Math Problem Statement

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A researcher found that a cigarette smoker smokes on average 32 cigarettes a day. She feels that this average is too high. She selected a random sample of 9 smokers and found that the mean number of cigarettes they smoked per day was 29. The sample standard deviation was 2.9. At α = 0.01, is there enough evidence to support her claim? Assume that the population is approximately normally distributed. Use the P-value method and a TI-83 Plus/TI-84 Plus calculator.

Solution

To help with this problem, let’s analyze the details and address Part 1: defining the hypotheses and identifying the type of test.

  1. Null Hypothesis H0H_0: The mean number of cigarettes smoked per day is equal to 32 (μ=32\mu = 32).
  2. Alternative Hypothesis H1H_1: The mean number of cigarettes smoked per day is less than 32 (μ<32\mu < 32), since the claim is that 32 is "too high."
  3. This is a left-tailed test since the alternative hypothesis suggests a value lower than 32.

Would you like further explanation on setting up the hypotheses or calculating the test statistic? Here are some follow-up questions:

  1. How do you calculate the test statistic in this scenario?
  2. What is the critical value for a left-tailed test at α=0.01\alpha = 0.01?
  3. How do you use a TI-83 calculator to find the p-value?
  4. What assumptions must be met for this hypothesis test?
  5. Why is the population distribution important in this context?

Tip: Always state your null and alternative hypotheses clearly to avoid confusion when performing statistical tests.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Mean and Standard Deviation
Normal Distribution

Formulas

Test Statistic: z = (x̄ - μ) / (σ / √n)
P-value for left-tailed test using z-distribution

Theorems

Central Limit Theorem
Properties of Normal Distribution

Suitable Grade Level

Undergraduate Level (Statistics Course)