Statistical Inference

Grade 11 · statistics · 111 practice problems · read aloud

🔊 Listen to this explanation

Statistical Inference: Drawing Conclusions from Data

🎯 What is Statistical Inference?

Statistical inference allows us to make predictions or draw conclusions about a population based on data from a smaller sample. Instead of surveying every student in your school (population), you could survey 100 students (sample) and infer things about the entire school.

📝 Step-by-Step Guide

  1. Define the Question: What do you want to know about the population? (e.g., average height).
  2. Collect Sample Data: Randomly select a sample from the population.
  3. Calculate a Statistic: Find a relevant value from your sample (e.g., sample mean, sample proportion).
  4. Build a Confidence Interval: Create a range of values that you're confident contains the true population parameter.
  5. Interpret the Results: State your conclusion in the context of the original problem.

🔍 Visual Examples

Example 1: Estimating a Population Mean

A sample of 50 11th graders has a mean math test score of 78 with a standard deviation of 5. Construct a 95% confidence interval for the true population mean score.

Step 1: Identify: n=50, x̄=78, s=5, CL=95% (z*=1.96)
Step 2: CI = x̄ ± z*(s/√n)
Step 3: CI = 78 ± 1.96*(5/√50) = 78 ± 1.96*(0.707) = 78 ± 1.39
Step 4: We are 95% confident the true mean math score for all 11th graders is between 76.61 and 79.39.

Example 2: Testing a Claim

A principal claims the average GPA is 3.2. You sample 30 students and find a mean GPA of 3.05. Is there evidence the true mean is less than 3.2?

Step 1: Hypotheses: H₀: μ=3.2, Hₐ: μ<3.2
Step 2: Calculate test statistic: t = (3.05-3.2)/(s/√n)
Step 3: Compare to critical value or find p-value
Step 4: Conclude: If p-value < 0.05, reject H₀.

⚠️ Common Mistakes

  • Confusing Population & Sample: μ (population mean) vs. x̄ (sample mean).
  • Misinterpreting Confidence: A 95% CI means 95% of such intervals contain the true parameter, NOT that there's a 95% chance the parameter is in your specific interval.
  • Using z instead of t: Use t-distribution when population standard deviation is unknown and sample size is small (n < 30).
  • Forgetting Random Sampling: Inference requires a random sample; convenience samples lead to biased results.

💡 Tips & Tricks

  • Memory Aid: "I'm confident about the parameter because I have a good statistic."
  • z* values: For common confidence levels: 90% = 1.645, 95% = 1.96, 99% = 2.576.
  • Check Conditions: Always verify: Random, Normal (n ≥ 30 or normal pop.), Independent (10% condition).
  • Context is Key: Always state your conclusion in the context of the original problem.

🏋️ Practice Suggestions

  • Start with confidence intervals for proportions (easier to calculate).
  • Practice writing null and alternative hypotheses in words and symbols.
  • Use real-world scenarios: estimate average phone usage in your class, or the proportion of students who prefer online vs. in-person learning.
  • Work with a partner to explain your reasoning out loud—this reveals gaps in understanding.

Practice problems

6 of the 111, worked through step by step — try them before opening the answer.

1 P(Z > 1.96) = ?

Hint: This involves finding the area under the standard normal curve beyond a given z-score. Consider using a standard normal distribution table or calculator to find the probability.

Show the answer

Answer: 0.025

  1. Understand the problem. We want the probability that Z is greater than 1.96. This is the area under the standard normal curve to the right of 1.96.
  2. Use symmetry and standard normal tables. Standard normal tables usually give P(Z <= z), i.e., the cumulative distribution function (CDF) from the left up to z. So, P(Z > 1.96) = 1 - P(Z <= 1.96).
  3. Look up P(Z <= 1.96) in the standard normal table. From the standard normal table, the value for z = 1.96 is 0.9750. That means P(Z <= 1.96) = 0.9750.
  4. Subtract from 1. P(Z > 1.96) = 1 - 0.9750 = 0.0250.
  5. Interpret the result. This means there is a 2.5% chance that a standard normal random variable exceeds 1.96. Final answer: 0.025

We are given: P(Z > 1.96) where Z is the standard normal random variable (mean 0, variance 1).

2 P(Z > 1.28) = ?

Hint: This involves finding the area under the standard normal curve beyond a given z-value. Consider how the total area under the curve relates to probability values.

Show the answer

Answer: 0.1003

  1. We need to find P(Z > 1.28), which is the area under the standard normal curve to the right of z = 1.28.
  2. Standard normal tables typically give P(Z < z), the area to the left of a z-value.
  3. Using a standard normal table, we find P(Z < 1.28) = 0.8997.
  4. Since the total area under the curve is 1, P(Z > 1.28) = 1 - P(Z < 1.28) = 1 - 0.8997 = 0.1003.
  5. The answer is 0.1003.

3 P(Z < 1.28) = ?

Hint: This involves finding the cumulative probability for a standard normal distribution. You'll need to use a z-table or standard normal distribution table to find the area to the left of the given z-score.

Show the answer

Answer: 0.8997

  1. We need to find P(Z < 1.28), which represents the cumulative probability up to z = 1.28 in a standard normal distribution.
  2. Look up z = 1.28 in a standard normal distribution table.
  3. The table shows that the area to the left of z = 1.28 is 0.8997.
  4. Therefore, P(Z < 1.28) = 0.8997.

The answer is 0.8997.

4 P(Z > 1.645) = ?

Hint: This involves finding the area under the standard normal curve beyond a given z-score. Consider what the z-score represents in terms of probability.

Show the answer

Answer: 0.05

  1. The problem asks for P(Z > 1.645), which is the probability that a standard normal random variable exceeds 1.645.
  2. For a standard normal distribution, P(Z > z) = 1 - P(Z ≤ z).
  3. Using a standard normal table or calculator, we find that P(Z ≤ 1.645) = 0.95.
  4. Therefore, P(Z > 1.645) = 1 - 0.95 = 0.05.

The answer is 0.05.

5 H₀: μ = 50, H₁: μ > 50, z = 2.33, α = 0.01; Reject H₀?

  1. A) yes
  2. B) no

Hint: Compare the test statistic to the critical value for a one-tailed test at the given significance level.

Show the answer

Answer: A) yes

  1. Identify the type of test. The alternative hypothesis is H₁: μ > 50, which means this is a right-tailed test.
  2. Determine the critical value for α = 0.01. For a right-tailed test at α = 0.01, we find the z-value such that the area to the right is 0.01. From the standard normal distribution table, the z-value for a right-tail area of 0.01 is z_critical = 2.33.
  3. Compare the test statistic to the critical value. The test statistic given is z = 2.33. For a right-tailed test, we reject H₀ if z > z_critical. Here, z = 2.33 and z_critical = 2.33.
  4. Make the decision. Since z is equal to the critical value, and for a right-tailed test we reject when z ≥ z_critical (depending on convention, but standard rule is reject if z > z_critical for strict inequality; however, at exactly the critical value, the p-value equals α). Let's check the p-value: For z = 2.33, the area to the right is 0.0099 ≈ 0.01. Since p-value ≤ α (0.01 ≤ 0.01), we reject H₀.
  5. Conclusion. Yes, we reject H₀ at α = 0.01.

6 H₀: μ = 100, H₁: μ ≠ 100, z = 2.58, α = 0.01; Reject H₀?

  1. A) no
  2. B) yes

Hint: Compare the test statistic to the critical value for a two-tailed test at the given significance level.

Show the answer

Answer: B) yes

  1. Identify the type of test. H₁: μ ≠ 100 indicates a two-tailed test.
  2. Determine the critical z-value for α = 0.01. For a two-tailed test, the area in each tail is α/2 = 0.005.
  3. The critical z-value for a two-tailed test with α = 0.01 is ±2.576.
  4. Compare the test statistic (z = 2.58) to the critical value (2.576). Since 2.58 > 2.576, the test statistic falls in the rejection region.
  5. Decision: Reject the null hypothesis H₀.

The answer is yes.

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