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Chapter 11: Problem 3

Assume that the populations are normally distributed. (a) Test whether \(\mu_{1}>\mu_{2}\) at the \(\alpha=0.1\) level of significance for the given sample data. (b) Construct a \(90 \%\) confidence interval about \(\mu_{1}-\mu_{2}\). $$ \begin{array}{lcc} & \text { Sample } \mathbf{1} & \text { Sample } \mathbf{2} \\ \hline n & 25 & 18 \\ \hline \bar{x} & 50.2 & 42.0 \\ \hline s & 6.4 & 9.9 \\ \hline \end{array} $$

Short Answer

Expert verified
a) Reject \( H_0 \); sufficient evidence that \( \mu_1 > \mu_2 \)b) \( 90% \) CI: \( 2.57 < \mu_1 - \mu_2 < 13.83 \)

Step by step solution

01

Identify the hypotheses

Define the null and alternative hypotheses based on the problem statement.i. Null hypothesis (\( H_0 \)): \( \mu_1 \leq \mu_2 \)ii. Alternative hypothesis (\( H_a \)): \( \mu_1 > \mu_2 \)
02

Determine the test statistic

Use the formula for the test statistic for two independent samples, assuming unequal variances (Welch's t-test):i. \( t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)ii. Plug in the provided values:\( \bar{x}_1 = 50.2 \), \( \bar{x}_2 = 42.0 \), \( s_1 = 6.4 \), \( s_2 = 9.9 \), \( n_1 = 25 \), \( n_2 = 18 \)ii. \( t = \frac{50.2 - 42.0}{\sqrt{\frac{6.4^2}{25} + \frac{9.9^2}{18}}} \)ii. Calculate the value:\( t \approx 3.10 \)
03

Determine the degrees of freedom

Use the formula for degrees of freedom for Welch's t-test:i. \[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]ii. Plug in the provided values: \( s_1 = 6.4 \), \( s_2 = 9.9 \), \( n_1 = 25 \), \( n_2 = 18 \)ii. Calculate the value: \( df \approx 30 \)
04

Determine the critical value

For a one-tailed test at \( \alpha = 0.1 \) level of significance and \( df \approx 30 \: t_{critical} \approx 1.31 \)
05

Make a decision

Compare the test statistic to the critical value:\( t \approx 3.10 > 1.31 \)The test statistic is greater than the critical value, so reject the null hypothesis. There is enough evidence to support that \( \mu_1 > \mu_2 \)
06

Construct the 90% Confidence Interval

Use the formula for the confidence interval for two independent samples with unequal variances:i. \[ (\bar{x}_1 - \bar{x}_2) \pm t_{\frac{\alpha}{2}, df} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]ii. Plug in the provided values and the critical value for a 90% confidence interval:\( \bar{x}_1 - \bar{x}_2 = 50.2 - 42.0 = 8.2 \), \( t_{0.05, 30} \approx 1.697 \)ii. Calculate the margin of error:\( ME = 1.697 \cdot \sqrt{\frac{6.4^2}{25} + \frac{9.9^2}{18}} \approx 5.63 \)ii. Construct the interval:\[ 8.2 \pm 5.63 \]The 90% confidence interval is \( 2.57 < \mu_1 - \mu_2 < 13.83 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistical inference. It allows us to make decisions or inferences about a population based on sample data. Hypothesis testing involves:
  • Formulating two opposing hypotheses: the null hypothesis (ull hypothesis \(H_0\)) and the alternative hypothesis (ull hypothesis \(H_a\)).
  • The null hypothesis typically represents no effect or status quo, while the alternative implies some effect or difference. For this problem:
    • Null hypothesis (ull hypothesis \(H_0\)): \( \mu_1 \leq \mu_2 \)
  • Alternative hypothesis (ull hypothesis \(H_a\)): \( \mu_1 > \mu_2 \)
  • Deciding on a significance level (\( \alpha \)), which is the probability of rejecting the null hypothesis when it's actually true. In this case, \( \alpha = 0.1 \).
  • Calculating a test statistic to compare against critical values from statistical distributions.
  • Using the result to either reject or fail to reject the null hypothesis.

    • This process helps determine if observed data supports one hypothesis over another.
    Confidence Intervals Explained
    A confidence interval gives us a range of values within which we expect the true population parameter to lie. Confidence intervals help assess the reliability of an estimate.
    • In this exercise, the confidence interval is constructed around the difference between two means (\( \mu_1 - \mu_2 \)).

    A 90% confidence interval means that if we repeated the sampling process multiple times, 90% of the calculated intervals would contain the true difference between the population means. The confidence interval is calculated using:
    • The sample means (\( \bar{x}_1 \) and (\( \bar{x}_2 \))
    • The standard deviations of the two samples (\( s_1 \) and (\( s_2 \)), the sample sizes (\( n_1 \) and (\( n_2 \)), and a critical value from the t-distribution.

    For this problem, the 90% confidence interval for \( \mu_1 - \mu_2 \) is approximately \(2.57 < \mu_1 - \mu_2 < 13.83\). This range reflects the possible values for the true difference between the means with 90% confidence.
    Degrees of Freedom in Context
    Degrees of freedom (df) relate to the number of values in a statistical calculation that are free to vary. It impacts the shape of the t-distribution used in hypothesis testing and confidence intervals.
    • For Welch's t-test, which is used when variances are assumed unequal, the formula for df is:
      • \( df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \)

      Substituting the given values, the df in this exercise is approximately 30.
      • This estimation allows us to use the t-distribution to find critical values and determine significance.

      Degrees of freedom are crucial because they influence the probability distribution and, therefore, the critical values used in tests.
    Significance Level Demystified
    The significance level (\( \alpha \)) is the threshold at which we decide whether to reject the null hypothesis. It quantifies the risk of concluding that there is an effect or difference when there's none.
    • Common values for \( \alpha \) are 0.01, 0.05, and 0.1, denoting 1%, 5%, and 10% risks respectively.

    • In our exercise, \( \alpha = 0.1 \), meaning we're willing to accept a 10% risk of mistakenly rejecting the null hypothesis.
    • A lower \( \alpha \) indicates stricter criteria for rejecting the null hypothesis
    , while a higher value reflects more lenient criteria.
    The chosen significance level directly affects the critical value from the t-distribution. For\( \alpha = 0.1 \), the critical value \( t_{critical} \) with df\( \approx 30 \) is approximately 1.31.
    • This critical value is then used to compare against our test statistic to make a decision.
    Calculating the Test Statistic
    The test statistic is a standardized value that allows us to compare our sample data to the null hypothesis.
    • For Welch's t-test, the test statistic formula is:
      • \( t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)

      • By plugging in our sample data, we calculate: \( t \approx 3.10 \).

      The test statistic measures how far our sample means are from each other, relative to the variability in the data. Comparing this value to the critical value (1.31 in our case):
      • Since 3.10 > 1.31, we reject the null hypothesis.
      • We conclude there is enough evidence to support that \( \mu_1 > \mu_2 \).
      Understanding and calculating the test statistic properly is essential to making valid inferences in hypothesis testing.

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    Most popular questions from this chapter

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