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Independent random samples of \(n_{1}=16\) and \(n_{2}=13\) observations were selected from two normal populations with equal variances: $$ \begin{array}{lrr} & & {\text { Population }} \\ { 2 - 3 } & 1 & 2 \\ \hline \text { Sample Size } & 16 & 13 \\ \text { Sample Mean } & 34.6 & 32.2 \\ \text { Sample Variance } & 4.8 & 5.9 \end{array} $$ a. Suppose you wish to detect a difference between the population means. State the null and alternative hypotheses for the test. b. Find the rejection region for the test in part a for \(\alpha=.01\) c. Find the value of the test statistic. d. Find the approximate \(p\) -value for the test. e. Conduct the test and state your conclusions.

Short Answer

Expert verified
Answer: To answer this question, perform a hypothesis test for the difference between population means with equal variances, following the steps of the solution. Calculate the pooled variance, the test statistic, and the p-value, then compare the p-value to the given significance level (0.01). If the p-value is less than or equal to 0.01, we can conclude that there is a significant difference between the population means. If the p-value is greater than 0.01, we cannot conclude there is a significant difference between the population means.

Step by step solution

01

a. State the null and alternative hypotheses

Let \(\mu_1\) be the mean of population 1 and \(\mu_2\) be the mean of population 2. The null hypothesis assumes that there is no difference between the population means: \(H_0: \mu_1 = \mu_2\) or \(\mu_1 - \mu_2 = 0\) The alternative hypothesis assumes that there is a significant difference between the population means: \(H_1: \mu_1 \neq \mu_2\) or \(\mu_1 - \mu_2 \neq 0\)
02

b. Find the rejection region for the test with \(\alpha=.01\)

Since we are given equal variances of the populations, we can use pooled variance to define our standard error. Let's first calculate the pooled variance: \(S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}\) where \(S_1^2\) and \(S_2^2\) are the sample variances, and \(n_1=16\) and \(n_2=13\). The test statistic follows a t-distribution with \(n_1 + n_2 - 2\) degrees of freedom. The rejection region will be determined by finding the critical t-value for a two-tailed test with significance level \(\alpha =(.01)\) and \(16 + 13 - 2 = 27\) degrees of freedom. \(t_{\alpha/2, 27} = \pm t_{0.005, 27}\)
03

c. Calculate the test statistic

The test statistic for the difference between population means with equal variances is given by: \(t = \frac{(\overline{X_1} - \overline{X_2}) - (\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\) Plug in the given sample means (\(\overline{X_1} = 34.6\) and \(\overline{X_2} = 32.2\)) and the previously calculated pooled variance \(S_p^2\). Under the null hypothesis, \((\mu_1 - \mu_2) = 0.\)
04

d. Find the approximate p-value for the test

The p-value can be found by comparing the calculated test statistic with the t-distribution with 27 degrees of freedom. Use the calculated test statistic value and find the probability of getting a t-value as extreme or more extreme than the calculated test statistic in a two-tailed test: \(p-value = 2 * P(t \geq |t|)\)
05

e. Conduct the test and state your conclusions

Compare the calculated p-value with the given significance level \(\alpha = .01\). If the p-value is less than or equal to \(\alpha\), then we reject the null hypothesis and conclude that there is a significant difference between the population means. If the p-value is greater than \(\alpha\), then we fail to reject the null hypothesis and cannot conclude there is a significant difference between the population means.

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Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It is commonly applied when the data follows a normal distribution and the sample sizes are relatively small. In the case of the independent samples t-test, which we discuss here, it's used for comparing the means from two different populations.
  • The t-test assumes that variances within the populations are equal.
  • It is often used when sample sizes are small, which makes it a perfect fit for the given problem, where the sample sizes are 16 and 13.
To calculate the t-statistic, we make use of the formula:\[ t = \frac{(\overline{X_1} - \overline{X_2}) - (\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \] where \( \overline{X_1} \) and \( \overline{X_2} \) are the sample means, and \( S_p \) is the pooled standard deviation. Calculating the pooled variance first is crucial to simplify computations. Once the t-value is computed, it determines how much the observed difference departs from the null hypothesis.
null hypothesis
The null hypothesis is a fundamental aspect of hypothesis testing that represents a default position or statement to be tested for potential rejection. In this exercise, the null hypothesis posits that there is no difference between the means of the two populations. We express this formally as:\[ H_0: \mu_1 = \mu_2 \] or equivalently, \( H_0: \mu_1 - \mu_2 = 0 \).
  • The null hypothesis assumes no effect or no difference in the context of a test.
  • It serves as the basis for statistical analysis, as our aim is to collect evidence against it.
The objective in hypothesis testing is to challenge this default assumption with data evidence. Depending on the results, we have the option to either reject or fail to reject the null hypothesis. If we find compelling evidence against the null hypothesis, we may consider the alternative, which suggests that there is a difference between the group means.
p-value
The p-value is an integral component in hypothesis testing, providing a measure of the evidence against the null hypothesis. In simple terms, it reflects how probable the observed data is under the assumption that the null hypothesis is true.
  • A smaller p-value indicates stronger evidence against the null hypothesis.
  • A large p-value suggests that the observed data is quite likely under the null hypothesis.
In this exercise, the p-value is calculated based on the t-statistic derived from the data. For a two-tailed test, which is what we have here, the formula for the p-value is: \[ p\text{-value} = 2 \times P(t \geq |t|) \] If the p-value obtained is less than the significance level \( \alpha \), typically 0.01 or 0.05, it suggests significant evidence to reject the null hypothesis. Otherwise, the null hypothesis is not rejected, meaning there's insufficient evidence of a difference between the population means.
rejection region
The rejection region is a range of values for the test statistic that leads us to reject the null hypothesis in favor of the alternative hypothesis. Its determination is crucial in hypothesis testing as it defines the critical boundaries before conducting the test.
  • The location and size of the rejection region depend on the significance level \( \alpha \).
  • For a two-tailed test with \( \alpha = 0.01 \), the rejection region is split equally at both ends of a t-distribution.
In this specific exercise, you calculate the critical t-value for \( \alpha = 0.01 \) using a t-distribution with 27 degrees of freedom. This corresponds to:\[ t_{\alpha/2, 27} = \pm t_{0.005, 27} \] Values of the test statistic that fall into this two-tailed area suggest significant evidence to reject the null hypothesis. Conversely, if the calculated statistic is outside of these bounds, the null hypothesis is not rejected.

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Most popular questions from this chapter

As part of a larger pilot study, students at a Riverside, California middle school, will compare the learning of algebra by students using iPads versus students using the traditional algebra textbook with the same author and publisher. \(^{20}\) To remove teacher-to-teacher variation, the same teacher will teach both classes, and the iPad and textbook material are both provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow. $$ \begin{array}{lcc} & \text { iPad } & \text { Textbook } \\ \hline \text { Mean } & 86.4 & 79.7 \\ \text { Standard Deviation } & 8.95 & 10.7 \\ \text { Sample Size } & 10 & 10 \end{array} $$ a. Use the summary data to test for a significant difference in advancement scores for the two groups using \(\alpha=.05 .\) b. Find a \(95 \%\) confidence interval for the difference in mean scores for the two groups. c. In light of parts a and b, what can we say about the efficacy of using an iPad versus a traditional textbook in learning algebra at the middle school level?

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