Foundations for Inference: Sampling Distributions, Point Estimates, Confidence Intervals, and Hypothesis Testing

Learning Objectives

Upon completing this study unit on foundations for statistical inference, students should be able to:

• Calculate and interpret point estimates and interval estimates for a population proportion.

• Verify whether sample size conditions are satisfied to justify using the normal approximation to the binomial distribution.

• Formulate and execute one-tailed and two-tailed statistical hypothesis tests for a single population proportion.

• Explain the concepts and implications of Type I and Type II errors within hypothesis testing.

• Determine the required sample size needed to estimate a population proportion within a specified margin of error.

Statistics and Sampling Distributions

• Sample Statistics: When selecting a random sample from a broader population, numerical descriptive measures computed from the sample (such as the sample mean $\bar{x}$, sample median $m$, or sample proportion $\hat{p}$) are called statistics.

• Variability of Statistics: Statistics vary from one random sample to another. Because their values depend on the specific sample drawn, statistics are classified as random variables.

• Sampling Distribution Definition: The probability distribution of a statistic obtained when random samples of size $n$ are repeatedly drawn from a target population.

  • It defines all possible values the statistic can assume.

  • It quantifies how frequently each possible value occurs under repeated sampling.

Methods for Finding a Sampling Distribution

1. Mathematical Derivation: Deriving the exact theoretical distribution using the mathematical laws of probability.

2. Simulation / Empirical Approximation: Drawing a large number of repeated samples of size $n$, computing the statistic for each sample, and organizing the values into a relative frequency histogram. As the number of simulated samples approaches infinity, the histogram converges to the true theoretical sampling distribution.

3. Statistical Theorems: Applying exact or asymptotic mathematical theorems (such as the Central Limit Theorem) to derive sampling distributions.

Exhaustive Example: Sampling Distribution for $N = 5$ Population

Consider a finite discrete population consisting of $N = 5$ distinct numbers: $3, 6, 9, 12, 15$

• Population Parameters:

  • Population Size: $N = 5$

  • Each value is equally likely with probability $p(x) = \frac{1}{5} = 0.20$

  • Population Mean: $\mu = \frac{3 + 6 + 9 + 12 + 15}{5} = \frac{45}{5} = 9$

  • Population Median: $M = 9$

• Sampling Scheme:

  • Random samples of size $n = 3$ are selected without replacement.

  • Total possible samples: $C_3^5 = \binom{5}{3} = \frac{5!}{3!2!} = 10$

  • Because sampling is simple random sampling, each sample has an equal probability of occurrence: $P(\text{Sample}) = \frac{1}{10} = 0.10$

• Exhaustive Enumeration of Samples and Statistics:

• Derived Sampling Distribution of the Sample Mean ($\bar{x}$):

• Derived Sampling Distribution of the Sample Median ($m$):

• Key Observations:

  • Certain sample statistic values occur with higher probability than others ($P(\bar{x} = 8) = \frac{2}{10} = 0.2$, whereas $P(m = 6) = \frac{3}{10} = 0.3$).

  • Both statistics produce distributions symmetric around the true population parameter value $9$.

Point Estimates and Sampling Variability

• Point Estimator: A rule or mathematical formula that describes how to compute a single numerical estimate of an unknown population parameter using sample data.

• Point Estimate: The specific single numerical value calculated from sample data using a point estimator.

• Estimation Error: The quantitative difference between the unobserved population parameter and the calculated sample statistic: $\text{Error} = |\text{Population Parameter} - \text{Sample Statistic}|$

• Unbiased Estimator: An estimator whose sampling distribution has a mean equal to the true value of the population parameter being estimated ($\mu_{\text{estimator}} = \text{Parameter}$). Otherwise, the estimator is biased.

• Bias: A systematic tendency of an estimator to over-estimate or under-estimate the true population parameter.

• Sampling Error: The inherent random variability that causes a sample statistic to differ from sample to sample.

Sampling Distribution of the Sample Proportion ($\hat{p}$)

In a binomial experiment consisting of $n$ independent trials where each trial results in either a success or failure with fixed probability $p$:

• Sample Proportion Formula: $\hat{p} = \frac{x}{n} = \frac{\text{Total number of successes}}{\text{Total number of trials}}$

• Mean of Sampling Distribution: $\mu_{\hat{p}} = p$

• Standard Deviation / Standard Error of $\hat{p}$: $SE(\hat{p}) = \sqrt{\frac{p q}{n}} = \sqrt{\frac{p(1 - p)}{n}}$

  • where $q = 1 - p$.

Central Limit Theorem (CLT) & Normal Approximation

• Statement: For a random sample of size $n$ drawn from a population with proportion parameter $p$, as sample size $n$ increases, the sampling distribution of the sample proportion $\hat{p}$ becomes approximately normal:     $\hat{p} \sim N\left(p, \frac{p(1 - p)}{n}\right)$

• Conditions for CLT / Normal Approximation:

  1. Independence: Sampled observations must be independent. This holds if:

     • Random sampling or random assignment is utilized.

     • When sampling without replacement, the sample size is less than 10% of the total population ($n < 0.10 N$).

  2. Success-Failure Condition: The sample size must be sufficiently large such that there are at least 10 expected successes and at least 10 expected failures:         $n p \ge 10 \quad \text{and} \quad n(1 - p) \ge 10$

• Impact of Sample Size and Parameter $p$ on Distribution Shape:

  • If the sampled population is strictly normal, the sampling distribution of the statistic is normal for any sample size $n$.

  • When $p$ is close to 0.50 (symmetric population), small sample sizes yield adequate normal approximations.

  • When $p$ is near extreme boundaries ($0$or$1$), the population is heavily skewed. Larger sample sizes ($n \ge 30$or larger) are required to satisfy$n p \ge 10$and$n(1 - p) \ge 10.\n * When conditions are violated (n p < 10$or$n(1 - p) < 10), the sampling distribution exhibits skewness and discrete gaps.\n\n## Standard Error Estimation when p is Unknown\n\nWhen the true population proportion p$is unknown, the sample proportion$\hat{p} is substituted into the standard error formula:\n\widehat{SE}(\hat{p}) = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\nCondition check under unknown p$: Ensure observed successes$x = n\hat{p} \ge 10$and observed failures$n - x = n(1 - \hat{p}) \ge 10.\n\n### Worked Example: E-Tolls Opinion Survey\n* **Scenario**: In a sample of n = 1200$South Africans,$x = 800 support scrapping e-tolls.\n* **Point Estimate**: \hat{p} = \frac{800}{1200} = 0.6667\n* **Estimated Standard Error**: SE = \sqrt{\frac{0.6667 (1 - 0.6667)}{1200}} = \sqrt{\frac{0.6667 \times 0.3333}{1200}} = \sqrt{\frac{0.2222}{1200}} = 0.0136\n\n# Confidence Intervals for Population Proportions\n\n* **Interval Estimator**: A rule or formula that calculates two boundaries—a Lower Confidence Limit (LCL) and an Upper Confidence Limit (UCL)—that define a range within which a population parameter is expected to fall with a specified probability.\n* **Confidence Coefficient**: The probability (1 - \alpha) that the calculated confidence interval captures the true population parameter value.\n* **Confidence Level**: Expressed as a percentage, (1 - \alpha) \times 100\%$. Standard choices include 90$\alpha)**: The probability that the confidence interval fails to contain the true population parameter.\n\n## Formula for Large-Sample Confidence Interval of p\n\n\text{Point Estimate} \pm z_{\alpha/2} \times SE\n\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\n\n* where z_{\alpha/2}$is the critical value from the standard normal distribution leaving an upper-tail area of$\alpha / 2\n\n### Standard Normal Critical Values (z^*)\n* **90% Confidence Interval** (\alpha = 0.10, \alpha/2 = 0.05$):$z^* = 1.645\n* **95% Confidence Interval** (\alpha = 0.05, \alpha/2 = 0.025$):$z^* = 1.96\n* **98% Confidence Interval** (\alpha = 0.022, \alpha/2 = 0.01$):$z^* = 2.33\n* **99% Confidence Interval** (\alpha = 0.01, \alpha/2 = 0.005$):$z^* = 2.576\n\n## Worked Example 1: Facebook Interest Categorization\n* **Scenario**: A representative sample of n = 850$South African Facebook users is surveyed;$67\%$($\hat{p} = 0.67) state that Facebook's interest categorization accurately reflects them.\n* **Condition Verification**:\n    \min(n\hat{p}, n(1 - \hat{p})) = \min(850 \times 0.67, 850 \times 0.33) = \min(569.5, 280.5) = 280.5 \ge 10\n* **Standard Error Calculation**:\n    SE = \sqrt{\frac{0.67 \times (1 - 0.67)}{850}} = \sqrt{\frac{0.67 \times 0.33}{850}} = \sqrt{0.0002601} = 0.0161\n* **95% Confidence Interval Construction**:\n    \text{Margin of Error (ME)} = z^* \times SE = 1.96 \times 0.0161 = 0.0316\n    \text{CI} = 0.67 \pm 0.0316 = (0.6384, 0.7016)\n* **Interpretation**: We are 95% confident that the true proportion of all South African Facebook users who think Facebook categorizes their interests accurately lies between 63.84% and 70.16%.\n\n## Worked Example 2: Experimental Design Intuition Survey\n* **Scenario**: In a sample of n = 670$Americans,$x = 571$answered questions on experimental design correctly ($\hat{p} = 0.85).\n* **Condition Check**:\n * Independence: Simple random sample, 670 < 10\% of all US adults.\n * Success-Failure: 571$successes,$99$failures, both$\ge 10.\n* **Standard Error Calculation**:\n    SE = \sqrt{\frac{0.85 \times 0.15}{670}} = 0.0138\n* **95% Confidence Interval Calculation**:\n    \text{CI} = 0.85 \pm 1.96(0.0138) = 0.85 \pm 0.0270 = (0.823, 0.877) \quad \text{or } (0.82, 0.88)\n\n## Interval Width and Trade-offs\n\n\n\n* **Confidence Level vs. Precision**: Increasing the confidence level increases the critical value z^$, making the interval wider. A wider interval provides higher certainty of capturing$p, but lower precision.\n* **Factors Affecting Margin of Error (ME = z^* \times SE)**:\n * **Confidence level 1 - \alpha$**: Higher confidence$\rightarrow$larger$z^*$$\rightarrow$larger$ME.\n * **Sample size n$**: Larger$n$$\rightarrow$smaller$SE$$\rightarrow$smaller$ME.\n * **Sample variability \hat{p}(1 - \hat{p})$**: Maximum standard error occurs at$p = 0.50.\n\n## Sample Size Determination for Estimating Proportions\n\nTo achieve a target margin of error ME$at a desired confidence level$(1 - \alpha)\times 100\%\nME = z^ \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\nME^2 = (z^)^2 \frac{\hat{p}(1 - \hat{p})}{n}\nn \ge \frac{(z^)^2 \hat{p}(1 - \hat{p})}{ME^2}\n\nIf no prior estimate for \hat{p}$is available, use the conservative value$\hat{p} = 0.50$, which maximizes$\hat{p}(1 - \hat{p}) = 0.25\n\n### Worked Example: Sample Size Calculation\n* **Scenario**: Target margin of error ME = 0.01$, confidence level = 95$z^* = 1.96$), estimated proportion$\hat{p} = 0.85\n* **Derivation**:\n    0.01 \ge 1.96 \times \sqrt{\frac{0.85 \times 0.15}{n}}\n    0.01^2 \ge 1.96^2 \times \frac{0.85 \times 0.15}{n}\n    n \ge \frac{1.96^2 \times 0.85 \times 0.15}{0.01^2} = \frac{3.8416 \times 0.1275}{0.0001} = \frac{0.489804}{0.0001} = 4898.04\n* **Conclusion**: Round up to the next integer: n \ge 4899 respondents.\n\n# Hypothesis Testing for Population Proportions\n\n## Five Core Elements of a Hypothesis Test\n\n1. **Null Hypothesis (H_0$)**: A statement asserting no effect, no difference, or status quo. Formulated as$H_0: p = p_0$, where$p_0 is the null value.\n2. **Alternative Hypothesis (H_a)**: The hypothesis representing the research claim to be tested:\n * Upper-tailed test: H_a: p > p_0\n * Lower-tailed test: H_a: p < p_0\n * Two-tailed test: H_a: p eq p_0\n3. **Test Statistic and p-Value**:\n * **Test Statistic (z$)**: Measures how many standard errors the observed sample statistic$\hat{p}$lies away from the hypothesized value$p_0\n        z = \frac{\hat{p} - p_0}{SE_0} = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\n * **p$-Value**: The probability, computed assuming$H_0 is true, of observing a test statistic value as extreme as or more extreme than the observed value.\n4. **Significance Level (\alpha) and Rejection Region**:\n * **Significance Level (\alpha$)**: Threshold for rejecting$H_0$(typically$0.05).\n * **Rejection Region**: Range of test statistic values leads to rejection of H_0.\n5. **Conclusion**: Final decision to either **Reject H_0$** or **Do Not Reject$H_0**, interpreted in context.\n\n\n\n## Three Equivalent Decision Approaches\n\n1. **p$-Value Approach**: Reject$H_0$if$p\text{-value} \le \alpha\n2. **Critical Value Approach**: Reject H_0 if:\n * z \ge z_{\alpha} (Upper-tailed)\n * z \le -z_{\alpha} (Lower-tailed)\n * z \ge z_{\alpha/2}$or$z \le -z_{\alpha/2} (Two-tailed)\n3. **Confidence Interval Approach** (for two-tailed tests at significance level \alpha$): Reject$H_0$if the null value$p_0$is not contained within the$(1 - \alpha)100\% confidence interval.\n\n## Four-Step Standard Operating Procedure\n\n1. **Prepare**: Identify the parameter p$, state hypotheses$H_0$and$H_a$, select significance level$\alpha$, record sample size$n$and observed sample proportion$\hat{p}.\n2. **Check**: Confirm independence and verify success-failure conditions under the null hypothesis: n p_0 \ge 10$and$n(1 - p_0) \ge 10\n3. **Calculate**: Compute standard error under null hypothesis SE_0 = \sqrt{\frac{p_0(1-p_0)}{n}}$, determine test statistic$z = \frac{\hat{p} - p_0}{SE_0}$, and find the corresponding$p\n-value.\n4. **Conclude**: Compare p\n-value to \alpha$(or$z$to$z^*), make a statistical decision, and state the real-world conclusion.\n\n## Decision Errors\n\n| Decision \\ Truth | Null Hypothesis (H_0$) is True | Alternative Hypothesis ($H_a) is True |\n| :--- | :--- | :--- |\n| **Do Not Reject H_0$** | Correct Decision ($1 - \alpha$) | **Type II Error ($\beta)** |\n| **Reject H_0$** | **Type I Error ($\alpha$)** | Correct Decision ($1 - \beta, Power) |\n\n* **Type I Error (False Positive)**: Rejecting the null hypothesis when it is true. The probability of committing a Type I error under H_0$is equal to the significance level$\alpha$:$P(\text{Type I Error} \mid H_0 \text{ is true}) = \alpha\n* **Type II Error (False Negative)**: Failing to reject the null hypothesis when the alternative hypothesis is true. Denoted by probability \beta.\n* **Error Rate Trade-off**: Decreasing \alpha$increases$\beta$. Set$\alpha$small (e.g.,$0.01$) when Type I error is severe, and higher (e.g.,$0.10) when Type II error is severe.\n\n## Worked Example 1: Physical Fitness Participation in Adults Over 40\n* **Scenario**: In Gauteng, 20% of adults overall participate in fitness activities twice weekly (p_0 = 0.20$). A sample of$n = 100$adults over 40 yields$x = 15$who participate ($\hat{p} = 0.15). Test if participation drops in adults over 40.\n* **Step 1 (Hypotheses)**:\n    H_0: p = 0.20\n    H_a: p < 0.20\n* **Step 2 (Conditions Check)**:\n    n p_0 = 100 \times 0.20 = 20 \ge 10\n    n(1 - p_0) = 100 \times 0.80 = 80 \ge 10\n* **Step 3 (Calculations)**:\n    SE_0 = \sqrt{\frac{0.20 \times 0.80}{100}} = \sqrt{\frac{0.16}{100}} = \sqrt{0.0016} = 0.04\n    z = \frac{0.15 - 0.20}{0.04} = \frac{-0.05}{0.04} = -1.25\n\n\n\n    p\text{-value} = P(z < -1.25) = 0.1056\n* **Step 4 (Conclusion)**: At \alpha = 0.05$(or$0.10$), since$p\text{-value} = 0.1056 > 0.05$, we **fail to reject$H_0$**. There is insufficient evidence to conclude that participation among adults over 40 is significantly below 20$n = 850$South African Facebook users shows$41\%$($\hat{p} = 0.41$) feel comfortable with Facebook auto-generating interest categories for them. Test whether the true proportion differs from 50$p_0 = 0.50).\n* **Step 1 (Hypotheses)**:\n    H_0: p = 0.50\n    H_a: p eq 0.50\n* **Step 2 (Conditions Check)**:\n    n p_0 = 850 \times 0.50 = 425 \ge 10\n    n(1 - p_0) = 850 \times 0.50 = 425 \ge 10\n* **Step 3 (Calculations)**:\n    SE_0 = \sqrt{\frac{0.50 \times 0.50}{850}} = \sqrt{\frac{0.25}{850}} = 0.0171\n    z = \frac{0.41 - 0.50}{0.0171} = \frac{-0.09}{0.0171} = -5.26\n    p\text{-value} = P(|z| > 5.26) < 0.0001\n* **Step 4 (Conclusion)**: Because p\text{-value} < 0.0001 < 0.05$, **reject$H_0$**. The sample data provides convincing evidence that the proportion of South African Facebook users comfortable with automated interest categories significantly differs from 50$n$and$p$, find mean$\mu_{\hat{p}}$and standard deviation$SE(\hat{p}):\n * a. n = 100, p = 0.3$:$\mu_{\hat{p}} = 0.3$,$SE(\hat{p}) = \sqrt{\frac{0.3 \times 0.7}{100}} = 0.0458\n * b. n = 400, p = 0.1$:$\mu_{\hat{p}} = 0.1$,$SE(\hat{p}) = \sqrt{\frac{0.1 \times 0.9}{400}} = 0.0150\n * c. n = 250, p = 0.6$:$\mu_{\hat{p}} = 0.6$,$SE(\hat{p}) = \sqrt{\frac{0.6 \times 0.4}{250}} = 0.0310\n\n* **Exercise 7.38**: Check normal approximation validity (np > 10$and$nq > 10):\n * a. n = 50, p = 0.05 \rightarrow np = 2.5 < 10 (Not appropriate).\n * b. n = 75, p = 0.10 \rightarrow np = 7.5 < 10 (Not appropriate).\n * c. n = 250, p = 0.99 \rightarrow nq = 250 \times 0.01 = 2.5 < 10 (Not appropriate).\n\n* **Exercise 7.39**: Binomial population with n = 75, p = 0.4\n * Mean \mu = 0.4$,$SE = \sqrt{\frac{0.4 \times 0.6}{75}} = 0.0566\n * a. P(\hat{p} \le 0.43)$:$z = \frac{0.43 - 0.40}{0.0566} = 0.53 \rightarrow P(z \le 0.53) = 0.7019\n * b. P(0.35 \le \hat{p} \le 0.43)$:$z_1 = \frac{0.35 - 0.40}{0.0566} = -0.88 \rightarrow P(z_1) = 0.1894$;$P = 0.7019 - 0.1894 = 0.5125\n\n* **Exercise 7.40**: Binomial population with n = 500, p = 0.1\n * a. np = 50 \ge 10$,$nq = 450 \ge 10$(Normal approximation appropriate).$SE = \sqrt{\frac{0.1 \times 0.9}{500}} = 0.0134\n * b. P(\hat{p} > 0.12)$:$z = \frac{0.12 - 0.10}{0.0134} = 1.49 \rightarrow P(z > 1.49) = 0.0681\n * c. P(\hat{p} < 0.10)$:$z = 0 \rightarrow P(z < 0) = 0.5000\n * d. P(0.08 \le \hat{p} \le 0.12)$:$z = \pm 1.49 \rightarrow P(-1.49 \le z \le 1.49) = 0.9319 - 0.0681 = 0.8638\n\n* **Exercise 7.41**: Compute SE(\hat{p})$for$n = 100$across values of$p:\n * a. p = 0.01 \rightarrow SE = \sqrt{\frac{0.01 \times 0.99}{100}} = 0.0099\n * b. p = 0.10 \rightarrow SE = \sqrt{\frac{0.10 \times 0.90}{100}} = 0.0300\n * c. p = 0.30 \rightarrow SE = \sqrt{\frac{0.30 \times 0.70}{100}} = 0.0458\n * d. p = 0.50 \rightarrow SE = \sqrt{\frac{0.50 \times 0.50}{100}} = 0.0500\n * e. p = 0.70 \rightarrow SE = \sqrt{\frac{0.70 \times 0.30}{100}} = 0.0458\n * f. p = 0.90 \rightarrow SE = \sqrt{\frac{0.90 \times 0.10}{100}} = 0.0300\n * g. p = 0.99 \rightarrow SE = \sqrt{\frac{0.99 \times 0.01}{100}} = 0.0099\n\n* **Exercise 7.43 (Losing Weight)**: Population p = 0.19$, sample$n = 100.\n * a. np = 19 \ge 10$,$nq = 81 \ge 10$. Approximate normal with$\mu = 0.19$,$SE = \sqrt{\frac{0.19 \times 0.81}{100}} = 0.0392.\n * b. P(\hat{p} > 0.25)$:$z = \frac{0.25 - 0.19}{0.0392} = 1.53 \rightarrow P(z > 1.53) = 0.0630.\n\n* **Exercise 7.45 (M&M's)**: p = 0.13$, package size$n = 55.\n * a. np = 55 \times 0.13 = 7.15 < 10$. Normal approximation condition fails; sampling distribution is slightly skewed discrete distribution with$\mu = 0.13$,$SE = \sqrt{\frac{0.13 \times 0.87}{55}} = 0.0454.\n * b. P(\hat{p} < 0.20)$:$z = \frac{0.20 - 0.13}{0.0454} = 1.54 \rightarrow P(z < 1.54) = 0.9382.\n * c. P(\hat{p} > 0.35)$:$z = \frac{0.35 - 0.13}{0.0454} = 4.85 \rightarrow P(z > 4.85) \approx 0.0000.\n * d. 95% range: \hat{p} \pm 1.96(0.0454) = 0.13 \pm 0.0890 = (0.0410, 0.2190).\n\n## Confidence Interval Exercises\n\n* **Exercise 8.25**: n = 300, x = 263 \rightarrow \hat{p} = 0.8767, SE = \sqrt{\frac{0.8767 \times 0.1233}{300}} = 0.0190$. 90$z^* = 1.645):\n    \text{CI} = 0.8767 \pm 1.645(0.0190) = 0.8767 \pm 0.0312 = (0.8455, 0.9079)\n\n* **Exercise 8.26**: n = 500, x = 27 \rightarrow \hat{p} = 0.054, SE = \sqrt{\frac{0.054 \times 0.946}{500}} = 0.0101$. 95$z^* = 1.96):\n    \text{CI} = 0.054 \pm 1.96(0.0101) = 0.054 \pm 0.0198 = (0.0342, 0.0738)\n    *Why narrower?* Larger sample size (n=500$vs$n=300$) and smaller variability ($p$near$0.05$vs$p$near$0.88) both reduce standard error.\n\n* **Exercise 8.32 (MP3 Players)**:\n * a. 12 to 17 years (n = 400, \hat{p} = 0.54$):$SE = \sqrt{\frac{0.54 \times 0.46}{400}} = 0.0249$. 95$0.54 \pm 1.96(0.0249) = (0.4912, 0.5888).\n * b. 18 to 34 years (n = 350, \hat{p} = 0.30$):$SE = \sqrt{\frac{0.30 \times 0.70}{350}} = 0.0245$. 95$0.30 \pm 1.96(0.0245) = (0.2520, 0.3480).\n\n* **Exercise 8.34 (Legal Abortions)**: Poll of n = 1002 adults.\n * a. Right-to-life (39\%$,$\hat{p} = 0.39$): 90$z^* = 1.645$):$SE = \sqrt{\frac{0.39 \times 0.61}{1002}} = 0.0154$.$\text{CI} = 0.39 \pm 1.645(0.0154) = (0.3647, 0.4153).\n * b. Pro-choice (53\%$,$\hat{p} = 0.53$):$SE = \sqrt{\frac{0.53 \times 0.47}{1002}} = 0.0158$.$\text{CI} = 0.53 \pm 1.645(0.0158) = (0.5040, 0.5560).\n\n* **Exercise 8.35 (SUVs in California)**: Sample n = 500$,$x = 68$SUVs ($\hat{p} = 0.136$).</p></li><li><p>95$SE = \sqrt{\frac{0.136 \times 0.864}{500}} = 0.0153$.$\text{CI} = 0.136 \pm 1.96(0.0153) = (0.1060, 0.1660)$.</p></li></ul></li></ul><h4 id="6ac98b43-cf40-4600-88f7-180a665d6c5d" data-toc-id="6ac98b43-cf40-4600-88f7-180a665d6c5d" collapsed="false" seolevelmigrated="true">Hypothesis Testing Exercises</h4><ul><li><p><strong>Exercise 9.30</strong>:$n = 1000, x = 279 \rightarrow \hat{p} = 0.279$.</p><ul><li><p>a.$H_0: p = 0.30$,$H_a: p < 0.30$.</p></li><li><p>b. Critical value for$\alpha = 0.05$(lower-tailed):$z^* = -1.645$.</p></li><li><p>c.$SE_0 = \sqrt{\frac{0.30 \times 0.70}{1000}} = 0.0145$. Test statistic$z = \frac{0.279 - 0.30}{0.0145} = -1.45$. Since$-1.45 > -1.645$, do not reject$H_0$. Data does not provide sufficient evidence that$p < 0.30$.</p></li></ul></li><li><p><strong>Exercise 9.31</strong>:$n = 1400, x = 529 \rightarrow \hat{p} = 0.3779$.</p><ul><li><p>a.$H_0: p = 0.40$,$H_a: p eq 0.40$.</p></li><li><p>b.$SE_0 = \sqrt{\frac{0.40 \times 0.60}{1400}} = 0.0131$. Test statistic$z = \frac{0.3779 - 0.40}{0.0131} = -1.69$. Two-tailed$p$-value$= 2 \times P(z < -1.69) = 0.0910$.</p></li><li><p>c. At$\alpha = 0.01$, since$p\text{-value} = 0.0910 > 0.01$, do not reject$H_0$. Insufficient evidence to indicate$p eq 0.40$.</p></li></ul></li><li><p><strong>Exercise 9.32</strong>:$n = 120, x = 72 \rightarrow \hat{p} = 0.60$.</p><ul><li><p>$H_0: p = 0.50$,$H_a: p > 0.50$.$\alpha = 0.05$.</p></li><li><p>$SE_0 = \sqrt{\frac{0.50 \times 0.50}{120}} = 0.0456$. Test statistic$z = \frac{0.60 - 0.50}{0.0456} = 2.19$.</p></li><li><p>$p\text{-value} = P(z > 2.19) = 0.0143$. Since$0.0143 < 0.05$, reject$H_0$. Sufficient evidence that$p > 0.50$.</p></li></ul></li><li><p><strong>Exercise 9.33 (Childhood Obesity)</strong>:$n = 750, x = 68 \rightarrow \hat{p} = 0.0907$. Claimed AOA figure$p_0 = 0.15$.</p><ul><li><p>a. Test$H_0: p = 0.15$vs$H_a: p < 0.15.

  • b & c. SE_0 = \sqrt{\frac{0.15 \times 0.85}{750}} = 0.0130$. Test statistic$z = \frac{0.0907 - 0.15}{0.0130} = -4.56$.$p\text{-value} = P(z < -4.56) < 0.0001$. Since$p\text{-value} < 0.05$, reject$H_0$. The proportion of parents describing their children as overweight is significantly less than 15$p_0 = 0.75$. Sample$n = 100, x = 58 \rightarrow \hat{p} = 0.58$.</p><ul><li><p>a.$H_0: p = 0.75$vs$H_a: p eq 0.75$.</p></li><li><p>b.$SE_0 = \sqrt{\frac{0.75 \times 0.25}{100}} = 0.0433$. Test statistic$z = \frac{0.58 - 0.75}{0.0433} = -3.93$.$p\text{-value} = 2 \times P(z < -3.93) < 0.0001$. Reject$H_0$at$\alpha = 0.01$. The data refutes the geneticist's 75$1/3$, survival rate baseline$p_0 = 2/3 = 0.6667$. Sample$n = 200, x = 164 \rightarrow \hat{p} = 0.82$.</p><ul><li><p>a.$H_0: p = 0.6667$(Screening program ineffective).</p></li><li><p>b.$H_a: p > 0.6667 (Screening program increases survival rate).

  • c & d. SE_0 = \sqrt{\frac{0.6667 \times 0.3333}{200}} = 0.0333$.$z = \frac{0.82 - 0.6667}{0.0333} = 4.60$.$p\text{-value} < 0.0001$. Reject$H_0$at$\alpha = 0.05$. Conclude screening program significantly improves survival rate.</p></li></ul></li><li><p><strong>Exercise 9.36 (Sweet Potato Whitefly)</strong>: Expected infestation rate$p_0 = 0.10$. Sample$n = 100, x = 25 \rightarrow \hat{p} = 0.25$.</p><ul><li><p>a.$H_0: p = 0.10$vs$H_a: p > 0.10$.$SE_0 = \sqrt{\frac{0.10 \times 0.90}{100}} = 0.0300$.$z = \frac{0.25 - 0.10}{0.0300} = 5.00$.$p\text{-value} < 0.0001$. Reject$H_0$at$\alpha = 0.05$. Infestation rate is significantly higher than expected.</p></li><li><p>b. <em>Practical Significance</em>: Indicates potential pest outbreak requiring agronomists to deploy pest control measures immediately to prevent crop loss.</p></li></ul></li><li><p><strong>Exercise 9.37 (Brown or Blue Eyes)</strong>: Claim$p_0 = 0.45$. Sample$n = 80, x = 32 \rightarrow \hat{p} = 0.40$.</p><ul><li><p>$H_0: p = 0.45$vs$H_a: p eq 0.45$.$\alpha = 0.01$.</p></li><li><p>$SE_0 = \sqrt{\frac{0.45 \times 0.55}{80}} = 0.0556$.$z = \frac{0.40 - 0.45}{0.0556} = -0.90$.$p\text{-value} = 2 \times P(z < -0.90) = 0.3681$. Since$p\text{-value} = 0.3681 > 0.01$, do not reject$H_0$. Insufficient evidence to dispute claim.</p></li></ul></li><li><p><strong>Exercise 9.38 (Colored Contacts)</strong>: Sample$n = 80$.</p><ul><li><p>a. Blue lenses ($p_0 = 0.25$,$x = 22 \rightarrow \hat{p} = 0.275$):$H_0: p = 0.25$vs$H_a: p eq 0.25$.$SE_0 = \sqrt{\frac{0.25 \times 0.75}{80}} = 0.0484$.$z = \frac{0.275 - 0.25}{0.0484} = 0.52$.$p\text{-value} = 0.6030 > 0.05$. Do not reject$H_0$.</p></li><li><p>b. Green lenses ($p_0 = 0.24$,$x = 15 \rightarrow \hat{p} = 0.1875$):$H_0: p = 0.24$vs$H_a: p eq 0.24$.$SE_0 = \sqrt{\frac{0.24 \times 0.76}{80}} = 0.0477$.$z = \frac{0.1875 - 0.24}{0.0477} = -1.10$.$p\text{-value} = 0.2713 > 0.05$. Do not reject$H_0$$.