A psychologist believes that it will take greater than an hour for certain disturbed children to learn a task. A random sample of 100 of these children results in a mean of 50 minutes to learn the task. Should the psychologist modify her belief at the 0.01 level if the population standard deviation can be assumed to be 15 minutes?
Answer (1)
To determine whether the psychologist should modify her belief, we will conduct a hypothesis test for the population mean using a z-test (since the population standard deviation is known).Step 1: Define HypothesesNull Hypothesis (H₀): The mean time to learn the task is ≤ 60 minutes (μ ≤ 60).Alternative Hypothesis (H₁): The mean time to learn the task is > 60 minutes (μ > 60).This is a right-tailed test since we are testing if the mean time is greater than 60 minutes.Step 2: Given DataSample size: n = 100Sample mean: x̄ = 50 minutesPopulation standard deviation: σ = 15 minutesSignificance level: α = 0.01Population mean under H₀: μ₀ = 60 minutesStep 3: Compute the Test Statistic (z-score)The formula for the z-test statistic is:z = \frac{x̄ - μ₀}{\frac{σ}{\sqrt{n}}}Substituting the values:z = \frac{50 - 60}{\frac{15}{\sqrt{100}}}z = \frac{-10}{\frac{15}{10}}z = \frac{-10}{1.5} = -6.67Step 4: Find the Critical ValueFor a right-tailed test at α = 0.01, the critical z-value (from a z-table) is +2.33.Step 5: Compare and Make a DecisionThe computed z = -6.67 is far less than the critical value +2.33.Since the test statistic does not fall in the rejection region (which is the right tail), we fail to reject the null hypothesis.Step 6: ConclusionAt the 0.01 level of significance, there is not enough evidence to support the psychologist’s belief that it takes greater than 60 minutes for the children to learn the task. Instead, the data suggests that the average learning time is significantly less than 60 minutes. Therefore, the psychologist should modify her belief based on this statistical analysis.
