Math Problem Statement

MS101-MIDTERM-REVIEWER.pdf

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You're a researcher investigating the mean height of adult males in a certain population. You have a hypothesis that the mean height is 175 cm. You collect a sample of 30 adult males, with a sample mean of 177 cm and a standard deviation of 5 cm. Test if this mean differs from 175 cm, using a significance level of -2.0452 to 2.0452.

Solution

Let's go through the quiz step by step, based on the scenario given:

  1. Null and Alternate Hypotheses:

    • Null Hypothesis (H₀): The mean height of adult males in this population is 175 cm, i.e., H0:μ=175cmH_0: \mu = 175 \, \text{cm}.
    • Alternative Hypothesis (H₁): The mean height of adult males is not 175 cm, i.e., H1:μ175cmH_1: \mu \neq 175 \, \text{cm} (this is a two-tailed test).
  2. Significance Level:

    • The significance level is not directly given, but typically it's α=0.05\alpha = 0.05, corresponding to a 95% confidence level. From the t-critical values given (-2.0452 to 2.0452), this aligns with a two-tailed test where α=0.05\alpha = 0.05.
  3. Degrees of Freedom (df):

    • The degrees of freedom for a one-sample t-test is calculated as: df=n1df = n - 1 where nn is the sample size. Since n=30n = 30, the degrees of freedom are: df=301=29df = 30 - 1 = 29
  4. Decision Rule:

    • The decision rule involves comparing the calculated t-statistic to the critical values. The critical t-values at α=0.05\alpha = 0.05 (two-tailed) are given as -2.0452 and 2.0452.
      • If tt-statistic < -2.0452 or tt-statistic > 2.0452, reject the null hypothesis.
      • Otherwise, do not reject the null hypothesis.
  5. Calculate the Test Statistic:

    • The t-statistic is calculated using the formula: t=xˉμsnt = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} where:

      • xˉ=177cm\bar{x} = 177 \, \text{cm} (sample mean),
      • μ=175cm\mu = 175 \, \text{cm} (hypothesized population mean),
      • s=5cms = 5 \, \text{cm} (sample standard deviation),
      • n=30n = 30 (sample size).

      Substituting the values: t=177175530=255.477=20.9129=2.19t = \frac{177 - 175}{\frac{5}{\sqrt{30}}} = \frac{2}{\frac{5}{5.477}} = \frac{2}{0.9129} = 2.19

  6. Result and Conclusion:

    • The calculated t-statistic is 2.19, which is greater than the critical value of 2.0452. Therefore, the test statistic falls in the rejection region.
    • Conclusion: Reject the null hypothesis. There is significant evidence to suggest that the mean height of the population is different from 175 cm at the 5% significance level.

Would you like further clarification or details?

Here are some questions to expand upon this topic:

  1. What are the differences between one-sample t-tests and paired t-tests?
  2. How does increasing the sample size affect the t-statistic in hypothesis testing?
  3. What is the impact of using a one-tailed test versus a two-tailed test?
  4. How would you calculate a confidence interval for the population mean in this scenario?
  5. What assumptions must be met for a one-sample t-test to be valid?

Tip: Always ensure that the sample data is approximately normally distributed when using a one-sample t-test, especially for smaller sample sizes.

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Math Problem Analysis

Mathematical Concepts

Hypothesis testing
T-distribution
Significance level
Degrees of freedom

Formulas

t = (x̄ - μ) / (s / sqrt(n))

Theorems

Central Limit Theorem
T-distribution

Suitable Grade Level

College/University