Question: Consider a quantitative variable $y$ measured on $n = 45$ cases that were chosen by simple random sampling from some population of interest. Here are some center-spread summaries of the sample so obtained: midrange = $54.5$ and range = $85$; $y_{0.5} = 51$ and IQR = $42$; $\bar{y} = 52$ and $s_y = 24$ Select each of the true statements from the following: a. The interval $[16, 88]$ is an extended Chebyshev 75% PI. b. The interval $[28, 76]$ is a 40% PI. c. The interval $[9, 93]$ is a 90% PI. d. The interval $[9, 93]$ is a 99% confidence interval for the population median. e. The estimated standard error of the sample mean is $3.58$. f. The interval $[9, 93]$ is a 95% confidence interval for the population mean. makng ststistical inferences

1. **State the problem:** We have a sample of size $n=45$ with given statistics: midrange $=54.5$, range $=85$, median $y_{0.5}=51$, IQR $=42$, sample mean $\bar{y}=52$, and sample standard deviation $s_y=24$. We need to evaluate which given intervals are valid prediction intervals (PI) or confidence intervals (CI) and verify the estimated standard error. 2. **Recall formulas and concepts:** - Range $= \text{max} - \text{min} = 85$. - Midrange $= \frac{\text{max} + \text{min}}{2} = 54.5$. - Chebyshev's inequality for any $k>1$: at least $1-\frac{1}{k^2}$ of data lies within $k$ standard deviations of the mean. - Prediction interval (PI) for a proportion $p$ using Chebyshev: $\bar{y} \pm k s_y$ where $1-\frac{1}{k^2} = p$. - Standard error of the mean (SEM): $\text{SEM} = \frac{s_y}{\sqrt{n}}$. - Confidence intervals for mean or median depend on distribution and sample size. 3. **Calculate min and max from midrange and range:** $$\text{min} = \text{midrange} - \frac{\text{range}}{2} = 54.5 - \frac{85}{2} = 54.5 - 42.5 = 12$$ $$\text{max} = \text{midrange} + \frac{\text{range}}{2} = 54.5 + 42.5 = 97$$ 4. **Check statement (a):** Interval $[16, 88]$ as extended Chebyshev 75% PI. - For 75% PI, $1 - \frac{1}{k^2} = 0.75 \Rightarrow k^2 = 4 \Rightarrow k=2$. - Chebyshev interval: $\bar{y} \pm 2 s_y = 52 \pm 48 = [4, 100]$. - Given interval $[16, 88]$ is inside $[4, 100]$, so it is a narrower interval. - Since Chebyshev guarantees at least 75% within $[4, 100]$, $[16, 88]$ is a valid 75% PI but not extended Chebyshev 75% PI (which is $[4,100]$). - So (a) is **false**. 5. **Check statement (b):** Interval $[28, 76]$ is a 40% PI. - For 40% PI, $1 - \frac{1}{k^2} = 0.4 \Rightarrow k^2 = \frac{1}{0.6} \approx 1.6667 \Rightarrow k \approx 1.29$. - Chebyshev interval: $52 \pm 1.29 \times 24 = 52 \pm 30.96 = [21.04, 82.96]$. - Given interval $[28, 76]$ is inside $[21.04, 82.96]$, so it is a valid 40% PI. - (b) is **true**. 6. **Check statement (c):** Interval $[9, 93]$ is a 90% PI. - For 90% PI, $1 - \frac{1}{k^2} = 0.9 \Rightarrow k^2 = 10 \Rightarrow k = \sqrt{10} \approx 3.16$. - Chebyshev interval: $52 \pm 3.16 \times 24 = 52 \pm 75.84 = [-23.84, 127.84]$. - Given interval $[9, 93]$ is inside $[-23.84, 127.84]$, so it is a valid 90% PI. - (c) is **true**. 7. **Check statement (d):** Interval $[9, 93]$ is a 99% confidence interval for the population median. - Confidence intervals for median require order statistics or bootstrap methods. - Given the wide interval and sample size, $[9, 93]$ is too wide and not justified as 99% CI for median. - (d) is **false**. 8. **Check statement (e):** Estimated standard error of the sample mean is 3.58. - Calculate SEM: $$\text{SEM} = \frac{s_y}{\sqrt{n}} = \frac{24}{\sqrt{45}} = \frac{24}{6.708} \approx 3.58$$ - (e) is **true**. 9. **Check statement (f):** Interval $[9, 93]$ is a 95% confidence interval for the population mean. - 95% CI for mean: $\bar{y} \pm t_{\alpha/2, n-1} \times \text{SEM}$. - For $n=45$, degrees freedom $=44$, $t_{0.025,44} \approx 2.015$. - Calculate margin of error: $$2.015 \times 3.58 \approx 7.21$$ - 95% CI: $52 \pm 7.21 = [44.79, 59.21]$. - Given interval $[9, 93]$ is much wider, so (f) is **false**. **Final answers:** (b), (c), and (e) are true.