One-Sample t and z Tests for Population Means

Hypothesis Tests for Population Means

Test 1: Sample of 10 — Mean < 550

(a) In a random sample of 10 people who take a test, the average score is 532.4, and the standard deviation is 28.3. Do the data provide significant evidence, at the 5% level, that the population mean score is less than 550?

1. Step 1: Hypotheses.

   H₀: μ = 550 vs. H_a: μ < 550 (1 point) Where μ is the population mean score.

2. Step 2: Significance level.

   α = 0.05 (1 point)

3. Step 3: Test statistic.

   Test Statistic: t_obs = (X̄ − μ₀)/(s/√n) = (X̄ − 550)/(s/√n) (1 point)

4. Step 4: Rejection region.

   The test rejects H₀ at level α in favor of H_a if t_obs < −t_α,df=9, where t is based on 9 degrees of freedom. (1 point)

5. Step 5: Calculation and conclusion.

   Test Statistic: t_obs = (532.4 − 550) / (28.3/√10) = −1.967 (1 point)

   t_α = t_{0.05, df=9} = 1.833 (1 point)

   Since t_obs (−1.967) < −1.833, reject the null and accept H_a at the 5% level. (1 point)

   Conclusion: The data provide significant evidence at the 5% level that the population mean score is less than 550. (1 point)

   DIFFERENT — = with slash.

Test 2: 40 Boxes — Mean < 16 Ounces

> In a random sample of 40 boxes, the average weight of contents is 15.8 ounces and the standard deviation is 0.5 ounces. Do the data provide significant evidence at the 1.4% level that the population mean weight of contents is less than 16 ounces?

1. Step 1: Hypotheses.

   H₀: μ = 16 vs. H_a: μ < 16 (1 point) Where μ is the population mean weight of contents of boxes of a specific brand of cereal. (1 point)

2. Step 2: Significance level.

   α = 0.014 (1 point)

3. Step 3: Test statistic.

   Test Statistic: z_obs = (X̄ − μ₀)/(σ/√n) = (X̄ − 16)/(σ/√n) (1 point)

4. Step 4: Rejection region.

   The test rejects H₀ at level α in favor of H_a if z_obs < −z_α. (1 point)

5. Step 5: Calculation and conclusion.

   Test Statistic: z_obs = (15.8 − 16) / (0.5/√40) = −2.530 (1 point)

   z_α = z_0.014 = 2.20 (1 point)

   [To find z_0.014: By definition: P(0 < z < Z_0.014) = 0.5 − 0.014 = 0.486. Searching the Normal table, z_0.014 = 2.20.]

   Since z_obs (−2.530) < −2.20, reject the null in favor of H_a at the 1.4% level. (1 point)

   Conclusion: The data provide significant evidence at the 1.4% level that the population mean weight of contents is less than 16 ounces. (1 point)

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