\). All expected counts are \(\ge 10\).
All conditions are met.
Do:
Calculate the z-test statistic:
\(z = \frac{(\hat{p}_{17} – \hat{p}_{14}) – 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)(\frac{1}{n_{17}} + \frac{1}{n_{14}})}}\)
\(z = \frac{0.385 – 0.197}{\sqrt{(0.2835)(0.7165)(\frac{1}{52} + \frac{1}{61})}} \approx \frac{0.188}{0.0851} \approx 2.21\)
Find the p-value for the one-sided test:
p-value = \(P(Z > 2.21) \approx 0.0135\)
Conclude:
Since the p-value (\(0.0135\)) is less than the significance level (\(\alpha = 0.05\)), we reject the null hypothesis.
There is convincing statistical evidence to conclude that there has been an increase in the proportion of all kochia plants in the western United States that are resistant to glyphosate between \(2014\) and \(2017\).
▶️ Answer/Explanation
(a)
We need to find the 25th percentile of the normal distribution for battery life span, which has a mean \(\mu = 30\) months and a standard deviation \(\sigma = 8\) months. First, we find the z-score corresponding to the 25th percentile of the standard normal distribution. This value is approximately \(z \approx -0.674\).
Next, we convert this z-score back to the original scale (in months) using the formula \(X = \mu + z\sigma\). \[ X = 30 + (-0.674)(8) \] \[ X = 30 – 5.392 = 24.608 \] So, it is expected that \(25\) percent of the batteries will no longer work within approximately \(24.6\) months.
(b)
We need to find the probability that a battery’s life span is less than \(24\) months, i.e., \(P(X < 24)\). We first calculate the z-score for \(X = 24\). \[ z = \frac{X – \mu}{\sigma} = \frac{24 – 30}{8} = \frac{-6}{8} = -0.75 \] Now, we find the probability \(P(Z < -0.75)\) using a standard normal table or calculator. \[ P(X < 24) = P(Z < -0.75) \approx 0.2266 \] The probability that a customer will require a replacement within \(24\) months is approximately \(0.2266\).
(c)
The expected value of the gain is calculated by summing the products of each outcome and its probability. Let \(G\) be the random variable for the company’s gain.
There are two outcomes:
- Replacement required: The battery fails within \(24\) months. The gain is \(-\$150\). The probability of this, from part (b), is \(P(X \le 24) = 0.2266\).
- No replacement required: The battery lasts longer than \(24\) months. The gain is \(\$50\). The probability is \(P(X > 24) = 1 – P(X \le 24) = 1 – 0.2266 = 0.7734\).
The expected value is: \[ E(G) = (-\$150) \times P(X \le 24) + (\$50) \times P(X > 24) \] \[ E(G) = (-\$150)(0.2266) + (\$50)(0.7734) \] \[ E(G) = -\$33.99 + \$38.67 = \$4.68 \] The expected value of the gain for the company for each warranty purchased is \(\$4.68\).
▶️ Answer/Explanation
(a)
Because the sample was taken from a Web site where people voluntarily list available apartments, the results can be generalized to the population of one-bedroom apartments listed on that particular Web site for that city.
(b)
The histogram shows that the distribution of rental prices is skewed to the right. In a right-skewed distribution, the mean is pulled toward the long tail of higher prices and will be greater than the median. This would lead to an overestimation of the typical rental price.
(c)
To create the theoretical sampling distribution, one would have to take every possible unique random sample of size \(50\) from the entire population of apartment listings on the Web site. Then, the median for each of these samples would be calculated. The distribution of all these sample medians would form the theoretical sampling distribution.
(d)
(i) The \(5^{th}\) percentile is at position \(15,000 \times 0.05 = 750\). Summing the frequencies from the start of the table, we find the \(750^{th}\) value falls within the group of medians with a value of \(\$2,500\).
(ii) The \(95^{th}\) percentile is at position \(15,000 \times 0.95 = 14,250\). Summing the frequencies, we find the \(14,250^{th}\) value falls within the group of medians with a value of \(\$2,950\).
(e)
We need to find the percentage of the \(15,000\) medians that are between \(\$2,500\) and \(\$2,950\), inclusive. The number of medians less than \(\$2,500\) is \(1+13+18+56+4+56+55+3+66+136 = 408\). The number of medians greater than \(\$2,950\) is \(93+6+65+12+1+6+2+3 = 188\).
The number of medians at or between these values is \(15,000 – 408 – 188 = 14,404\).
Percentage = \(\frac{14,404}{15,000} \times 100\% \approx 96.03\%\).
(f)
An approximate \(96\%\) confidence interval for the median rental price is \((\$2,500, \$2,950)\).
Interpretation: We are approximately \(96\%\) confident that the true median rental price of all one-bedroom apartments listed on this Web site for this city is between \(\$2,500\) and \(\$2,950\).