Statistics and Inference

1. a) By the empirical rule (68-95-99.7 rule):

$59 = 64 - 2(2.5) = \mu - 2\sigma$ $69 = 64 + 2(2.5) = \mu + 2\sigma$

Approximately 95% of women are between $59$ and $69$ inches tall (within 2 standard deviations of the mean).

b) The $z$-score for $68$ inches:

$z = \frac{x - \mu}{\sigma} = \frac{68 - 64}{2.5} = \frac{4}{2.5} = 1.6$

From the standard normal table, $P(Z < 1.6) = 0.9452$.

Approximately 94.52% of women are shorter than $68$ inches.

c) Using $z = 1.28$:

$x = \mu + z\sigma = 64 + 1.28(2.5) = 64 + 3.2 = 67.2 \text{ inches}$

A woman at the 90th percentile is approximately $67.2$ inches ($5$ feet $7.2$ inches) tall.

2. a) $z$-scores:

$z_{9.90} = \frac{9.90 - 10.00}{0.05} = \frac{-0.10}{0.05} = -2$

$z_{10.10} = \frac{10.10 - 10.00}{0.05} = \frac{0.10}{0.05} = 2$

b) The percentage of acceptable bolts is:

$P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544$

Approximately 95.44% of bolts are acceptable.

c) The percentage of defective bolts is $1 - 0.9544 = 0.0456 = 4.56\%$.

$5{,}000 \times 0.0456 = 228$

We would expect approximately 228 defective bolts in a batch of 5,000.

3. a) Random sampling is important because it gives every student an equal chance of being selected, which minimizes selection bias. Without random sampling, the sample may not be representative of the population, making the results unreliable.

b) No. She can generalize to the population of $800$ students at her school (the sampling frame), but not to all high school students in the country. Students at one school may not be representative of the national population — differences in school policies, demographics, curriculum, and community norms could all affect homework time.

c) Volunteers in the library are likely students who study more than average. This voluntary response bias (or self-selection bias) would produce an overestimate of the average homework time. Students who rarely do homework are unlikely to be in the library and would be underrepresented in the sample.

4. a) No. This is an observational study, not an experiment. It shows an association (correlation) between eating breakfast and test scores, but it does not prove that breakfast causes higher scores. Correlation does not imply causation — there may be other factors that explain the difference.

b) Possible confounding variables include:

• Socioeconomic status: Students from higher-income families may be more likely to eat breakfast and have access to better educational resources, tutoring, and stable home environments.
• Sleep habits and overall health: Students who eat breakfast may also get more sleep and have better overall health habits, which independently affect test performance.
• Parental involvement: Parents who ensure their children eat breakfast may also be more involved in their education (helping with homework, reading with them, attending school events).

c) A randomized controlled experiment would be needed. For example: randomly assign a large group of students to two groups — one group is provided breakfast every school day for a semester, and the other is not (the control group). Both groups take the same standardized test at the end. Random assignment ensures that confounding variables are approximately balanced between the groups, so any difference in scores can be attributed to the breakfast intervention.

(Ethical note: in practice, withholding food from students raises ethical concerns, so researchers might compare "breakfast program provided" vs. "no program, but students may eat on their own.")

5. a) Compute the test statistic:

$z = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{75 - 72}{9 / \sqrt{36}} = \frac{3}{9/6} = \frac{3}{1.5} = 2.0$

b) For a one-tailed test at $\alpha = 0.05$, the critical value is $z^* = 1.645$. Since $z = 2.0 > 1.645$, we reject Claim A.

The data provides sufficient evidence at the 5% significance level to conclude that the tutoring program raises the average score above $72$.

c) In plain language: If the tutoring program had no effect (Claim A were true), the probability of observing a sample mean as high as $75$ (or higher) from a sample of $36$ students is small (less than 5%). Because this outcome is unlikely under the "no effect" assumption, we conclude that the tutoring program likely does raise scores.

Limitations:

• This analysis assumes the sample is randomly selected from the population of interest. If students self-selected into the tutoring program (e.g., more motivated students chose to participate), the improvement could be due to motivation rather than the tutoring itself.
• "Statistically significant" does not mean "practically significant." A 3-point increase may or may not be meaningful depending on the context and scale of the test.
• The analysis does not account for other variables (teacher quality, student background, time spent studying independently) that could influence the results.
• With a larger sample, even a trivially small effect could become "statistically significant," so effect size should also be considered.