Introductory Statistics: Learning Objectives 1, 2, 3

Use the shape (skewness and symmetry), the center (mean/median), and the spread/range of the distributions displayed in the histograms to determine which boxplots share those characteristics. You may need to use process of elimination.

1. C
2. B
3. A
4. D

• Histogram 1 corresponds to Boxplot C. The histogram shows a roughly uniform or flat distribution with data is spread evenly between 0 and 100. Boxplot C reflects this with a very wide interquartile range (the box) and whiskers that extend across most of the range. The median is near 50, as expected for a symmetric distribution on [0, 100].


• Histogram 2 corresponds to Boxplot B. The histogram shows a symmetric, unimodal (bell-shaped) distribution centered near 65. Likewise, Boxplot B displays a symmetric distribution with the median line around 65 and whiskers of similar length.


• Histogram 3 corresponds to Boxplot A. The histogram is clearly left-skewed, with data concentrated between measurements of 80 to 90. The box in Boxplot A contains left skew because the median is closer to the third quartile (top of box) than the first quartile (bottom of box). Further, the bottom whisker extends further than the top whisker and there are several small-value outliers indicated, both consistent with left-skew.


• Histogram 4 corresponds to Boxplot D. The histogram is multi-modal, which boxplots are not usually effective at displaying. Measurements extend the full range from 0 to 100 which is consistent with the large variability (wide box) in Boxplot D.

a) For City A, compare the sample mean and the sample median. If the mean is greater than the median, what does that suggest about the shape? For City B, compare the distances between the median (Q2) and the other quartiles (Q1, Q3), and between the median and the min/max values.
b) For City A, consider what a large standard deviation relative to the mean implies, especially when the data (lead concentration) cannot be negative. For City B, use the 1.5×IQR rule to check if the minimum or maximum values are outliers.

1. City A: Right-skewed. City B: Symmetric.
2. City A: The combination of right skew and large standard deviation relative to the mean suggests the potential for extreme measurments. City B: No, the minimum and maximum measurements are not outliers based on the 1.5xIQR rule.

a) Shape of Distribution:

• City A: The mean (7.3) is substantially greater than the median (4.9) given the scale of the measurements. This indicates that the distribution is likely right-skewed. Larger measurements are pulling the mean up, away from the center of the data (the median).
• City B: The distribution appears symmetric. We can check the distances from the center:
  • Q2 to Q1: $5.3 - 4.0 = 1.3$
  • Q2 to Q3: $6.5 - 5.3 = 1.2$ (very similar)
  • Median to Min: $5.3 - 2.4 = 2.9$
  • Median to Max: $8.3 - 5.3 = 3.0$ (very similar)
  The symmetry in these distances supports a symmetric distribution.

b) Outliers:

• City A: We should be concerned about potential outliers. The right-skewness (mean > median) suggests the potential for outliers. Furthermore, the sample standard deviation ($s=5.87$) is larger than the sample median ($\bar{x}=4.9$). Since lead measurements cannot be negative, a standard deviation this large relative to the center of the data indicates that there must be some large extreme values to create such a wide spread.
• City B: We can check for outliers using the 1.5×IQR rule. The Interquartile Range (IQR) is $Q3 - Q1 = 6.5 - 4.0 = 2.5$.
  • Lower bound: $Q1 - 1.5 \times IQR = 4.0 - 1.5 \times 2.5 = 4.0 - 3.75 = 0.25$. The minimum value (2.4) is above this bound.
  • Upper bound: $Q3 + 1.5 \times IQR = 6.5 + 1.5 \times 2.5 = 6.5 + 3.75 = 10.25$. The maximum value (8.3) is below this bound.
  Since both the minimum and maximum are within the bounds, we do not need to be concerned about outliers for City B.

a) Calculate these directly from the 10 data points.
b) The null hypothesis is the statement of "no difference" from the proposed value (8 hours). The alternative is what the test is trying to find evidence for (i.e., the mean is not 8 hours).
c) Plug your values from part (a) into the t-statistic formula. Compare your calculated t-statistic to the given critical values.
d) What are properties of the true sleep times of students and and the specific sample collected by the professor that would limit our ability to detect the difference?

1. Sample mean $\bar{x}=7.43$ and sample standard deviation $s \approx 1.15$.
2. Let $\mu$ be the mean sleep time. $H_0: \mu = 8$ vs. $H_1: \mu \neq 8$. In words, the null hypothesis is that students sleep on average 8 hours per night, and the alternative is that they do not.
3. The test statistic is $t \approx -1.57$. Since $-2.262 < -1.57 < 2.262$, we fail to reject the null hypothesis.
4. Potential reasons for failing to detect a true difference include the small sample size (N=10), large variability in sleep times between students and/or a small difference in the true average sleep time and the null value of 8.

a) Sample Mean and Standard Deviation:

• Sample Mean ($\bar{X}$): $\frac{9.2+5.6+7.5+8.9+7.0+7.1+7.9+7.7+7.6+5.8}{10} = \frac{74.3}{10} = 7.43$
• Sample Standard Deviation ($s$): $s = \sqrt{\frac{\sum(x_i - \bar{X})^2}{n-1}} \approx 1.15$

b) Hypotheses:

• Symbols: $H_0: \mu = 8$; $H_1: \mu \neq 8$.
• Words: The null hypothesis ($H_0$) is that the true population mean sleep time for students is equal to 8 hours. The alternative hypothesis ($H_1$) is that the true population mean sleep time is not equal to 8 hours.

c) Test Statistic and Conclusion:

• Calculate the t-statistic: $$ t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}} = \frac{7.43 - 8}{1.15 / \sqrt{10}} = \frac{-0.57}{1.15 / 3.162} = \frac{-0.57}{0.364} \approx -1.57 $$
• Conclusion: This one-sample t-test has N-1=9 degrees of freedom. We compare our calculated t-statistic (-1.57) to the critical values of $\pm 2.262$. Since our t-statistic is not more extreme than the critical values (it falls between -2.262 and +2.262), we fail to reject the null hypothesis. There is not enough statistical evidence to conclude that the average sleep time is different from 8 hours.

d) The ability to detect a true difference (power) is a function of the effect size (difference between true average sleep time and 8), the variability of sleep times among students, and sample size. The larger the sample size, the greater the ability to overcome a small effect size and/or large measurement variability. In this case, 10 samples was not sufficient to overcome the effect size and variability inherent in the data.

a) The null hypothesis should state that there is no difference between the population means. The alternative should state that there is a difference.
b) Identify all the values from the table ($ \bar{X}_Y, \bar{X}_N, s_Y, s_N, n_Y, n_N $) and plug them into the formula.
c) If your calculated test statistic is more extreme (further from zero) than the critical value, you reject the null hypothesis.

1. $H_0: \mu_Y = \mu_N$ vs. $H_1: \mu_Y \neq \mu_N$.
2. $t \approx 6.60$
3. Since $6.60 > 1.96$, we reject the null hypothesis. There is statistically significant evidence for a difference in mean income between the two groups.

a) Hypotheses: We are testing for any difference between the two groups, so it is a two-tailed test.

• Null Hypothesis ($H_0$): $\mu_Y = \mu_N$. (There is no difference in the population mean incomes).
• Alternative Hypothesis ($H_1$): $\mu_Y \neq \mu_N$. (There is a difference in the population mean incomes).

b) T-statistic Calculation: From the problem description, we have:

• $\bar{X}_Y = 63.4$, $s_Y = 12.2$, $n_Y = 50$
• $\bar{X}_N = 42.8$, $s_N = 18.4$, $n_N = 50$

Now, plug these values into the formula: $$ t = \frac{\bar{X}_Y - \bar{X}_N}{\sqrt{s_Y^2/n_Y+s_N^2/n_N}} = \frac{63.4 - 42.8}{\sqrt{12.2^2/50+18.4^2/50}} $$ $$ t = \frac{20.6}{\sqrt{148.84/50+338.56/50}} = \frac{20.6}{\sqrt{2.9768+6.7712}} $$ $$ t = \frac{20.6}{\sqrt{9.748}} = \frac{20.6}{3.122} \approx 6.60 $$ c) Decision: We compare our calculated t-statistic ($t \approx 6.60$) to the critical value of $\pm 1.96$.

Since our observed t-statistic of 6.60 is much larger than the critical value of 1.96, it falls in the rejection region for the test. Therefore, we reject the null hypothesis. We have strong evidence to conclude that there is a statistically significant difference in mean income between those with steady employment and those without.

What assumptions are being made about the data when performing a two-sample t-test? Are they satisfied in this scenario? Consider the relationship between pre- and post-filtering measurements.

The two-sample t-test assumes independence between measurements. Because the same water samples are used for pre- and post-filtering, the set of pre- and post-filtering measurements are not independent. Therefore, a two-sample t-test is not the appropriate test to determine if the filtering is removing lead in this study design.

A two-sample t-test would not be appropriate. The two-sample t-test requires the measurements in the two groups being compared to be independent. Although the individual water samples are independent, the lead measurements in the pre- and post-filtering groups are not. The post-filtering measurement for each sample is dependent upon the initial pre-filtering measurement because they both come from the same water sample.

Because of the paired study design, the correct statistical test is a paired t-test. This test works by first calculating the difference in lead for each water sample (e.g., Pre - Post) and then performing a one-sample t-test on those differences to see if the mean difference is significantly different from zero.