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Magazine Chapter 8: Problem 16
In Exercises 11-16, test the claim about the difference between two population means \(\mu_{1}\) and \(\mu_{2}\) at the level of significance \(\alpha\). Assume the samples are random and independent, and the populations are normally distributed. Claim: \(\mu_{1}>\mu_{2} ; \alpha=0.10\). Assume \(\sigma_{1}^{2} \neq \sigma_{2}^{2}\) Sample statistics: \(\bar{x}_{1}=520, s_{1}=25, n_{1}=7\) and \(\bar{x}_{2}=500, s_{2}=55, n_{2}=6\)
Short Answer
Expert verified
We fail to reject the null hypothesis, so there is no evidence that \( \mu_1 > \mu_2 \).
Step by step solution
01
State the Hypotheses
To test the claim, we need to set up the null and alternative hypotheses. Since the claim is that \( \mu_1 > \mu_2 \), our hypotheses will be : - Null Hypothesis \( H_0: \mu_1 \leq \mu_2 \) - Alternative Hypothesis \( H_a: \mu_1 > \mu_2 \)
02
Determine the Test Statistic
Since \( \sigma_1^2 eq \sigma_2^2 \), we use the formula for the independent samples t-test:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the given values:\[ t = \frac{520 - 500}{\sqrt{\frac{25^2}{7} + \frac{55^2}{6}}} \]
03
Calculate the t-Statistic Value
Calculate the standard error (SE):\[ SE = \sqrt{\frac{25^2}{7} + \frac{55^2}{6}} = \sqrt{\frac{625}{7} + \frac{3025}{6}} = \sqrt{89.29 + 504.17} = \sqrt{593.46} \approx 24.37 \]Now calculate the t-statistic:\[ t = \frac{520 - 500}{24.37} \approx \frac{20}{24.37} \approx 0.82 \]
04
Find the Critical Value
For a one-tailed test at \(\alpha = 0.10\) and degrees of freedom (df) approximated using:\[ df \approx \text{min}(n_1 - 1, n_2 - 1) = \text{min}(6, 5) = 5 \]Using a t-distribution table, the critical value for df=5 at \(\alpha = 0.10\) is approximately 1.476.
05
Make a Decision
Compare the calculated \( t \)-statistic to the critical value:- \( t = 0.82 \) is less than critical value \(1.476\).Since \( t \) is not greater than the critical value, we fail to reject the null hypothesis.
06
State the Conclusion
At the \( \alpha = 0.10 \) level of significance, we do not have enough evidence to support the claim that \( \mu_1 > \mu_2 \). Therefore, the sample data does not provide sufficient evidence to reject the null hypothesis in favor of the alternative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Null Hypothesis
In statistical testing, the null hypothesis (denoted as \( H_0 \)) is a statement that there is no effect or no difference. It acts as a starting point for testing against the alternative hypothesis. In essence, it assumes that any observed effect is due to sampling variability and not a true effect. For example, in our provided exercise, the null hypothesis is set as \( H_0: \mu_1 \leq \mu_2 \). This means we assume that the mean of population 1 is less than or equal to the mean of population 2. The null hypothesis is typically what we aim to test against and potentially reject with statistical evidence.
- The role of the null hypothesis is crucial since it provides the basis for the testing procedures.
- Rejecting or failing to reject the null hypothesis affects the conclusion we draw from our analysis.
- It is important to note that failing to reject the null hypothesis does not prove it true; it merely shows there isn't strong enough evidence against it.
Alternative Hypothesis
The alternative hypothesis, denoted by \( H_a \), is a statement that indicates the presence of an effect or difference. It suggests that any observed difference in sample data is not due to random chance. In our example, the alternative hypothesis is \( H_a: \mu_1 > \mu_2 \). Here, it represents the claim we are testing: the mean of population 1 is greater than that of population 2. Choosing the correct alternative hypothesis is pivotal, as it directs us in the appropriate testing approach.
- The alternative hypothesis can be one-tailed or two-tailed, depending on the nature of the test.
- In a one-tailed test like ours, we are interested in whether one mean is greater than another.
- Evidence from data that supports the alternative hypothesis allows us to reject the null hypothesis.
Level of Significance
The level of significance, represented by \( \alpha \), is a probability threshold below which the null hypothesis is rejected. It defines how much uncertainty we are willing to accept when deciding whether to reject the null hypothesis. In statistical terms, it is the probability of committing a Type I error, which is rejecting the null hypothesis when it is true.
In our scenario, \( \alpha = 0.10 \) indicates a 10% risk of making this error. A common practice is to set \( \alpha \) at 0.05 or 0.01, but different fields may have different standards based on the consequences of errors.
In our scenario, \( \alpha = 0.10 \) indicates a 10% risk of making this error. A common practice is to set \( \alpha \) at 0.05 or 0.01, but different fields may have different standards based on the consequences of errors.
- A smaller \( \alpha \) value means a stricter criterion for rejecting the null hypothesis.
- Deciding on \( \alpha \) is essential before conducting the test to avoid biased outcomes.
- Implementing a significance level helps ensure that conclusions drawn are valid.
t-statistic
The t-statistic measures the size of the difference relative to the variation in your sample data. It is computed by assessing the difference between the sample means divided by the sample standard deviation. For our test, we calculated the t-statistic using the formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] With the given sample statistics, this computes to a t-value of approximately 0.82. This value helps us determine how far our sample mean difference is from zero, given the variation in the data.
- It serves as a standardized score indicating how proper the difference between means is.
- A larger t-statistic indicates a greater likelihood that the difference is not due to chance.
- T-statistics form the basis for hypothesis testing, allowing us to compare against critical values.
Critical Value
The critical value is a point on the t-distribution that represents the threshold for significance. It delineates the boundary beyond which we would reject the null hypothesis. When performing a t-test, the calculated t-statistic is compared against this critical value.
In our case, the critical value at \( \alpha = 0.10 \) with approximated degrees of freedom (\( df = 5 \)) is approximately 1.476. Since the calculated t-statistic of 0.82 is less than the critical value, we do not reject the null hypothesis.
In our case, the critical value at \( \alpha = 0.10 \) with approximated degrees of freedom (\( df = 5 \)) is approximately 1.476. Since the calculated t-statistic of 0.82 is less than the critical value, we do not reject the null hypothesis.
- Critical values are derived from a statistical table or calculator for the t-distribution.
- Comparison of the t-statistic and critical value helps make the decision about the null hypothesis.
- The critical value depends on both the significance level and degrees of freedom used.
